262659041 Math 29 Problem Set Compilation FIXED

EXERCISE 9.1

1.

BASIC INTEGRATION FORMULAS

6𝑥 2 − 4𝑥 + 5 𝑑𝑥 =

6𝑥 3 3

−

4𝑥 2 2

7.

+ 5𝑥 + 𝑐

= Factor, (x-c), c = 2 P(c) = 0 – the (x-c ) is the factor P(c) = 0 2 1 0 0 -8 2 4 8 12 4 0

= 𝟐𝒙𝟑 − 𝟐𝒙𝟐 + 𝟓𝒙 + 𝒄

3.

=

𝑥( 𝑥 − 1)𝑑𝑥 =

= 𝑥 𝑑𝑥 − 𝑥𝑑𝑥

5.

𝟐 𝟓 𝒙𝟐 𝟓

𝟏 𝟐 𝒙 𝟐

−

(𝑋 2 +2𝑋+4)(𝑋−2) (𝑋−2)

= (𝑥 2 + 2𝑥 + 4)𝑑𝑥

𝑥 𝑥 − 𝑥 𝑑𝑥 3 2

=

𝑥 3 −8 𝑑𝑥 𝑥−2

+𝒄

9.

2𝑥 2 +4𝑥−3 𝑑𝑥 𝑥2

=

𝑥3 3

=

𝒙𝟑 + 𝟑

+

2𝑥 2 2

+ 4𝑥 + 𝑐

𝒙𝟐 + 𝟒𝒙 + 𝒄

𝑥 4 − 2𝑥 3 + 𝑥 2 𝑑𝑥 2

=

2+

4 𝑋

= 2𝑑𝑥 + = 2𝑥 + 4

−

3 𝑋2

4 𝑑𝑥 𝑥 𝑑𝑥 𝑥

𝑑𝑥 −

3 𝑑𝑥 𝑥2

=

𝒙𝟑 𝟑

𝟓

−

𝟔𝟑 𝟓

+

𝒙𝟐 𝟐

+𝑪

3𝑥 −1 𝑑𝑥 −1

−

= 𝟐𝒙 + 𝟒𝒍𝒏𝒙 +

= 𝑥 2 𝑑𝑥 − 2𝑥 3 𝑑𝑥 + 𝑥𝑑𝑥

𝟑 𝒙

+𝒄

DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy

1

EXERCISE 9.2

1.

INTEGRATION BY SUBSTITUTION

2 − 3𝑥 𝑑𝑥

(2𝑥+3)𝑑𝑥 𝑥 2 +3𝑥+4

5.

𝑑𝑢

Let u = 2 - 3x𝑑𝑥 = −3

Let u = 𝑥 2 + 3𝑥 + 4 𝑑𝑥 = 2𝑥 + 3

−𝑑𝑢 = 𝑑𝑥 3

𝑑𝑢 = (2𝑥 + 3)𝑑𝑥

= 𝑢

1 2

𝑑𝑢 𝑢

=

𝑑𝑢 (− 3 )

1 =− 3

𝑑𝑢

= 𝑙𝑛𝑢 + 𝑐

1 2

𝑢 𝑑𝑢

= 𝐥𝐧 𝒙𝟐 + 𝟑𝒙 + 𝟒 + 𝒄

3

1 2𝑥 2 3

= −3

+𝑐 𝟑

=−

𝟐 𝟐−𝟑𝒙 𝟐 𝟗

2

𝑥 2 𝑑𝑥 (𝑥 3 −1)4

7. +𝒄

3

𝑑𝑢

Let u = 𝑥 3 − 1 𝑑𝑥 = 3𝑥 2 4

3. 𝑥 (2𝑥 − 1) 𝑑𝑥 Let u = 2𝑥 3 − 1

𝑑𝑢 = 𝑥 2 𝑑𝑥 3 𝑑𝑢 3 𝑥4

=

𝑑𝑢 = 6𝑥 2 𝑑𝑥

=

1 3

𝑑𝑢 = 𝑥 2 𝑑𝑥 6

=

1 𝑢 −3 3 −3 𝑢 −3 −9

𝑢−4 +𝑐

=

𝑥 2 (2𝑥 3 − 1)4 𝑑𝑥

=

=

𝑑𝑢 (𝑢 )( ) 6

= − 𝟗(𝒙𝟑−𝟏)𝟑 + 𝒄

=

1 (𝑢4 )𝑑𝑢 6

=

1 𝑢5 +𝑐 6 5

4

+𝑐 𝟏

𝑢5

= 30 + 𝑐 =

(𝟐𝒙𝟑 − 𝟏)𝟓 +𝒄 𝟑𝟎


2

EXERCISE 9.2

INTEGRATION BY SUBSTITUTION 13. cos4 𝑥 sin 𝑥𝑑𝑥

𝑑𝑥 𝑥𝑙𝑛 2 𝑥

9.

𝑑𝑢

1

Let u = 𝑙𝑛𝑥 𝑑𝑥 = 𝑥

= 𝑢4 −𝑑𝑢

=

1 𝑑𝑥 ( ) 𝑙𝑛 2 𝑥 𝑥

= - 𝑢4 𝑑𝑢

=

1 𝑑𝑢 𝑢2

=−

𝑢5 5

=

𝑢−2 𝑑𝑢

=−

𝐜𝐨𝐬 𝟓 𝒙 𝟓

=

𝑢 −1 −1

= −sin 𝑥

𝑑𝑢 = −sin 𝑥 𝑑𝑥

𝑑𝑥 𝑥

𝑑𝑢 =

𝑑𝑢 𝑑𝑥

Let u = cos 𝑥

+𝑐 +𝒄

+𝑐

𝟏

= − 𝒍𝒏𝒙 + 𝒄 15. 𝑑𝑥 𝑒 𝑥 −1

11.

Let u = 3𝑥

Let u = 𝑒 𝑥 𝑑𝑢 = 𝑒 𝑥 𝑑𝑥 1 𝑢−1

= ( =

1 𝑢

− )𝑑𝑢

1 𝑑𝑢 𝑢−1

1 𝑢

−

𝑑𝑢

𝑑𝑢 =3 𝑑𝑥 𝑑𝑢 = 𝑑𝑥 3

=

𝑑𝑣 = 𝑑𝑢 𝑢 1 𝑢

𝑑𝑣 −

𝑑𝑢

=

Let v = 𝑢 − 1

=

1 + 2 sin 3𝑥 𝑐𝑜𝑠3𝑥𝑑𝑥

1 + 2 sin 𝑢 𝑐𝑜𝑠𝑢 ( 3 ) 1 3

1 + 2 sin 𝑢 𝑐𝑜𝑠𝑢𝑑𝑢

Let v =1 + 2 sin 𝑢 1 𝑑𝑣 𝑢

= 𝑙𝑛𝑢 − 𝑙𝑛𝑢 + 𝑐 𝑢 =𝑢−1

𝑑𝑣 𝑑𝑢

=

; 𝑢 = 𝑒𝑥

= ln⁡|𝑢 − 1| − ln⁡|𝑒 𝑥 | + 𝑐 = ln⁡ (1 − 𝑒 𝑥 )

𝑑𝑣 2

= 2𝑐𝑜𝑠𝑢 ;

= 𝑐𝑜𝑠𝑢𝑑𝑢

1 + 2 sin 3𝑥 𝑐𝑜𝑠3𝑥𝑑𝑥 1

=

1 [ 3

𝑑𝑣

=

1 2𝑣 2 6 3

=

(𝟏+𝟐𝒔𝒊𝒏𝒙)𝟐 𝟗

𝑣 2 ( 2 )] 3

= 𝐥𝐧 𝟏 − 𝒆𝒙 − 𝒙 + 𝒄

+𝑐 𝟑

+𝒄


3

EXERCISE 9.2


𝑠𝑒𝑐 2 𝑥𝑑𝑥 ` 𝑎+𝑏 𝑡𝑎𝑛𝑥

17.

3𝑥 2 +14𝑥+14 𝑑𝑥 𝑥+4

21.

𝑓 (𝑥) 𝑔 (𝑥)

Let u = 𝑎 + 𝑏 𝑡𝑎𝑛𝑥

=

𝑑𝑢 𝑑𝑥

* using synthetic division

=𝑏

;

𝑑𝑢 𝑏

= 𝑠𝑒𝑐 2 𝑥𝑑𝑥

𝑑𝑢 𝑏

=

𝑅 𝑥 𝑔 𝑥

𝑑(𝑥)

-4 3 14 13

𝑢

1 =𝑏

=

sin 𝑥 𝑐𝑜𝑠𝑥

= 𝑄 𝑥 𝑑 𝑥 +

-12 -8

𝑑𝑢 𝑢 𝟏 𝐥𝐧 𝒃

3 2 5 - R(x) 𝒂 + 𝒃𝒕𝒂𝒏𝒙 + 𝒄

𝑄 𝑥 = 3𝑥 + 2 𝑥 + 4 = 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑔(𝑥) 5 𝑑𝑥 𝑥+4

= (3𝑥 + 2) 𝑑𝑥 +

For the second integral : 2

19.

𝑡𝑎𝑛3𝑥 𝑠𝑒𝑐 3𝑥𝑑𝑥 𝑙𝑒𝑡 𝑢 = 𝑥 + 4

Let u = 𝑡𝑎𝑛3𝑥 𝑑𝑢 𝑑𝑥

= 3𝑠𝑒𝑐 2 3𝑥 ;

= 𝑢

1 2

𝑑𝑢 3

=

1 [ 3 3

3𝑥 2 2

=

1 3

=

𝟐(𝐭𝐚𝐧 𝟑𝒙)𝟐 𝟗

=

]+𝑐

𝑑𝑦 = 1 ; 𝑑𝑢 = 𝑑𝑥 𝑑𝑥

= (3𝑥 + 2)𝑑𝑥 + 5 =[

𝑑𝑢 (3)

3 2𝑢 2

= 𝑠𝑒𝑐 2 3𝑥𝑑𝑥

;

𝟑𝒙𝟐 𝟐

𝑑𝑢 𝑢

+ 2𝑥 + 5𝑙𝑛𝑢 + 𝑐]

+ 𝟐𝒙 + 𝟓𝐥𝐧⁡ (𝒙 + 𝟒) + 𝒄

3

2tan 3𝑥 2 3

+𝑐

𝟑

+𝒄


4

EXERCISE 9.2


𝑥 5 −2𝑥 3 −2𝑥 𝑥 2 +1

23.

𝑑𝑥

𝑥 3 − 3𝑥 𝑥 2 + 1 𝑥 5 − 2𝑥 3 − 2𝑥 𝑥5 + 𝑥3 −3𝑥 3 − 2𝑥 −3𝑥 3 − 3𝑥 𝑥 𝑓(𝑥) dx = 𝑔(𝑥)

𝑄 𝑥 𝑑𝑥 +

= 𝑥 3 − 3𝑥 𝑑𝑥 + 𝑥4

=4 −

3𝑥 2 + 2

𝑅(𝑥) 𝑑𝑥 𝑔(𝑥) 𝑥 𝑑𝑥 𝑥 2 +1

𝑥 𝑑𝑥 𝑥 2 +1

For the 2nd term Let u = x2+1 𝑑𝑢 = 2𝑥 𝑑𝑥 𝑑𝑢 = 𝑥𝑑𝑥 2 𝑥4 =4

=

3𝑥 2 − 2

𝒙𝟒 𝟒

−

+

𝟑𝒙𝟐 𝟐

𝑑𝑢 2

𝑢 𝟏

+ 𝟐 𝐥𝐧 𝒙𝟐 + 𝟏 + 𝒄


5

EXERCISE 9.3 1.

INTEGRATION OF TRIGONOMETRIC FUNCTIONS

𝑠𝑒𝑐5𝑥𝑡𝑎𝑛5𝑥𝑑𝑥 𝐿𝑒𝑡 𝑢 = 5𝑥 𝑑𝑢 𝑑𝑥

𝑑𝑢 5

=5

=

cos 3 𝑥𝑑𝑥 1+𝑠𝑖𝑛𝑥 . 1−𝑠𝑖𝑛𝑥 1+𝑠𝑖𝑛𝑥

7. =

(cos 3 𝑥) 1+𝑠𝑖𝑛𝑥 𝑑𝑥 (1−𝑠𝑖𝑛𝑥 )(1+𝑠𝑖𝑛𝑥 )

=

(co s 3 𝑥+cos 3 𝑥𝑠𝑖𝑛𝑥 )𝑑𝑥 1−sin 2 𝑥

=

cos 3 𝑥 1+𝑠𝑖𝑛𝑥 𝑑𝑥 cos 3 𝑥

= 𝑑𝑥

𝑠𝑒𝑐𝑢𝑡𝑎𝑛𝑢

𝑑𝑢 5

1

= 5 𝑠𝑒𝑐𝑢𝑡𝑎𝑛𝑢𝑑𝑢 1 =5

𝑠𝑒𝑐𝑢 + 𝑐

𝟏 =𝟓

𝒔𝒆𝒄𝟓𝒙 + 𝒄

= 𝑐𝑜𝑠𝑥 1 + 𝑠𝑖𝑛𝑥 𝑑𝑥 = 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥𝑑𝑥 𝐿𝑒𝑡 𝑢 = 𝑠𝑖𝑛𝑥 𝑑𝑢 𝑑𝑥

𝑠𝑖𝑛𝑥 +𝑐𝑜𝑠𝑥 𝑠𝑖𝑛 2 𝑥

3.

𝑑𝑥

=

𝑠𝑖𝑛𝑥 𝑠𝑖𝑛 2 𝑥

=

1 𝑠𝑖𝑛𝑥

=

𝑐𝑠𝑐𝑑𝑥 +

= 𝑐𝑜𝑠𝑥 ; 𝑑𝑢 = 𝑐𝑜𝑠𝑥𝑑𝑥

= 𝑠𝑖𝑛𝑥 + 𝑢𝑑𝑢 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛 2 𝑥

𝑑𝑥 +

𝑑𝑥

𝑑𝑥 + 𝑐𝑜𝑡𝑥𝑐𝑠𝑐𝑥𝑑𝑥

= 𝑠𝑖𝑛𝑥 +

𝑢2 2

= 𝒔𝒊𝒏𝒙 +

𝐬𝐢𝐧𝟐 𝒙 𝟐

+𝑐 +𝒄

𝑐𝑜𝑡𝑥𝑐𝑠𝑐𝑥𝑑𝑥

= − 𝒍𝒏 𝒄𝒔𝒄𝒙 + 𝒄𝒐𝒕𝒙 − 𝒄𝒔𝒄𝒙 + 𝒄

9.

1 + 𝑡𝑎𝑛𝑥 2 𝑑𝑥 = (1 + 2𝑡𝑎𝑛𝑥 + tan2 𝑥)𝑑𝑥

5.

𝑑𝑥 1 2

1

; Let u= 2 𝑥

1 2

sin 𝑥 cot 𝑥 𝑑𝑢 𝑑𝑥

=

= =2 =2 =2

1 2

2𝑑𝑢 = 𝑑𝑥

= [2𝑡𝑎𝑛𝑥 + (1 + tan2 𝑥)]𝑑𝑥 = 2 𝑡𝑎𝑛𝑥𝑑𝑥 +

sec 2 𝑥𝑑𝑥

= 2 −𝑙𝑛|𝑐𝑜𝑠𝑥| + 𝑡𝑎𝑛𝑥 + 𝑐

2𝑑𝑢 𝑠𝑖𝑛𝑢𝑐𝑜𝑡𝑢

= −𝟐 𝒍𝒏 |𝒄𝒐𝒔𝒙| + 𝒕𝒂𝒏𝒙 + 𝒄

𝑑𝑢 𝑠𝑖𝑛𝑢 𝑐𝑜𝑡𝑢 𝑑𝑢 𝑠𝑖𝑛𝑢 ( 1 𝑐𝑜𝑠𝑢

𝑐𝑜𝑠𝑢 𝑠𝑖𝑛𝑢

)

(𝑑𝑢)

= 2 𝑠𝑒𝑐𝑢𝑑𝑢 = 𝟐𝒍𝒏 𝒄𝒔𝒄𝒙 + 𝒄𝒐𝒕𝒙 + 𝒄 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy

6

EXERCISE 9.3

INTEGRATION OF TRIGONOMETRIC FUNCTIONS

𝑐𝑜𝑠 6𝑥𝑑𝑥 cos 2 3𝑥

11.

Let u = 3x ; 2u = 6x 𝑑𝑢 𝑑𝑥

=3 ;

𝑑𝑢 3

𝑐𝑜𝑠 2𝑢

𝑑𝑢 3

= =

= 𝑑𝑥

cos 2 𝑢 2 3

4 sin 2 𝑥𝑐𝑜 𝑠 2 𝑥 𝑠𝑖𝑛 2𝑥𝑐𝑜𝑠 2𝑥

15.

1 𝑐𝑜𝑠𝑢

𝑐𝑜𝑠𝑢 𝑐𝑜𝑠𝑢

𝑑𝑢

=

(4𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 )(𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 ) 𝑑𝑥 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 𝑐𝑜𝑠 2𝑥

=

2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 𝑐𝑜𝑠 2𝑥

=

𝑠𝑖𝑛 2𝑥 𝑐𝑜𝑠 2𝑥

𝑑𝑢 𝑑𝑥

2

𝑑𝑥 𝑐𝑜𝑠𝑥

=

1 𝑐𝑜𝑠𝑥

= 𝑑𝑥

𝑑𝑢 2

𝑡𝑎𝑛𝑢𝑑𝑢

𝟏 = − 𝐥𝐧 𝒄𝒐𝒔𝟐𝒙 + 𝒄 𝟐

𝑠𝑖𝑛 2𝑥𝑑𝑥 2𝑠𝑖𝑛𝑥𝑐𝑜 𝑠 2 𝑥

=

.

1 2

=

2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥𝑑𝑥 (2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 )𝑐𝑜𝑠𝑥

𝑑𝑥

𝑑𝑢 2

=2

𝑠𝑖𝑛𝑢 𝑐𝑜𝑠𝑢

𝟐

= 𝟑 𝐥𝐧 𝒔𝒆𝒄𝟑𝒙 + 𝒕𝒂𝒏𝟑𝒙 + 𝒄

=

𝑑𝑥

Let u = 2x

= 3 𝑠𝑒𝑐𝑢𝑑𝑢

13.

𝑑𝑥

𝑑𝑥 𝑠𝑖𝑛 3𝑥𝑡𝑎𝑛 3𝑥

17.

Let u = 3x 𝑑𝑥

= 𝑠𝑒𝑐𝑥𝑑𝑥 = 𝒍𝒏 𝒔𝒆𝒄𝒙 + 𝒕𝒂𝒏𝒙 + 𝒄

𝑑𝑢 𝑑𝑥

𝑑𝑢 3

=3

= 𝑑𝑥

𝑑𝑢 3

=

𝑠𝑖𝑛𝑢𝑡𝑎𝑛𝑢 1

=3 𝑐𝑠𝑐𝑢 + 𝑐 𝟏

= − 𝟑 𝒄𝒔𝒄𝟑𝒙 + 𝒄


7

EXERCISE 9.4

1.

𝑑𝑥 𝑒 2𝑥

=

𝑒 −2𝑥 dx

INTEGRATION OF EXPONENTIAL FUNCTIONS

𝑑𝑢 𝑑𝑢 𝑙𝑒𝑡 𝑢 = 2𝑥 ; = −2 ; − = 𝑑𝑥 𝑑𝑥 2 = 𝑒 𝑢 (− =−

1 2

2𝑑𝑢 ) 3

2

𝑒 𝑢 𝑑𝑢

= 3 𝑒 𝑢 𝑑𝑢

1

1

𝟏

=− 𝟐 (𝒆−𝟐𝒙 ) + 𝒄

𝑒 𝑠𝑖𝑛 4𝑥 𝑐𝑜𝑠4𝑥𝑑𝑥

2 3

=

𝟐 𝒆𝟑𝒙 𝟑

𝑑𝑢 4

1 4

=

𝑒

𝑠𝑖𝑛𝑢

1 4

)

1 =4

=

𝑒𝑣 + 𝑐

𝒆𝒔𝒊𝒏𝟒𝒙 𝟒

53−2𝑥 𝑑𝑥

= 5𝑢 (−

𝑐𝑜𝑠𝑢𝑑𝑢

𝑒 𝑣 𝑑𝑣

+𝒄

𝑙𝑒𝑡 𝑢 = 3 − 2𝑥 ;

𝑑𝑣 𝑙𝑒𝑡 𝑣 = 𝑠𝑖𝑛𝑢 ; = cos 𝑢 ; 𝑑𝑣 = 𝑐𝑜𝑠𝑢𝑑𝑢 𝑑𝑢 =

𝑒2 +𝑐

7.

𝑑𝑢 𝑑𝑢 𝑙𝑒𝑡 𝑢 = 4𝑥 ; =4; = 𝑑𝑥 𝑑𝑥 4 𝑒 𝑠𝑖𝑛𝑢 𝑐𝑜𝑠𝑢(

3𝑥

=

=− 2𝑒 𝑢 + 𝑐

=

3𝑥 2𝑑𝑢 ; = 𝑑𝑥 2 3

𝑙𝑒𝑡 𝑢 = = 𝑒𝑢 (

𝑑𝑢 ) 2

=− 2 𝑒 𝑢 + 𝑐

3.

3𝑥

𝑒 3𝑥 𝑑𝑥 = 𝑒 2 𝑑𝑥

5.

𝑑𝑢 𝑑𝑢 = −2 ; − = 𝑑𝑥 𝑑𝑥 2

𝑑𝑢 ) 2

1

= − 2 5𝑢 𝑑𝑢 1 53−2𝑥 𝑙𝑛 5

= −2 =−

𝟓𝟑−𝟐𝒙 𝒍𝒏𝟐𝟓

+𝑐

+𝒄

+𝒄 3𝑥 2𝑥 𝑑𝑥

9.

𝑎 𝑥 𝑏 𝑥 = (𝑎𝑏)𝑥 = 6𝑥 𝑑𝑥 𝟔𝒙

= 𝒍𝒏𝟔 + 𝒄


8

EXERCISE 9.5

1.

INTEGRATION OF HYPERBOLIC FUNCTIONS

𝑠𝑖𝑛𝑕 3𝑥 − 1 𝑑𝑥

𝑠𝑒𝑐 𝑕 2 𝑙𝑛𝑥 𝑑𝑥 𝑥

5.

Let u = 3𝑥 − 1 𝑑𝑢 𝑑𝑥

𝑙𝑒𝑡 𝑢 = 𝑙𝑛𝑥 ; 𝑑𝑢 3

= 3 ; 𝑑𝑥 =

=

𝑑𝑢

= 1 3

=

1 𝑐𝑜𝑠𝑕 𝑢𝑑𝑢 3

=

𝟏 𝒄𝒐𝒔𝒉 𝟑

𝑠𝑒𝑐𝑕2 𝑢𝑑𝑢

= 𝑡𝑎𝑛𝑕𝑢 + 𝑐

𝑠𝑖𝑛𝑕 𝑢 ( 3 )

=

𝑑𝑢 1 𝑑𝑥 = ; 𝑑𝑢 = 𝑑𝑥 𝑥 𝑥

= 𝒕𝒂𝒏𝒉⁡ (𝒍𝒏𝒙) + 𝒄

𝑠𝑖𝑛𝑕 𝑢𝑑𝑢 +c

𝟑𝒙 − 𝟏 + 𝒄

7.

1

1

𝑐𝑠𝑐𝑕 2 𝑥 𝑐𝑜𝑡𝑕 2 𝑥𝑑𝑥 1

Let u = 2 𝑥 ;

𝑑𝑢 𝑑𝑥

=

1 2

; 2𝑑𝑢 = 𝑑𝑥

= 2 𝑐𝑠𝑐𝑕𝑢 𝑐𝑜𝑡𝑕𝑢𝑑𝑢 3.

𝑐𝑠𝑐𝑕2 1 − 𝑥 2 𝑥𝑑𝑥

= 2(−𝑐𝑠𝑐𝑕𝑢 + 𝑐)

Let u=1 − 𝑥 2

= −𝟐𝒄𝒔𝒄𝒉 𝒙 + 𝒄

𝑑𝑢 𝑑𝑥

−

𝟏 𝟐

= -2𝑥

𝑑𝑢 = 𝑥𝑑𝑥 2 = 𝑐𝑠𝑐𝑕2 𝑢(− =−

1 2

𝑑𝑢 ) 2

𝑐𝑠𝑐𝑕2 𝑢𝑑𝑢

1

=− 2 (−𝑐𝑜𝑡𝑕𝑢 + 𝑐) =

𝟏 𝒄𝒐𝒕𝒉 𝟐

𝟏 − 𝒙𝟐 + 𝒄


9

EXERCISE 9.6

APPLICATION OF INDEFINITE INTEGRATION

1. Given slope 3𝑥 2 + 4 𝑑𝑦 = 3𝑥 2 + 4 𝑑𝑥 𝑑𝑦 = 3𝑥 2 + 4 𝑑𝑥

𝑦=

𝒚 = 𝟑𝒙𝟐 + 𝟒𝒙 + 𝒄

𝑥+1 𝑦−1

𝑑𝑦 𝑥 + 1 = 𝑑𝑥 𝑦 − 1

𝑑𝑦 = 𝑦2

𝑦 2 − 2𝑦 =

𝑑𝑥 𝑥

− ln 𝑥 −

1 =𝑐 4

− ln 1 −

1 =𝑐 4

𝑐=−

1 4

− ln 𝑥 −

𝑦 − 1 𝑑𝑦 = 𝑥2 2

through 1,4

1 − = 𝑙𝑛𝑥 + 𝑐 4

+ 4𝑥 + 𝑐

3. Given slope

𝑦2 , 𝑥

𝑑𝑦 𝑦 2 = 𝑑𝑥 𝑥

3𝑥 2 + 4 𝑑𝑥

𝑑𝑦 = 3𝑥 3 3

7. Given slope

𝑥 + 1 𝑑𝑥 +𝑥+𝑐 2

1 1 + = 0 4𝑦 𝑦 4

−4𝑦 ln 𝑥 − 4 + 𝑦 = 0 𝟒𝒚 𝐥𝐧 𝒙 − 𝒚 + 𝟒 = 𝟎

𝑦 2 − 2𝑦 = 𝑥 2 + 2𝑥 + 2𝑐 𝒙𝟐 − 𝒚𝟐 + 𝟐𝒚 + 𝟐𝒙 + 𝟐𝒄 = 𝟎

1 5. Given slope 𝑥𝑦

𝑑𝑦 1 = 𝑑𝑥 𝑥𝑦

𝟐

=

𝑑𝑦 = 𝑑𝑥

𝑦

1

𝑦 −2 𝑑𝑦 =

𝑑𝑥

1

𝑦2 𝑑𝑥 𝑥

𝑦𝑑𝑦 = 𝑦2 2

9. Given slope 𝑦, through 1,1

ln 𝑥 2 2 𝟐

+𝑐 2

𝒚 = 𝒍𝒏𝒙 + 𝟐𝒄

1 2

2𝑦

=𝑥+𝑐 1

2

=𝑥+𝑐

When 𝑥 = 1 , 𝑦 = 1 2 1 =1+𝑐 ; 𝑐 =1 2𝑦

1

2

=𝑥+𝑐

𝟒𝒚 = 𝒙 + 𝟏

2

𝟐


10

EXERCISE 9.6


11. Given slope 𝑥 −2 , through 1,2

v = -32t + vo

𝑑𝑦 1 = 2 𝑑𝑥 𝑥

when t = 1 sec, s=h=48ft 𝑑𝑥 𝑥2

𝑑𝑦 =

1 𝑦 =− +𝑐 𝑥

h=-16t2+ vot + c1 48 = -16(1)2 + vo(1) + c2 64 - vo = c2 When t = 0, s = 0, c2 = 0

1 2=− +𝑐 1

s = -16t2 + vot

2 = −1 + 𝑐

when t = 1 sec, s = 48

𝑐=3

s = -16t2 + c1t 1 𝑥

𝑦 = − +3 x 𝑥𝑦 = −1 + 3𝑥 𝒙𝒚 − 𝟑𝒙 + 𝟏 = 𝟎

c1=64 s=-16t2 + 64t v = -32t + 64 @ max, v = 0

13. a=-32 ft/sec2 a=-2

0 = -32t + 64 32t=64 t = 2 sec

𝑑𝑦 = −32 𝑑𝑡 𝑑𝑣 =

48 = -16(1)2 + c1(1)

s = -16t2 + 64t −32𝑑𝑡

s = -16(2)2 + 64(2) s = 64ft

v=-32t+c 𝑑𝑠 = −32𝑡 + 𝑐1 𝑑𝑡 𝑑𝑠 =

(−32𝑡 + 𝑐1 )𝑑𝑡

s=16t2 + c1t + c2 when t = 0, v = vo v=-32t + c1 vo= -32(0) + c1 vo =c1 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy

11

EXERCISE 9.6


15. a = 32ft/sec2 a = 32 𝑑𝑣 = 32 𝑑𝑡 𝑑𝑣 =

32𝑑𝑡

v = 32t + c1 𝑑𝑠 = 32𝑡 + 𝑐1 𝑑𝑡 𝑑𝑠 =

32𝑡 + 𝑐1 𝑑𝑡

S = 16t2 + c1 + c2 when t = 0, v = 0 c1 = 0 v = 32t when t = 0 , s = 0 c2 = 0 s = 16t2 400 16

𝑡=

𝑡=

20 4

t = 5 sec v = vt *since it is a free falling body, its velocity is ( - ) vt = -32t vt = -32(5) vt = -160 ft/sec


12

EXERCISE 10.1

PRODUCT OF SINES AND COSINES

1. ʃ sin 5𝑥 sin 𝑥 𝑑𝑥 =

5. ʃ cos 3𝑥 − 2𝜋 cos 𝑥 + 𝜋 𝑑𝑥 1 = ʃ[cos 𝑢 + 𝑣 + cos⁡ (𝑢 − 𝑣)]𝑑𝑥 2

2 sin 𝑢 sin 𝑣 𝑑𝑥

𝑙𝑒𝑡 𝑢 = 3𝑥 − 2𝜋

= [cos 𝑢 − 𝑥 − cos(𝑢 + 𝑣)]𝑑𝑥 𝑢 = 5𝑥

𝑣 =𝑥+𝜋

𝑣=𝑥

1 = ʃ[cos 5𝑥 − 𝑥 − cos⁡ (5𝑥 + 𝑥)]𝑑𝑥 2 1 = ʃ[cos 4𝑥 − cos 6𝑥]𝑑𝑥 2

= =

𝒔𝒊𝒏 𝟒𝒙 𝒔𝒊𝒏𝟔𝒙 − +𝑪 𝟖 𝟏𝟐

−

1 sin 6𝑥 6

𝑢 + 𝑣 = 3𝑥 − 2𝜋 + 𝑥 + 𝜋 = 4𝑥 − 𝜋



1 = [ʃ cos 4𝑥𝑑𝑥 − ʃ cos 6𝑥𝑑𝑥 2 1 1 [ sin 4𝑥 2 4



𝑢 − 𝑣 = 3𝑥 − 2𝜋 − 𝑥 + 𝜋 = 2𝑥 − 3𝜋 1 = ʃ[cos 4𝑥 − 𝜋 + cos(2𝑥 − 3𝜋)]𝑑𝑥 2

]+𝐶

𝑓𝑜𝑟 cos 4𝑥 − 𝜋 = cos 4𝑥𝑐𝑜𝑠𝜋 + 𝑠𝑖𝑛4𝑥𝑠𝑖𝑛𝜋 = −𝑐𝑜𝑠4𝑥 𝑓𝑜𝑟 cos 2𝑥 − 3𝜋 = cos 2𝑥𝑐𝑜𝑠 3𝜋 + sin 2𝑥𝑠𝑖𝑛 3𝜋 = − cos 2𝑥

3. ʃ sin 9𝑥 − 3 cos 𝑥 + 5 𝑑𝑥 1 = ʃ [sin 9x − 3 + x + 5 + sin 9𝑥 − 3 − 𝑥 − 5 𝑑𝑥 2

1 = ʃ[sin 5𝑥 + 2 + sin(3𝑥 − 8)]𝑑𝑥 2 𝑙𝑒𝑡 𝑧 = 5𝑥 + 2 𝑑𝑧 =5 𝑑𝑥 𝑑𝑧 = 𝑑𝑥 5 1

= 2 [− cos 𝑧 =−

1 5

1 2

= ʃ(cos 4𝑥 − cos 2𝑥)𝑑𝑥 1

1

1

= 2 [− 4 sin 4𝑥 − 2 sin 2𝑥] + 𝐶 𝟏 𝟏 = − 𝐬𝐢𝐧 𝟒𝒙 − 𝐬𝐢𝐧 𝟐𝒙 + 𝑪 𝟖 𝟒

; 𝑙𝑒𝑡 𝑤 = 3𝑥 − 8 𝑑𝑤 ; =3 𝑑𝑥 𝑑𝑤 ; = 𝑑𝑥 3

1

− 3 𝑐𝑜𝑠𝑤] + 𝐶

𝟏 𝟏 𝒄𝒐𝒔 𝟓𝒙 + 𝟐 − 𝒄𝒐𝒔 𝟑𝒙 − 𝟖 + 𝑪 𝟏𝟎 𝟔


13

EXERCISE 10.1

7.

PRODUCT OF SINES AND COSINES

𝜋

4 𝑠𝑖𝑛 8𝑥 𝑐𝑜𝑠3𝑥𝑑𝑥

= 2ʃ[sin 8𝑥 + 3𝑥 + 𝑠𝑖𝑛𝑥 8𝑥 − 3𝑥 𝑑𝑥

5 = ʃ[cos 𝑢 − 𝑣 − cos⁡ (𝑢 + 𝑣)]𝑑𝑥 2

= 2ʃ[𝑠𝑖𝑛11𝑥 + sin 5𝑥]𝑑𝑥

𝑙𝑒𝑡 𝑢 = 4𝑥 +

𝑙𝑒𝑡 𝑢 = 11𝑥 ; 𝑙𝑒𝑡 𝑣 = 5𝑥

1

𝑑𝑢 = 11 ; 𝑑𝑥

𝑑𝑣 =5 𝑑𝑥

𝑑𝑢 = 𝑑𝑥 ; 11

𝑑𝑣 = 𝑑𝑥 5

1

= 2[− 11 cos 11𝑥 − 5 cos 5𝑥 ] + 𝐶 =−

𝜋

9. 5 𝑠𝑖𝑛 4𝑥 + 3 𝑠𝑖𝑛 2𝑥 − 6 𝑑𝑥

𝜋 3

;

𝜋

𝜋

𝜋

𝜋



𝑢 − 𝑣 = 4𝑥 + 3 − 2𝑥 − 6 = 2𝑥 + 𝜋/2 𝜋 𝑣 = 2𝑥 − 6



𝑢 + 𝑣 = 4𝑥 + 3 + 2𝑥 − 6 = 6𝑥 + 𝜋/6

𝟐 𝟐 𝐜𝐨𝐬 𝟏𝟏𝒙 − 𝒄𝒐𝒔𝟓𝒙 + 𝑪 𝟏𝟏 𝟓

5 𝜋 𝜋 = ʃ[𝑐𝑜𝑠 2𝑥 + − 𝑐𝑜𝑠 6𝑥 + ]𝑑𝑥 2 2 6 

𝜋

𝑓𝑜𝑟 cos 2𝑥 + 2 = cos 2𝑥𝑐𝑜𝑠 = −𝑠𝑖𝑛2𝑥



𝑓𝑜𝑟 cos 6𝑥 +

𝜋 𝜋 − sin 2𝑥 sin 2 2

𝜋 6

𝜋 𝜋 − 𝑠𝑖𝑛 6𝑥 𝑠𝑖𝑛 6 6 3 1 = 𝑐𝑜𝑠 6𝑥 − 𝑠𝑖𝑛 6𝑥 2 2 = 𝑐𝑜𝑠 6𝑥𝑐𝑜𝑠

5 3 1 = ʃ[− 𝑠𝑖𝑛 2𝑥 − 𝑐𝑜𝑠 6𝑥 + 𝑠𝑖𝑛 6𝑥 ]𝑑𝑥 2 2 2 5 1

3

2 2

12

= [ 𝑐𝑜𝑠 2𝑥 −

=

𝟓 𝒄𝒐𝒔 𝟐𝒙 𝟒

−

𝑠𝑖𝑛 6𝑥 −

𝟓 𝟑 𝒔𝒊𝒏 𝟔𝒙 𝟐𝟒

1 12

𝑠𝑖𝑥 6𝑥 + 𝐶 𝟓

− 𝟏𝟐 𝒔𝒊𝒏 𝟔𝒙 + 𝑪


14

EXERCISE 10.2

POWER OF SINES AND COSINES

1. 𝑠𝑖𝑛3 𝑥𝑐𝑜𝑠 4 𝑥𝑑𝑥; 𝑏𝑦 𝐶𝑎𝑠𝑒 𝐼 = =

=

1 𝑢5 𝑢7 − +𝐶 3 5 7

=

1 5 𝑢 15

=

𝟏 𝟏 𝒔𝒊𝒏𝟓 𝟑𝒙 − 𝒔𝒊𝒏𝟕 𝟑𝒙 + 𝑪 𝟏𝟓 𝟐𝟏

𝑠𝑖𝑛4 𝑥𝑐𝑜𝑠 4 𝑥𝑠𝑖𝑛𝑥𝑑𝑥 (1 − 𝑐𝑜𝑠 2 𝑥)2𝑐𝑜𝑠 4 𝑥𝑠𝑖𝑛𝑥𝑑𝑥

=

(1 − 2𝑐𝑜𝑠 2 𝑥 + 𝑐𝑜𝑠 4 𝑥)𝑐𝑜𝑠 4 𝑥𝑠𝑖𝑛𝑥𝑑𝑥

=

(𝑐𝑜𝑠 4 𝑥 − 2𝑐𝑜𝑠 6 𝑥 + 𝑐𝑜𝑠 8 𝑥)𝑠𝑖𝑛𝑥𝑑𝑥

1

− 21 𝑢7 + 𝐶

5.

𝑠𝑖𝑛4 𝑥𝑐𝑜𝑠 2 𝑥𝑑𝑥

Let u = cosx

=

(𝑠𝑖𝑛2 𝑥)2𝑐𝑜𝑠 2 𝑥𝑑𝑥

𝑑𝑢 = −𝑠𝑖𝑛𝑥 𝑑𝑥

=

1 − 𝑐𝑜𝑠2𝑥 2 1 + 𝑐𝑜𝑠2𝑥 ( ) 𝑑𝑥 2 2

-𝑑𝑢 = 𝑠𝑖𝑛𝑥𝑑𝑥 =

= - (𝑢4 − 2𝑢6 + 𝑢8 )𝑑𝑢 = =

2𝑢 6 7

𝑢5 5

𝑢9 9

+𝐶

=

𝟐 𝟏 𝟏 𝒄𝒐𝒔𝟕 𝒙 − 𝒄𝒐𝒔𝟓 𝒙 − 𝒄𝒐𝒔𝟗 𝒙 + 𝑪 𝟕 𝟓 𝟗

=

−

−

=

3. 𝑠𝑖𝑛4 3𝑥𝑐𝑜𝑠 3 3𝑥𝑑𝑥 ; 𝑏𝑦 𝐶𝑎𝑠𝑒 𝐼𝐼 =

4

2

𝑠𝑖𝑛 3𝑥𝑐𝑜𝑠 3𝑥𝑐𝑜𝑠3𝑥𝑑𝑥

1 1 1 1 𝑐𝑜𝑠2𝑥 − ( 𝑐𝑜𝑠2𝑥 + 𝑐𝑜𝑠 2 2𝑥) + 𝑑𝑥 4 2 4 2 2

1 8

1 − 2𝑐𝑜𝑠2𝑥 + 𝑐𝑜𝑠 2 2𝑥 1 + 𝑐𝑜𝑠2𝑥 𝑑𝑥

1 (1 − 2𝑐𝑜𝑠2𝑥 + 𝑐𝑜𝑠 2 𝑥 + 𝑐𝑜𝑠2𝑥 − 2𝑐𝑜𝑠 2 2𝑥 + 𝑐𝑜𝑠 3 2𝑥)𝑑𝑥 8

=

1 8

𝑠𝑖𝑛4 3𝑥 1 − 𝑠𝑖𝑛2 3𝑥 𝑐𝑜𝑠3𝑥𝑑𝑥 (𝑠𝑖𝑛4 3𝑥 − 𝑠𝑖𝑛6 3𝑥)𝑐𝑜𝑠3𝑥𝑑𝑥

4

(𝑐𝑜𝑠 2 2𝑥 1 + 𝑐𝑜𝑠2𝑥 𝑑𝑥 2

1 (1 − 𝑐𝑜𝑠2𝑥 − 𝑐𝑜𝑠 2 2𝑥 + 𝑐𝑜𝑠 3 2𝑥)𝑑𝑥 8

= =

1 4

=

=

=

1 − 2𝑐𝑜𝑠2𝑥 +

1 8

𝑑𝑥 −

𝑐𝑜𝑠 2 2𝑥𝑑𝑥 +

𝑐𝑜𝑠2𝑥𝑑𝑥 −

𝑐𝑜𝑠 3 2𝑥𝑑𝑥

1

1

1

1

1

2

2

8

2

6

[𝑥 − 𝑠𝑖𝑛2𝑥 − ( 𝑥 + 𝑠𝑖𝑛4𝑥 + 𝑠𝑖𝑛2𝑥 − 𝑠𝑖𝑛3 2𝑥]

𝒙 𝒔𝒊𝒏𝟒𝒙 𝒔𝒊𝒏𝟑 𝟐𝒙 − − +𝑪 𝟏𝟔 𝟔𝟒 𝟒𝟖

Let u = sin3x 𝑑𝑢 𝑑𝑥

=

= 3𝑐𝑜𝑠3𝑥 ; ( 𝑢4 − 𝑢6 )

𝑑𝑢 3

= 𝑐𝑜𝑠3𝑥𝑑𝑥

𝑑𝑢 3 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy

15

EXERCISE 10.2

POWER OF SINES AND COSINES

7. ( 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)2dx =

(𝑠𝑖𝑛𝑥 + 2 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 + 𝑐𝑜𝑠 2 𝑥)𝑑𝑥

=

𝑠𝑖𝑛𝑥𝑑𝑥 + 2

=

𝑠𝑖𝑛𝑥𝑑𝑥 + 2 𝑠𝑖𝑛2 𝑐𝑜𝑠𝑥𝑑𝑥 + (

1

𝑐𝑜𝑠 2 𝑥𝑑𝑥

𝑠𝑖𝑛2 𝑐𝑜𝑠𝑥𝑑𝑥 + 1

1+𝑐𝑜𝑠2𝑥 )𝑑𝑥 2

Let u = sinx

=

𝑠𝑖𝑛2 2𝑥 𝑠𝑖𝑛2𝑥 𝑑𝑥

=

1 − 𝑐𝑜𝑠 2 2𝑥 𝑠𝑖𝑛2𝑥 𝑑𝑥

𝑙𝑒𝑡 𝑢 = 𝑐𝑜𝑠 2𝑥

= -𝑐𝑜𝑠𝑥 + 2

𝑢 𝑑𝑢 +

2 3 𝑢2 3 𝟑

𝟒

1 𝑑𝑥 + 2

1 2

𝑥

+2+ 𝒙

= -𝒄𝒐𝒔𝒙 + 𝟑 𝒔𝒊𝒏𝟐 + 𝟐 +

𝑠𝑖𝑛 2𝑥 4

𝒔𝒊𝒏𝟐𝒙 𝟒

𝑐𝑜𝑠2𝑥 𝑑𝑥 2

1

−𝟏 𝟏 𝒄𝒐𝒔𝟐𝒙 + 𝒄𝒐𝒔𝟑 𝟐𝒙 + 𝒄 𝟐 𝟔

= 𝒙−

𝑥

𝒄𝒐𝒔𝒙

+𝑐

𝑠𝑖𝑛7 𝑥 𝑐𝑜𝑠 2 𝑥 𝑑𝑥

=

𝑠𝑖𝑛7 𝑥 𝑐𝑜𝑠 2 𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥

=

𝑠𝑖𝑛7 𝑥 1 − 𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠𝑥𝑑𝑥

=

𝑠𝑖𝑛7 𝑥 − 𝑠𝑖𝑛9 𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥 u=sinx du=cosxdx

=

𝑢7 − 𝑢9 𝑑𝑢

𝑐𝑜𝑠 2 2𝑥 𝑑𝑥

1

= 2 − 2 𝑠𝑖𝑛6𝑥 − 5 𝑐𝑜𝑠 5𝑥 − 𝑐𝑜𝑠𝑥 + 2 + 8 𝑠𝑖𝑛4𝑥 + 𝑐 𝒔𝒊𝒏𝟔𝒙 𝒄𝒐𝒔𝟓𝒙 − 𝟓 − 𝟏𝟐

𝑢−

+𝑪

𝑠𝑖𝑛3𝑥𝑐𝑜𝑠2𝑥 𝑑𝑥 +

1

𝑑𝑢 2

=

𝟏𝟓.

= 𝑠𝑖𝑛2 3𝑥 + 2𝑠𝑖𝑛3𝑥𝑐𝑜𝑠2𝑥 + 𝑐𝑜𝑠 2 2𝑥 𝑑𝑥

𝑥

−

−1 2

+𝐶

9. (𝑠𝑖𝑛3𝑥 + 𝑐𝑜𝑠2𝑥)2 𝑑𝑥

= 𝑠𝑖𝑛2 3𝑥 𝑑𝑥 + 2

𝑢3 3

; 𝐷𝑢 = −2𝑠𝑖𝑛2𝑥 𝑑𝑥

=

𝑑𝑢 = 𝑐𝑜𝑠𝑥𝑑𝑥 𝑠𝑖𝑛𝑥𝑑𝑥 + 2

1 − 𝑢2

=

𝑑𝑢 = 𝑐𝑜𝑠𝑥 𝑑𝑥

=

𝑠𝑖𝑛3 2𝑥 𝑑𝑥

𝟏𝟑.

𝒔𝒊𝒏𝟒𝒙 + 𝟖 +

𝒄

= =

𝑢8 𝑢10 − +𝑐 8 10 𝟏 𝟏 𝒔𝒊𝒏𝟖 𝒙 − 𝟏𝟎 𝒔𝒊𝒏𝟏𝟎 𝒙 + 𝟖

𝒄

11. 𝑐𝑜𝑠 2 4𝑥 𝑑𝑥 1 + 𝑐𝑜𝑠8𝑥 𝑑𝑥 2

= 1

=2 =

1 + 𝑐𝑜𝑠8𝑥 𝑑𝑥

𝒙 𝒔𝒊𝒏𝟖𝒙 + +𝒄 𝟐 𝟏𝟔 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy

16

EXERCISE 10.3

POWER OF TANGENTS AND SECANTS

1. 𝑡𝑎𝑛2 2𝑥𝑠𝑒𝑐 4 2𝑥𝑑𝑥

5. ____ 2 𝑥𝑑𝑥 → 𝑎𝑛𝑠. 𝑦 = 3 𝑡𝑎𝑛

= 𝑡𝑎𝑛2 2𝑥𝑠𝑒𝑐 2 2𝑥𝑠𝑒𝑐 2 2𝑥𝑑𝑥

𝐹𝑖𝑛𝑑 𝑡𝑕𝑒 𝑚𝑖𝑠𝑠𝑖𝑛𝑔 𝑡𝑒𝑟𝑚:

= 𝑡𝑎𝑛2 2𝑥(1 + 𝑡𝑎𝑛2 2𝑥)𝑠𝑒𝑐 2 2𝑥𝑑𝑥

𝑑𝑦 𝑥 𝑥 𝑥 = 𝑡𝑎𝑛2 𝑠𝑒𝑐 2 − 𝑠𝑒𝑐 2 + 1 𝑑𝑥 2 2 2

1

= (𝑡𝑎𝑛2 2𝑥 + 𝑡𝑎𝑛4 2𝑥)𝑠𝑒𝑐 2 2𝑥𝑑𝑥

𝑑𝑢

=

1 𝑢3 2 3

=

+

𝒕𝒂𝒏𝟑 𝟐𝒙 𝟔

+

2

𝑥 𝑥 = 𝑡𝑎𝑛2 (𝑡𝑎𝑛2 ) 2 2

(𝑢2 + 𝑢4 )𝑑𝑢 𝑢5 5

𝑥

− 2𝑡𝑎𝑛 + 𝑥 + 𝑐

𝑥 𝑥 = 𝑡𝑎𝑛2 (𝑠𝑒𝑐 2 − 1) 2 2

= (𝑢2 + 𝑢4 )( 2 ) =

2

𝑥 𝑥 𝑥 = 𝑡𝑎𝑛2 𝑠𝑒𝑐 2 − 𝑡𝑎𝑛2 2 2 2

𝑑𝑢 = 𝑠𝑒𝑐 2 2𝑥𝑑𝑥 2

1 2

3𝑥

𝑥 𝑥 𝑥 = 𝑡𝑎𝑛2 𝑠𝑒𝑐 2 − (𝑠𝑒𝑐 2 − 1) 2 2 2

𝑑𝑢 = 2𝑠𝑒𝑐 2 2𝑥 𝑑𝑥

𝑙𝑒𝑡 𝑢 = 𝑡𝑎𝑛2𝑥 ;

2

𝑥 2

𝑑𝑦 = 𝑡𝑎𝑛4 dx

+𝑐

= 𝒕𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, 𝒕𝒉𝒆 𝒎𝒊𝒔𝒔𝒊𝒏𝒈 𝒕𝒆𝒓𝒎 𝒊𝒔 "𝒕𝒂𝒏𝟒 "

𝒕𝒂𝒏𝟓 𝟐𝒙

+𝒄

𝟏𝟎

7. (𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛 𝑥)2 𝑑𝑥 3.

𝑡𝑎𝑛𝑥 𝑠𝑒𝑐 6 𝑥𝑑𝑥 ; 𝐶𝐴𝑆𝐸 𝐼

= (𝑠𝑒𝑐 2 𝑥 + 2𝑠𝑒𝑐𝑥 𝑡𝑎𝑛 𝑥 + 𝑡𝑎𝑛2 𝑥) 𝑑𝑥

1

= 𝑡𝑎𝑛𝑥 + 2𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛2 𝑥 𝑑𝑥

= 𝑡𝑎𝑛2𝑥 𝑠𝑒𝑐 4 𝑥𝑠𝑒𝑐 2 𝑥𝑑𝑥 1

= 𝑡𝑎𝑛𝑥 + 2𝑠𝑒𝑐𝑥 + (𝑠𝑒𝑐 2 𝑥 − 1) 𝑑𝑥

1

= 𝑡𝑎𝑛𝑥 + 2𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 − 𝑥 + 𝑐

= 𝑡𝑎𝑛2 𝑥(1 + 𝑡𝑎𝑛2 𝑥)2 𝑠𝑒𝑐 2 𝑥𝑑𝑥 = 𝑡𝑎𝑛2 𝑥(1 + 2𝑡𝑎𝑛2 𝑥 + 𝑡𝑎𝑛4 𝑥)𝑠𝑒𝑐 2 𝑥𝑑𝑥 1 2

5 2

= 𝟐𝒕𝒂𝒏𝒙 + 𝟐𝒔𝒆𝒄𝒙 − 𝒙 + 𝒄

9 2

2

= (𝑡𝑎𝑛 𝑥 + 2𝑡𝑎𝑛 𝑥 + 𝑡𝑎𝑛 𝑥)𝑠𝑒𝑐 𝑥𝑑𝑥 𝑙𝑒𝑡 𝑢 = 𝑡𝑎𝑛𝑥 ; 1

𝑑𝑢 = 𝑠𝑒𝑐 2 𝑥 ; 𝑑𝑢 = 𝑠𝑒𝑐 2 𝑥 𝑑𝑥 𝑑𝑥

5

9

= (𝑢2 𝑥 + 2𝑢2 𝑥 + 𝑢2 𝑥)𝑑𝑢 3

7

4𝑢 2 + 7

11

=

2𝑢 2 3

=

𝟐𝒕𝒂𝒏𝟐 𝒙 𝟒𝒕𝒂𝒏𝟐 𝒙 + 𝟑 𝟕

𝟑

+

2𝑢2 11 𝟕

+𝑐 𝟏𝟏

𝟐 𝒕𝒂𝒏 𝟐 𝒙 + 𝟏𝟏

+𝒄 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy

17

EXERCISE 10.3

POWER OF TANGENTS AND SECANTS

𝑠𝑒𝑐 3𝑥

9. (𝑡𝑎𝑛 3𝑥 )4 𝑑𝑥 𝑠𝑒𝑐 4 3𝑥 𝑡𝑎𝑛 4 3𝑥

=

1

𝑑𝑥

= 𝑡𝑎𝑛3 𝑥𝑠𝑒𝑐 −2 𝑥𝑑𝑥

= 𝑠𝑒𝑐 4 3𝑥𝑡𝑎𝑛−4 3𝑥𝑑𝑥

3

= 𝑡𝑎𝑛2 𝑥𝑡𝑎𝑛𝑥𝑠𝑒𝑐 −2 𝑥𝑠𝑒𝑐𝑥𝑑𝑥

= 𝑠𝑒𝑐 2 3𝑥𝑠𝑒𝑐 2 3𝑥𝑡𝑎𝑛−4 3𝑥𝑑𝑥

3

= (𝑠𝑒𝑐 2 𝑥 − 1)𝑡𝑎𝑛𝑥𝑠𝑒𝑐 −2 𝑥𝑠𝑒𝑐𝑥𝑑𝑥

= 𝑠𝑒𝑐 2 3𝑥(1 + 𝑡𝑎𝑛2 3𝑥)𝑡𝑎𝑛−4 3𝑥𝑑𝑥 = (𝑡𝑎𝑛−4 3𝑥 + 𝑡𝑎𝑛−2 3𝑥)𝑠𝑒𝑐 2 3𝑥𝑑𝑥 𝑑𝑢 𝑙𝑒𝑡 𝑢 = 𝑡𝑎𝑛 3𝑥 ; = 3𝑠𝑒𝑐 2 3𝑥 𝑑𝑥 𝑑𝑢 = 𝑠𝑒𝑐 2 3𝑥𝑑𝑥 3

1

=

−

𝑢 −2 3

1 𝑡𝑎𝑛 −3 3𝑥 3 −3

= −

−

𝑑𝑢 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥𝑑𝑥 1

3

3

=

+𝑐

2𝑢 2 3

1

− 2𝑢−2 + 𝑐 𝟑

𝑡𝑎𝑛 −1 3𝑥 3

𝒄𝒐𝒕𝟑 𝟑𝒙 𝒄𝒐𝒕𝟑𝒙 − 𝟑 + 𝟗

𝑑𝑢 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 𝑑𝑥

𝑙𝑒𝑡 𝑢 = 𝑠𝑒𝑐 𝑥 ;

1

1 𝑢 −3 −3

3

= (𝑠𝑒𝑐 2 𝑥 − 𝑠𝑒𝑐 −2 𝑥)𝑡𝑎𝑛𝑥𝑠𝑒𝑐𝑥𝑑𝑥

= (𝑢2 − 𝑢−2 )𝑑𝑢

=3 (𝑢−4 + 𝑢−2 ) 𝑑𝑢 =3

𝑡𝑎𝑛 3 𝑥 𝑑𝑥 𝑠𝑒𝑐𝑥

11.

+𝑐

=

𝟐𝒔𝒆𝒄𝟐 𝒙 − 𝟑

𝟐 𝟏

+𝒄

𝒔𝒆𝒄𝟐 𝒙

𝒄


18

EXERCISE 10.4

POWER OF COTANGENTS AND COSECANTS

1. 𝑐𝑜𝑡 4 𝑥𝑐𝑠𝑐 4 𝑥𝑑𝑥

= 𝑐𝑜𝑡 4 𝑥(1 + 𝑐𝑜𝑡 2 𝑥)𝑐𝑠𝑐 2 𝑥𝑑𝑥

1

= 𝑐𝑜𝑡 2 3𝑥 𝑐𝑠𝑐 2 3𝑥 𝑐𝑠𝑐 2 3𝑥 𝑑𝑥

= (𝑐𝑜𝑡 4 𝑥 + 𝑐𝑜𝑡 6 𝑥)𝑐𝑠𝑐 2 𝑥𝑑𝑥 𝑙𝑒𝑡 𝑢 = 𝑐𝑜𝑡 𝑥 ;

1

= 𝑐𝑜𝑡 2 3𝑥 1 + 𝑐𝑜𝑡 2 3𝑥 𝑐𝑠𝑐 2 3𝑥 𝑑𝑥

𝑑𝑢 = −𝑐𝑠𝑐 2 𝑥 ; 𝑑𝑢 = −𝑐𝑠𝑐 2 𝑥𝑑𝑥 𝑑𝑥

= − (𝑢4 + 𝑢6 )𝑑𝑢 =-

𝑢5 5

= -

3.

+

𝑢7 7

5

𝑑𝑢 𝑑 𝑥 = −3 𝑐𝑠𝑐 2 3𝑥 𝑑𝑥 𝑑𝑥

c −

𝑐𝑜𝑡 5 4𝑥𝑑𝑥

= 𝑐𝑜𝑡 3 4𝑥𝑐𝑜𝑡 2 4𝑥𝑑𝑥 3

1

= 𝑐𝑜𝑡 2 3𝑥 + 𝑐𝑜𝑡 2 3𝑥 𝑐𝑠𝑐 2 3𝑥 𝑑𝑥 𝑙𝑒𝑡 𝑢 = 𝑐𝑜𝑡 3𝑥

+c

𝒄𝒐𝒕𝟓 𝒙 𝒄𝒐𝒕𝟕 𝒙 + + 𝟓 𝟕

𝑐𝑜𝑠 3𝑥 𝑐𝑠𝑐 4 3𝑥 𝑑𝑥

5.

2

= 𝑐𝑜𝑡 4𝑥(𝑐𝑠𝑐 4𝑥 − 1)𝑑𝑥

𝑑𝑢 = 𝑐𝑠𝑐 2 3𝑥 𝑑𝑥 3

=−

1 3

=−

1 3

1

5

𝑢2 + 𝑢2 𝑑𝑢 3

𝑢2 3 2

7

+

𝑢2 7 2

+𝑐

𝟑 𝟕 𝟐 𝟐 = − 𝒄𝒐𝒕𝟐 𝟑𝒙 − 𝒄𝒐𝒕𝟐 𝟑𝒙 + 𝒄 𝟗 𝟐𝟏

= (𝑐𝑜𝑡 3 4𝑥𝑐𝑠𝑐 2 4𝑥 − 𝑐𝑜𝑡 3 4𝑥)𝑑𝑥 = [𝑐𝑜𝑡 3 4𝑥𝑐𝑠𝑐 2 4𝑥 − (𝑐𝑠𝑐 2 4𝑥 − 1)𝑐𝑜𝑡4𝑥]𝑑𝑥 =

𝑐𝑜𝑡 3 4𝑥𝑐𝑠𝑐2 4𝑥𝑑𝑥 − 𝑐𝑜𝑡4𝑥𝑐𝑠𝑐2 4𝑥𝑑𝑥 − 𝑐𝑜𝑡4𝑥𝑑𝑥

𝑙𝑒𝑡 𝑢 = 𝑐𝑜𝑡 4𝑥 ; 1

𝑑𝑢 = 𝑐𝑠𝑐 2 4𝑥𝑑𝑥 −4

1

1

=− 4 𝑢3 𝑑𝑢 − 4 𝑢𝑑𝑢 + 4 𝑙𝑛⁡(𝑐𝑜𝑠4𝑥) 1 𝑢4 𝑑𝑥 4

=− 4

= −

−

𝑢2 2

1 4

+ 𝑙𝑛 𝑐𝑜𝑠4𝑥 + 𝑐

𝒄𝒐𝒕𝟒 𝟒𝒙 𝒄𝒐𝒕𝟐 𝟒𝒙 + 𝟏𝟔 𝟖

𝟏

+ 𝟒 𝒍𝒏 𝒄𝒐𝒔𝟒𝒙 + 𝒄


19

EXERCISE 10.4

POWER OF COTANGENTS AND COSECANTS

7.

𝑐𝑜𝑠 5 2𝑥𝑑𝑥 𝑠𝑖𝑛 8 2𝑥

𝑐𝑜𝑠 5 2𝑥𝑑𝑥 𝑠𝑖𝑛 5 2𝑥

=

𝑐𝑜𝑡 5 2𝑥 𝑐𝑠𝑐 3 2𝑥 𝑑𝑥 \

=

𝑐𝑜𝑡 4 2𝑥 𝑐𝑠𝑐 2 2𝑥 𝑐𝑠𝑐 2𝑥 𝑐𝑜𝑡 2𝑥 𝑑𝑥

=

𝑐𝑠𝑐 2 2𝑥 − 1

=

2

1 𝑠𝑖𝑛 3 2𝑥

𝑑𝑥

𝑐𝑠𝑐 2 2𝑥 𝑐𝑠𝑐 2𝑥 𝑐𝑜𝑡 2𝑥 𝑑𝑥

=

𝑐𝑠𝑐 4 2𝑥 − 2 𝑐𝑠𝑐2 2𝑥 + 1 𝑐𝑠𝑐 2 2𝑥 𝑐𝑠𝑐 2𝑥 𝑐𝑜𝑡 2𝑥 𝑑𝑥

=

𝑐𝑠𝑐 6 2𝑥 − 2 𝑐𝑠𝑐 4 2𝑥 + 𝑐𝑠𝑐 2 2𝑥 𝑐𝑠𝑐 2𝑥 𝑐𝑜𝑡 2𝑥 𝑑𝑥

𝑙𝑒𝑡 𝑢 = 𝑐𝑠𝑐 2𝑥

=−

1 2

𝑢6 − 2𝑢4 + 𝑢2 𝑑𝑢

=−

1 𝑢7 2 7

−

= −

𝑑𝑥

=

𝑐𝑜𝑡 −6 𝑥 𝑐𝑠𝑐 2 𝑥 𝑐𝑠𝑐 2 𝑥 𝑑𝑥

=

𝑐𝑜𝑡 −6 𝑥 1 + 𝑐𝑜𝑡 2 𝑥 𝑐𝑠𝑐 2 𝑥 𝑑𝑥

=

𝑐𝑜𝑡 −6 𝑥 + 𝑐𝑜𝑡 −4 𝑥 𝑐𝑠𝑐 2 𝑥 𝑑𝑥

𝑙𝑒𝑡: 𝑢 = 𝑐𝑜𝑡 𝑥 𝑑𝑢 = − 𝑐𝑠𝑐 2 𝑥 𝑑𝑥 𝑑𝑥

= −1

𝑑𝑢 = 𝑐𝑠𝑐 2𝑥 𝑐𝑜𝑡 2𝑥 𝑑𝑥 2

2𝑢 5 5

𝑐𝑠𝑐 4 𝑥 𝑐𝑜𝑡 6 𝑥

−𝑑𝑢 = 𝑐𝑠𝑐 2 𝑥𝑑𝑥

𝑑𝑢 𝑑(𝑥) = −2 𝑐𝑠𝑐 2𝑥 𝑐𝑜𝑡 2𝑥 𝑑𝑥 𝑑𝑥 −

9.

+

𝑢3 3

𝑢−6 + 𝑢−4 𝑑𝑥

= −1 − =

+𝑐

𝒄𝒔𝒄𝟕 𝟐𝒙 𝒄𝒔𝒄𝟓 𝟐𝒙 𝒄𝒔𝒄𝟑 𝟐𝒙 + − +𝒄 𝟏𝟒 𝟓 𝟔

𝑐𝑜𝑡 −5 𝑥 5

𝑢−5 𝑢−3 − +𝑐 5 3

+

𝑐𝑜𝑡 −3 𝑥 3

+𝑐

𝒕𝒂𝒏𝟓 𝒙 𝒕𝒂𝒏𝟑 𝒙 = + +𝒄 𝟓 𝟑


20

EXERCISE 10.5

1.

TRIGONOMETRIC SUBSTITUTIONS

𝑥 2 𝑑𝑥

3.

4−𝑥 2

𝑑𝑥

𝑥 = 𝑢 ; 𝑎 = 2 ; 22 − 𝑥 2

𝑢 = 3𝑥

𝑢 = 𝑎 𝑠𝑖𝑛 𝜃 𝑥 𝑠𝑖𝑛 𝜃 = 2 𝑥 = 2𝑠𝑖𝑛 𝜃 𝑑𝑥 = 2𝑐𝑜𝑠𝜃𝑑𝜃

𝑎=2

= =

𝑢 = 𝑎𝑡𝑎𝑛𝜃 3𝑥 = 2𝑡𝑎𝑛𝜃 2 3𝑥 𝑥 = 𝑡𝑎𝑛𝜃 ; 𝑡𝑎𝑛𝜃 = 3 2

𝑥 2 𝑑𝑥 4 − 𝑥2

2 𝑑𝑥 = 𝑠𝑒𝑐 2 𝜃𝑑𝜃 3

4 𝑠𝑖𝑛 𝜃 2𝑐𝑜𝑠𝜃𝑑𝜃 2𝑐𝑜𝑠𝜃 𝑠𝑖𝑛2 𝜃𝑑𝜃

=4

1 − 𝑐𝑜𝑠2𝜃𝑑𝜃 2 𝑑𝜃 −

𝑥 2

= 𝟐𝒂𝒓𝒄𝒔𝒊𝒏

1 2

𝑑𝑥

=

𝑥 9𝑥 2 + 4 2 𝑠𝑒𝑐 2 𝜃𝑑𝜃 3

=

2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 ]2 + C

𝒙 𝒙 −𝟐 𝟐 𝟐

9𝑥 2 + 4

2𝑠𝑒𝑐𝜃 =

1 𝑥 𝑠𝑖𝑛2𝜃 + 𝐶 ; 𝜃 = 𝑠𝑖𝑛−( ) 2 2

= 2[𝑎𝑟𝑐𝑠𝑖𝑛 −

9𝑥 2 + 4 2

𝑠𝑒𝑐𝜃 =

=4

=2

;

𝑥 9𝑥 2 +4

𝟒 − 𝒙𝟐 +𝑪 𝟐

2 𝑡𝑎𝑛𝜃2𝑠𝑒𝑐𝜃 3

𝑠𝑒𝑐𝜃𝑑𝜃 2𝑡𝑎𝑛𝜃

= 1 = 2

1 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

𝑑𝜃

=

1 2

1 𝑑𝜃 𝑠𝑖𝑛𝜃

=

1 2

𝑐𝑠𝑐𝜃𝑑𝜃

=

1 [-𝑙𝑛⁡|𝑐𝑠𝑐𝜃 2

= −

𝟏 𝒍𝒏 𝟐

+ 𝑐𝑜𝑡𝜃|] + 𝐶

𝟗𝒙𝟐 + 𝟒 𝟐 − +𝑪 𝟑𝒙 𝟑𝒙


21

EXERCISE 10.5

5.

𝑥 2 𝑑𝑥 3 9−𝑥 2 2

7.

𝑥 2 𝑑𝑥

=

=


9−

9 − 𝑥2

3 𝑑𝑥 = 𝑐𝑜𝑠𝜃𝑑𝜃 2

𝑥 = 3𝑠𝑖𝑛𝜃 𝑑𝑥 = 3𝑐𝑜𝑠𝜃

=

= =

2𝑥

𝜃 = 𝑠𝑖𝑛−1 ( )

(3𝑠𝑖𝑛𝜃)2 3 𝑐𝑜𝑠 𝜃 9 − (3𝑠𝑖𝑛𝜃)2 3 𝑐𝑜𝑠 𝜃 9𝑠𝑖𝑛2 𝜃 1 − 𝑠𝑖𝑛2 𝜃

3

= =

𝑠𝑖𝑛2 𝜃 (1 − 𝑠𝑖𝑛2 𝜃)

=

𝑠𝑖𝑛2 𝜃 𝑑𝜃 𝑐𝑜𝑠 2 𝜃 𝑡𝑎𝑛2 𝜃𝑑𝜃 −

3 𝑥 = 𝑠𝑖𝑛𝜃 2 2𝑥 = 𝑠𝑖𝑛𝜃 3

𝑢 = 𝑎𝑠𝑖𝑛𝜃

=

𝑑𝑥

𝑢 = 𝑎𝑠𝑖𝑛𝜃 ; 2𝑥 = 3𝑠𝑖𝑛𝜃

𝑥 2 𝑑𝑥

𝑢=𝑥 ; 𝑎=3

=

𝑥2

𝑎 = 3 ; 𝑢 = 2𝑥

𝑥2 3

9 − 𝑥2

9−4𝑥 2

3 2

3𝑐𝑜𝑠𝜃 ( 𝑐𝑜𝑠𝜃𝑑𝜃 ) 3 2

( 𝑠𝑖𝑛𝜃 )2 3𝑐𝑜𝑠𝜃 (3𝑐𝑜𝑠𝜃𝑑𝜃 ) 3 2

2( 𝑠𝑖𝑛𝜃 )2 9𝑐𝑜𝑠 2 𝜃𝑑𝜃 2(

9 ) 4𝑠𝑖𝑛 2 𝜃

𝑠𝑒𝑐 2 𝜃𝑑𝜃

= 𝑡𝑎𝑛𝜃 − 𝜃 =

𝒙

𝒙 − 𝑨𝒓𝒄𝒔𝒊𝒏 + 𝒄 𝟑 𝟗 − 𝒙𝟐


22

EXERCISE 10.5

𝑑𝑥 𝑥 2 +4 2

9.


𝑥 = 2 𝑡𝑎𝑛𝜃 ;

𝑑𝑥

11.

; 𝑤𝑕𝑒𝑟𝑒: 𝑢 = 𝑥 , 𝑎 = 2

𝑥 𝑥 2 −9

𝑎 = 3 ;𝑢 = 𝑥

𝑥 𝑡𝑎𝑛𝜃 = 2

𝑢 = 𝑎𝑠𝑒𝑐𝜃 𝑥 = 3𝑠𝑒𝑐𝜃 ; 𝑑𝑥 = 3𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃 𝑥 𝑥 𝑠𝑒𝑐𝜃 = ; 𝜃 = 𝐴𝑟𝑐𝑠𝑒𝑐 3 3

𝑥

𝐷𝑥 = 2 𝑠𝑒𝑐 2 𝜃𝑑𝜃 ; 𝜃 = 𝑎𝑟𝑐𝑡𝑎𝑛 2

x

𝑥2 + 4

x 𝑥2 − 9

2 𝑥2 + 4 2

𝑠𝑒𝑐𝜃 =

𝑥2 + 4

2𝑠𝑒𝑐𝜃 =

3

2

4 𝑠𝑒𝑐 2 𝜃 = 𝑥 2 + 4 𝑡𝑎𝑛𝜃 =

2

2 𝑠𝑒𝑐 𝜃𝑑𝜃 4 𝑠𝑒𝑐 2 𝜃 2

=

2 𝑠𝑒𝑐 2 𝜃𝑑𝜃 = 16 𝑠𝑒𝑐 4 𝜃

= 1 8

𝑐𝑜𝑠 2 𝜃𝑑𝜃

=

1 8

1 + 𝑐𝑜𝑠2𝜗 𝑑𝜃 2

=

1 8

=

1 𝑑𝜃 + 2

1 1 𝜃 2

=8 =

𝑑𝜃 1 = 2 8 𝑠𝑒𝑐 𝜃 8

𝑐𝑜𝑠2𝜃 𝑑𝜃 2

𝑑𝜃 𝑠𝑒𝑐 2 𝜃

= =

𝑥2 − 9 ; 3𝑡𝑎𝑛𝜃 = 3 𝑑𝑥

𝑥 𝑥2 − 9

=

𝑥2 − 9

3𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃 3𝑠𝑒𝑐𝜃(3𝑡𝑎𝑛𝜃)

𝑑𝜃 3

=

1 𝜃 3

=

𝟏 𝒙 𝑨𝒓𝒄𝒔𝒆𝒄 + 𝒄 𝟑 𝟑

1

+ 4 (2)𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 + 𝑐

𝟏 𝟏 𝜽+ 𝒔𝒊𝒏𝜽𝒄𝒐𝒔𝜽 + 𝒄 𝟏𝟔 𝟏𝟔


23

EXERCISE 10.5


𝑑𝑥

3

𝑥 2 − 16 2 𝟏𝟑. ( ) 𝑥3

𝟏𝟓.

𝑢 = 𝑥; 𝑎 = 4

5 − 12𝑥 + 4𝑥 2 = 2𝑥 − 9 − 4

𝑢 = 𝑠𝑒𝑐∅ ; 𝑥 = 4𝑠𝑒𝑐∅ ; 𝑠𝑒𝑐∅ =

𝑥 4

𝑢 = 𝑎𝑠𝑒𝑐∅ ; 2𝑥 − 3 = 2𝑠𝑒𝑐∅ 2𝑥 − 3 = 2𝑠𝑒𝑐∅ ; 2𝑥 = 2𝑠𝑒𝑐∅ + 3

𝑥 3 = 64 sec 3 ∅; 𝑑𝑥 = 4𝑠𝑒𝑐∅𝑡𝑎𝑛∅𝑑∅ 𝑥 2 − 16 ; 4𝑡𝑎𝑛∅ = 4

𝑥 2 − 16

=4 =4

2𝑑𝑥 = 2𝑠𝑒𝑐∅𝑡𝑎𝑛∅ 𝑑𝑥 = 𝑠𝑒𝑐∅𝑡𝑎𝑛∅ 𝑠𝑒𝑐∅ =

3

=

5 − 12𝑥 + 4𝑥 2

𝑎 = 2 ; 𝑢 = 2𝑥 − 3

𝑥 ∅ = 𝑎𝑟𝑐𝑠𝑒𝑐∅ 4

𝑡𝑎𝑛∅ =

2𝑥 − 3

( 4𝑡𝑎𝑛∅ (4𝑠𝑒𝑐∅𝑡𝑎𝑛∅𝑑∅)) (64sec3 )

2𝑥 − 3 2

∅ = 𝑎𝑟𝑐𝑠𝑒𝑐

tan4 ∅𝑑∅ sec⁡^2∅

𝑡𝑎𝑛∅ =

(sec 2 −1)^2𝑑∅ sec 2 ∅

2𝑡𝑎𝑛∅ =

2𝑥 − 3 2

2𝑥 − 3 2

2𝑥 − 3

=4

sec 4 −2 sec 2 ∅ + 1 sec 2 ∅

=

(𝑠𝑒𝑐∅𝑡𝑎𝑛∅) 2𝑠𝑒𝑐∅2𝑡𝑎𝑛∅

=4

sec 4 ∅ − 2 sec 2 ∅ + 1 sec 2 ∅

=

1 4

𝑠𝑒𝑐∅𝑡𝑎𝑛∅ 𝑠𝑒𝑐∅𝑡𝑎𝑛∅

=4

sec 2 ∅ − 2 + 1/ sec 2 ∅𝑑∅

=

1 4

𝑑∅

1

= 4(𝑡𝑎𝑛∅ − 2∅ + 2 ∅ + 𝑠𝑖𝑛∅𝑐𝑜𝑠∅ =

1 4

= ∅; ∅ = 𝑎𝑟𝑐𝑠𝑒𝑐

𝒙 𝟖 𝒙𝟐 − 𝟏𝟔 𝒙𝟐 − 𝟏𝟔 − 𝟔 𝐚𝐫𝐜𝐬𝐞𝐜 + +𝒄 𝟒 𝒙𝟐

=

2

−4 2

−4

2𝑥−3 2

𝟏 𝟐𝒙 − 𝟑 𝒂𝒓𝒄𝒔𝒆𝒄 + 𝒄 𝟒 𝟐


24

EXERCISE 10.6

𝟏.

𝑥2

ADDITIONAL STANDARD FORMULAS

𝑑𝑥 + 25

36 − 9𝑥 2 𝑑𝑥

𝟕.

Let: 𝑢 = 𝑥

Let: 𝑎 = 6

𝑎=5

𝑢 = 3𝑥

𝑑𝑢 = 𝑑𝑥

𝑑𝑢 = 𝑑𝑥 3

=

𝟏 𝒙 𝑨𝒓𝒄𝒕𝒂𝒏 + 𝒄 𝟓 𝟓

𝑥𝑑𝑥

𝟑.

1 − 𝑥4

=

1 3𝑥 1 3𝑥 36 − 9𝑥 2 + 8𝐴𝑟𝑐𝑠𝑖𝑛 +𝑐 3 2 3 6 1 3𝑥 2

1

=

Let: 𝑢 = 𝑥 2

𝑥

36 − 9𝑥 2 + 3 8𝐴𝑟𝑐𝑠𝑖𝑛 2 + 𝑐

=3

𝒙 𝒙 𝟑𝟔 − 𝟗𝒙𝟐 + 𝟔𝑨𝒓𝒄𝒔𝒊𝒏 + 𝒄 𝟐 𝟐

𝑎=1 𝑑𝑢 = 𝑑𝑥 2

𝟗.

1 𝑥2 = 𝐴𝑟𝑐𝑠𝑖𝑛 + 𝑐 2 1

Let: 𝑎 = 5

𝟏 𝟐

= 𝑨𝒓𝒄𝒔𝒊𝒏𝒙𝟐 + 𝒄

16𝑥 2 + 25𝑑𝑥

𝑢 = 4𝑥 𝑑𝑢 = 𝑑𝑥 4

𝟓.

𝑑𝑥 49 − 25𝑥 2

Let: 𝑎 = 7

=

=

1 4𝑥 4

2

16𝑥 2 + 25 +

1 52 4

2

𝑙𝑛 4𝑥 + 16𝑥 2 + 25 + 𝑐

𝟏 𝟐𝟓 𝟏𝟔𝒙𝟐 + 𝟐𝟓 + 𝒍𝒏 𝟒𝒙 + 𝟏𝟔𝒙𝟐 + 𝟐𝟓 + 𝒄 𝟐𝒙 𝟖

𝑢 = 5𝑥 𝑑𝑢 2

= 𝑑𝑥

=

𝟏 𝟓𝒙 − 𝟕 𝒍𝒏 +𝒄 𝟕𝟎 𝟓𝒙 + 𝟕


25

EXERCISE 10.7

INTEGRANDS INVOLVING QUADRATIC EQUATIONS

𝟑.

𝑑𝑥 𝑥 2 −3𝑥+2

1.

𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑖𝑛𝑔 𝑡𝑕𝑒 𝑠𝑞𝑢𝑎𝑟𝑒

2𝑥 2

𝑑𝑥

=

1 2

𝑥−2

2

𝑥 − 3𝑥 = −2 𝑥 2 − 3𝑥 = 3 2

2

3 𝑥− 2

2

𝑥−

9 9 = −2 + 4 4

=

(𝑥 −

=

𝑢2 1

−

2

1 2

= 𝒍𝒏

𝑑𝑢 1 𝑢−𝑎 = 𝑙𝑛 +𝑐 2 −𝑎 2𝑎 𝑢+𝑎

𝑙𝑛

1 2

1 4

1 2

=

𝑑𝑢 + 𝑎2

= 𝑎𝑟𝑐𝑡𝑎𝑛

3 𝑢=𝑥− 2 𝑎=

𝑢2

1 1 𝑎= , 𝑢=𝑥− 2 2

1 − 4 3 2 ) 2

=

1

+4

1 𝑢 = 𝑎𝑟𝑐𝑡𝑎𝑛 + 𝑐 2 𝑎

1 4

𝑑𝑥

=

𝑑𝑥 − 2𝑥 + 1

3 1 2 2 3 1 𝑥− + 2 2

𝑥− −

+𝑐

𝒙−𝟐 +𝒄 𝒙−𝟏

1 2

𝑥− 1 2

+𝑐

=

𝟏 𝒂𝒓𝒄𝒕𝒂𝒏𝟐𝒙 − 𝟏 + 𝒄 𝟐

𝟓.

3 − 2𝑥 − 𝑥 2

=

4− 𝑥+1

2

𝑢 = 𝑥 + 1, 𝑎 = 2 𝑎2

= =

𝑥+1 2

=

−

𝑢2

𝑢 𝑎2 𝑢 2 2 = 𝑎 − 𝑢 + 𝑎𝑟𝑐𝑠𝑖𝑛 + 𝑎 2 𝑎 4

3 − 2𝑥 − 𝑥 2 + 2 𝑎𝑟𝑐𝑠𝑖𝑛

𝑥+1 2

+𝑐

𝒙+𝟏 𝒙+𝟏 𝟑 − 𝟐𝒙 − 𝒙𝟐 + 𝟐𝒂𝒓𝒄𝒔𝒊𝒏 +𝒄 𝟐 𝟐


26

EXERCISE 10.7 𝟕.

𝑥2

INTEGRANDS INVOLVING QUADRATIC EQUATIONS

𝑑𝑥 − 8𝑥 + 7

2𝑥−3𝑑𝑥 4𝑥 2 −1

11.

2𝑥𝑑𝑥 4𝑥 2 −1

Completing the square

=

𝑥 2 − 8𝑥 = −7

=2

𝑥 2 − 8𝑥 + 16 = −7 + 16 𝑥−4

2

=9

𝑥−4

2

−9=0

𝑥𝑑𝑥 4𝑥 2 −1

−3

𝑑𝑥 4𝑥 2 −1

𝑙𝑒𝑡 𝑢 = 4𝑥 2 − 1 ; 𝑑𝑢 8

=2

𝑑𝑥 (𝑥 − 4)2 + 9

=

3𝑑𝑥 4𝑥 2 −1

−

=

𝑢

1

− 3[2 𝑙𝑛

𝟏 𝒍𝒏|𝟒𝒙𝟐 𝟒

𝑑𝑢 = 𝑥𝑑𝑥 8 4𝑥 2 −1 4𝑥 2 +1 𝟑

− 𝟏| − 𝟒 𝒍𝒏

+ 𝑐] 𝟐𝒙−𝟏 𝟐𝒙+𝟏

+𝒄

𝑎 = 3 ;𝑢 = 𝑥 −4 = =

𝑢2

1 𝑢−𝑎 𝑙𝑛 +𝑐 2𝑎 𝑢+𝑎 1

= 6 𝑙𝑛 =

𝑑𝑢 − 𝑎2

𝑥−4−3 𝑥−4+3

+𝑐

=

𝟏 𝒙−𝟕 𝒍𝒏 +𝒄 𝟔 𝒙−𝟏

(2𝑥+7)𝑑𝑥 𝑥 2 +2𝑥+5

13.

=

2𝑥+2 +5𝑑𝑥 𝑥 2 +2𝑥+5 2𝑥+2 𝑥 2 +2𝑥+5

+5

𝑑𝑥 (𝑥+1)2 +4

𝑙𝑒𝑡 𝑢 = 𝑥 2 + 2𝑥 + 5 ; 𝑑𝑢 = (2𝑥𝑡2)𝑑𝑥 =

3+2𝑥 𝑑𝑥 𝑥 2 +9

9. = =3

2𝑥𝑑𝑥 𝑥 2 +9

𝑑𝑥 + 𝑥 2 +9

1

+ 2 𝐴𝑟𝑐𝑡𝑎𝑛

𝑥+1 2

+𝑐 𝟏

=

3𝑑𝑥 + 𝑥 2 +9

𝑑𝑢 𝑢

2

⁡𝒍𝒏|𝒙𝟐 + 𝟐𝒙 + 𝟓| + 𝟐 𝑨𝒓𝒄𝒕𝒂𝒏

𝒙+𝟏 + 𝟐

𝒄

𝑥𝑑𝑥 𝑥 2 +9

𝑙𝑒𝑡 𝑢 = 𝑥 2 + 9 ; 𝑑𝑢 = 2𝑥𝑑𝑥 1

𝑥

= 33 𝐴𝑟𝑐𝑡𝑎𝑛 3 + 2

𝑑𝑢 2

𝑢

𝒙

= 𝑨𝒓𝒄𝒕𝒂𝒏 𝟑 + 𝒍𝒏 𝒙𝟐 + 𝟗 + 𝒄


27

EXERCISE 10.7 (𝑥−3)𝑑𝑥

15. = =

INTEGRANDS INVOLVING QUADRATIC EQUATIONS 19.

4𝑥−𝑥 2 𝑥−2 −1𝑑𝑥

=2

4𝑥−𝑥 2 𝑥−2 𝑑𝑥 4𝑥−𝑥 2

𝑙𝑒𝑡 𝑢 = 𝑑𝑢 =

4𝑥−𝑥 2

4𝑥 −

𝑥2

−2(𝑋−2)𝑑𝑥 2 4𝑥−𝑥 2

+

17 2

𝑑𝑥 𝑥 2 −4𝑥+20

𝑥 2 − 4𝑥 + 20 = 𝑥 − 2

4 − 2𝑥

; 𝑑𝑢 =

2 4𝑥 − 𝑥 2

𝑑𝑥 = 2[

; 4𝑥 − 𝑥 2 = 4 − (2 − 𝑥)2

𝑑𝑢 𝑢

+

17 2

2

𝟐−𝒙 + 𝟐

+ 16

𝑑𝑥 ] 𝑥−2 2 +16

= 2[𝑙𝑛 𝑥 2 − 4𝑥 + 20 +

17 1 𝑥−2 ( )Arctan 4 2 4

= 𝟐 𝒍𝒏 𝒙𝟐 − 𝟒𝒙 + 𝟐𝟎 +

4−(2−𝑥)2

= − 𝟒𝒙 − 𝒙𝟐 − 𝑨𝒓𝒄𝒔𝒊𝒏

𝟏𝟕 𝒙−𝟐 Arctan + 𝟒 𝟒

+ 𝑐] 𝒄

𝒄

𝑥+3 𝑑𝑥

17.

8𝑥−𝑥 2 𝑥−4 +7𝑑𝑥

=

8𝑥−𝑥 2 𝑥−4 𝑑𝑥 8𝑥−𝑥 2

𝑙𝑒𝑡 𝑢 = 𝑑𝑢 = =

2𝑥+4𝑑𝑥 𝑥 2 −4𝑥+20

2(2𝑥+4+17)𝑑𝑥 𝑥 2 −4𝑥+20

𝑑𝑢

= 𝑑𝑢 −

=

=

𝑙𝑒𝑡 𝑢 = 𝑥 2 − 4𝑥 + 20 ; 𝑑𝑢 = (2𝑥 − 4)𝑑𝑥

𝑑𝑥

−

(4𝑥+9)𝑑𝑥 𝑥 2 −4𝑥+20

+7

𝑑𝑥 8𝑥−𝑥 2

8𝑥 − 𝑥 2 ; 𝑑𝑢 =

−2(𝑥 − 4)𝑑𝑥 2 8𝑥

− 𝑥2

8 − 2𝑥 2 8𝑥 − 𝑥 2

𝑑𝑥

; 8𝑥 − 𝑥 2

16 − (4 − 𝑥)2

=− 𝑑𝑢 + 7

𝑑𝑢 16−(4−𝑥)2

= - 𝟖𝒙 − 𝒙𝟐 + 𝟕𝑨𝒓𝒄𝒔𝒊𝒏

𝟒−𝒙 + 𝟒

𝒄


28

EXERCISE 10.8

ALGEBRAIC SUBSTITUTION

𝑑𝑥

1.

5

𝑧=

3

𝑥

5

=

𝑧 2 𝑑𝑧 𝑧3 − 𝑧2

=3

3

9𝑥 6 −10𝑥 4 30 𝟏

+𝑐

𝟏

𝟏

𝒙𝟐 (𝟗𝒙𝟑 − 𝟏𝟎𝒙𝟒 ) = +𝒄 𝟑𝟎

𝑑𝑧 𝑧−1

=3

7

3𝑥 6 𝑥 4 = − +𝑐 10 3

2 𝑥−𝑥 3

𝑢 = 𝑧−1 𝑑𝑢 = 𝑑𝑧 5.

𝑑𝑢 𝑢

=3

𝑑𝑥 𝑥+2

= 3 𝑙𝑛 |𝑧 − 1 | + 𝑐

𝑧 = 𝑥+2 𝑧4 = 𝑥 + 2 𝑥 = 2 − 𝑧4 𝑑𝑥 = −4𝑧 3 𝑑𝑧

𝟑

= 𝟑 𝒍𝒏 | 𝒙 − 𝟏 | + 𝒄

1

1

𝑥+2 2

4

= 3 𝑙𝑛 𝑢 + 𝑐

= −4

𝑧 3 𝑑𝑧 𝑧3 − 𝑧2

= −4

𝑧𝑑𝑧 𝑧−1

1

(𝑥 3 −𝑥 4 𝑑𝑥

3.

3 4−

𝑢 = 𝑧−1

1

4𝑥 2

𝑧=

12

𝑑𝑢 = 𝑑𝑧

𝑥

𝑧 = 𝑢+1

𝑑𝑥 = 12𝑧11 𝑑𝑧 5

=3

(𝑧 4 −𝑧 3 ) 𝑧 11 𝑑𝑧 𝑧8

= 3 (𝑧 9 − 𝑧 8 ) 𝑑𝑥 = 3[ 𝑧 9 𝑑𝑧 −

𝑧 8 𝑑𝑧]

= −4

𝑢 + 1 𝑑𝑢 𝑢

= −4[

𝑢 𝑑𝑢 + 𝑢

𝑑𝑢 ] 𝑢

= −4[𝑢 + 𝑙𝑛 𝑢 + 𝑐 = −𝟒[𝒛 − 𝟏 + 𝒍𝒏 𝒛 − 𝟏 + 𝒄]

10

9

𝑧 𝑧 = 3[ − + 𝑐] 10 9 =

3𝑧10 𝑧 9 − +𝑐 10 3 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy

29

EXERCISE 10.8

7.


1

4 + 𝑥𝑑𝑥 ;

9. 𝑥 𝑥 + 4 3 𝑑𝑥 𝑧 = 𝑥+4

𝑧 =(4+ 𝑥)1/2

12

𝑥 (𝑧 2 − 4)2= 𝑥𝑑𝑥 = 4𝑧 3 − 16𝑧 𝑑𝑧

=

𝑧 4𝑧 3 − 16𝑧 𝑑𝑧

=

4𝑧 4 − 16𝑧 2 𝑑𝑧

=4

𝑧5 𝑧3 − 16 +𝐶 5 3

=

4 (4 + 5

= 4+ 𝑥

= 4+ 𝑥

= 4+ 𝑥

𝑥)5/2−

4 4+ 𝑥 5

3 2

3 2

3 2

=

4 15

=

𝟒 𝟒+ 𝒙 𝟓

4+ 𝑥

16 (4 + 3

; 𝑧3 = 𝑥 + 4

𝑥 = 𝑧 3 − 4 ; 𝑑𝑥 = 3𝑧 3 𝑑𝑧

𝑧 2 − 4 = 𝑥𝑧 4 − 8𝑧 2 + 16 = 𝑥 𝑧=

1 3

𝑧 3 − 4 𝑧 3𝑧 2 𝑑𝑧

=

𝑧 6 𝑑𝑧 − 4

=3

𝑧 3 𝑑𝑧

7

4 3𝑧 3 = − 3𝑧 3 + 𝑐 7

3 𝑥+4 = 7

𝑥)3/2+C

7 3

−3 𝑥+4

4 3

+𝑐

4

16 − +𝐶 3

12 4 + 𝑥 − 80 +𝐶 15

=

3 𝑥+4 3 7

𝑥+4−7 +𝑐 1 3

𝑥 𝑥 + 4 𝑑𝑥 =

𝟑 𝒙+𝟒

𝟒 𝟑

𝟕

𝒙−𝟑

+𝒄

48 + 12 𝑥 − 80 +𝐶 15 3 2

𝟑 𝟐

12 + 3 𝑥 − 20 + 𝐶 𝟑 𝒙−𝟖 +𝑪


30

EXERCISE 10.8

11.

4− 2𝑥+1 1−2𝑥


𝑑𝑥

𝑧 = 2𝑥 + 1 ; 𝑧 2 = 2𝑥 + 1 𝑧2 − 1 2𝑥 = 𝑧 2 − 1 ; 𝑥 = ; 𝑑𝑥 = 𝑧𝑑𝑧 2 =

= = = =

𝟏𝟑.

x 5 4 + x 3 dx

𝑧=

4 + 𝑥3𝑧2 = 4 + 𝑥3 ; 𝑥 =

𝑑𝑥 =

1 4 − 𝑧2 3

=−

4 − 𝑧 𝑧𝑑𝑧 1−2

𝑧 2 −1 2

𝑑𝑧 − 2

=𝑧−2

2𝑧𝑑𝑧 − 𝑧2 − 2

= 𝑧 − 2 𝑙𝑛 𝑧 2 − 2 +

𝑑𝑧 2 𝑧 −2 1 𝑙𝑛 2

= 𝟐𝒙 + 𝟏 − 𝟐 𝒍𝒏 𝟐𝒙 − 𝟏 +

𝟏 𝟐

2𝑥𝑡1− 2 2𝑥+1+ 2

𝒍𝒏

+𝑐

𝟐𝒙𝒕𝟏 − 𝟐 𝟐𝒙 + 𝟏 + 𝟐

+𝒄

2 3

4 − 𝑧2

5

(𝑧)

=

1 3

=

1 2𝑧 5 8𝑧 3 − +𝑐 3 5 3

=

2𝑧 5 8𝑧 3 − +𝑐 15 9

=

−2𝑧𝑑𝑧 3 4 − 𝑧2

2 3

−8𝑧 2 + 2𝑧 4 𝑑𝑧 3

=

2𝑧 − 1 𝑑𝑧 𝑧2 − 2

3 4 − 𝑧2

4 − 𝑧 2 𝑧 −2𝑧𝑑𝑧 3

=

4𝑧 − 2𝑑𝑧 𝑧2 − 2

−2𝑧𝑑𝑧

2𝑧𝑑𝑧

3

=

4𝑧 − 2𝑑𝑧 1+ − 𝑧2 − 2 1−

6

2𝑧 4 𝑑𝑧 − 8𝑧 2 𝑑𝑧

4 + 𝑥 3 − 40 4 − 𝑥 3 +𝑐 45

=

4 + 𝑥 3 6 4 + 𝑥 3 − 40 +𝑐 45

=

4 + 𝑥 3 24 + 6𝑥 3 − 40 +𝑐 45

= =

4 − 𝑧2

𝑥 5 4 + 𝑥 3 𝑑𝑥

=

4𝑧 − 𝑧 2 𝑑𝑧 2 − 𝑧2

2 3

3

4+𝑥 3 −16+6𝑥 3 45

+𝑐

𝟐 𝟒 + 𝒙𝟑 𝟑𝒙𝟑 − 𝟖 +𝒄 𝟒𝟓


31

EXERCISE 10.8


3

1

15. 𝑥 3 (4 + 𝑥 2 )2 𝑑𝑥

17. ʃ

3+1 =2 2

𝑥 = 𝑡𝑎𝑛 𝑢 ; 𝑑𝑥 = 𝑠𝑒𝑐 2 𝑢 𝑑𝑢

𝑧2 = 4 + 𝑥2

= ʃ 𝑐𝑜𝑡 3 𝑢 𝑐𝑠𝑐 𝑢 𝑑𝑢

X = 4 − 𝑧2

= ʃ 𝑐𝑜𝑡 𝑢 𝑐𝑠𝑐 𝑢 𝑐𝑠𝑐 2 ( 𝑢 − 1)𝑑𝑢 1

1

dx = 2 4 − 𝑧 2 −2 (-2zdz)

= 𝑠𝑢𝑏𝑠. 𝑠 = 𝑐𝑠𝑐 𝑢 𝑎𝑛𝑑 𝑑𝑠 = − 𝑐𝑜𝑡 𝑢 𝑐𝑠𝑐 𝑢 𝑑𝑢

= −ʃ 𝑠 2 − 1 𝑑𝑠

𝑧𝑑𝑧

1 (4−𝑧 2 )2

= ʃ1𝑑𝑠 − ʃ𝑠 2 𝑑𝑠

= (4 − 𝑧 2 )(𝑧 3 )(−𝑧𝑑𝑧) = ʃ𝑠 − = =

𝑑𝑥

𝑡𝑕𝑒𝑛 𝑥 2 + 1 = 𝑡𝑎𝑛2 𝑢 + 1 = 𝑠𝑒𝑐 𝑢 & 𝑢 = 𝑡𝑎𝑛−1 𝑥

Z = 4 + 𝑥2

=-

𝑥 4 𝑥 2 +1

−4𝑧 4 + 𝑧 6 𝑑𝑧

𝑠3 3

= 𝑐𝑠𝑐 𝑢 −

4𝑧 5 𝑧 7 − + +𝐶 5 7

=

=

28𝑧 5 + 5𝑧 7 +𝐶 35

=

−28( 4+𝑥 2 )5 +5( 4+𝑥 2 )7 +C 35

𝑐𝑠𝑐 3 𝑢 3

+𝐶

𝒙𝟐 + 𝟏 𝟐𝒙𝟐 − 𝟏 +𝑪 𝟑𝒙𝟑

5

= =

=

4 + 𝑥 2 (−28 + 5(4 + 𝑥 2 ) +𝐶 35 4+𝑥 2

5

(−28+20+5𝑥 2 ) 35

𝟒 + 𝒙𝟐

𝟓

+𝐶

(𝟓𝒙𝟐 − 𝟖)

𝟑𝟓

+𝑪


32

EXERCISE 10.8

𝟏𝟗. ( 𝑥=


𝑑𝑥 𝑥 2 (81 +

𝑥4

)

1 1 ; 𝑑𝑥 = − 2 𝑑𝑧 𝑧 𝑧

1 𝑧2

−𝑑𝑧 𝑧2

81 +

𝑙𝑒𝑡 𝑥 = 1 𝑧

=

1

3 4

1 𝑥

𝑑𝑧

− 𝑥3

− 𝑧2

1 𝑧4

𝑑𝑧

− 𝑧2

1/𝑧 4

1

(𝑧 2 −1) 3 𝑧 𝑧4

=

𝑙𝑒𝑡 𝑢 = 81𝑧 4 + 1 ; 𝑑𝑢 = 324𝑧 3 𝑑𝑧

−1 81

𝑧=

1

81𝑧 4 + 1

1 324

1 , 𝑧2

𝑧 2 −1 3 𝑧3

𝑧3

=

=

(𝑥−𝑥 3 )1/3 𝑑𝑥 𝑥4

=

1 𝑍4

−𝑑𝑧 𝑧2 81𝑧 4 +1 𝑧6

=

21.

(−

𝑑𝑧 ) 𝑧2

1

𝑧 2 − 1 3 𝑧𝑑𝑧

=−

𝑑𝑢 𝑙𝑒𝑡 𝑢 = 𝑧 2 − 1

3

𝑢4 81𝑧 4 + 1

1 4

+𝑐

=−

1 2

1

𝑢3 𝑑𝑢 4

𝟏 𝟖𝟏 + 𝒙𝟒 =− +𝒄 𝟖𝟏 𝒙𝟒

=

1 𝑢3 − 2 ( 4 )+c 3

=−

3 2 𝑧 −1 8 3

= −8

1 𝑥2

−1

4 3

4 3

+𝑐 +𝑐

4

=

3 1-x2 3 − 8 x2 +c


33

EXERCISE 10.9

INTEGRATION OF RATIONAL FUNCTIONS OF SINES AND COSINES

𝐷𝑥 1+𝑐𝑜𝑠𝑥

1.

𝑑𝑥 = 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 + 3

𝟓.

2𝑑𝑧 1+𝑧 2 2𝑧 1−𝑧 2 + +3 1+𝑧 2 1+𝑧 2

2 𝐷𝑧

=

1+𝑧 2 1−𝑧 2 1+𝑧 2

1+

2𝑑𝑧

=

1+𝑧 2 1+𝑧 2 + 1−𝑧 2 1+𝑧 2

=

2𝑑𝑧 1 2

𝑧+2

𝑧+2 2 𝑢 2 = 𝑎𝑟𝑐𝑡𝑎𝑛 + 𝑐 = 𝑎𝑟𝑐𝑡𝑎𝑛 + 𝑐 7 7 𝑎 𝑎 2

= 𝑑𝑥 4+2 𝑠𝑖𝑛𝑥

2 𝑑𝑧

=

=

1+ 𝑧 2 2𝑧 1+𝑧 2

1 2𝑧+1 𝑎𝑟𝑐𝑡𝑎𝑛 7 7

𝟏 𝟕

4+2

2𝑑𝑧

=

1+𝑧 2

𝑑𝑧

=

=

1 2 2

+

3

= 2

𝑢2

3 2

𝑎𝑟𝑐𝑡𝑎𝑛

𝟏 𝟑

𝑧+2 3 2

𝑑𝑢 2 𝑢 = 𝑎𝑟𝑐𝑡𝑎𝑛 + 𝑐 2 +𝑎 𝑎 𝑎

4

1

2

𝒙

𝒂𝒓𝒄𝒕𝒂𝒏

𝟐𝒕𝒂𝒏 𝟐 + 𝟏 𝟕

+ 𝒄

1+𝑧 2

𝟕.

𝑠𝑒𝑐𝑥𝑑𝑥 = 2𝑑𝑧 =2 1 − 𝑧2

= 𝑧+

+ 𝐶

4+4𝑧 2 + 4𝑧 1+𝑧 2

2 𝑑𝑧 1 3 ; 𝑤𝑕𝑒𝑟𝑒: 𝑢 = 𝑧 + ; 𝑎 = 4𝑧 2 + 4𝑧 + 4 2 2

=2

2

2𝑑𝑧

=

4𝑧 4+ 1+𝑧 2

=

7 4

1

𝒙 = 𝒕𝒂𝒏 + 𝒄 𝟐

3.

+

𝑑𝑢 7 1 ∶ 𝑤𝑕𝑒𝑟𝑒 𝑎 = , 𝑢 = 𝑧 + 𝑢2 + 𝑎2 2 2

=2

𝑑𝑧 = 𝑧 + 𝑐

1+2𝑧−2𝑧 2 +3+3𝑧 2 1+𝑧 2

2𝑑𝑧 = 4 + 2𝑧 + 2𝑧 2

=

2𝑑𝑧 2

=

1+𝑧 2

1+2𝑧−2𝑧 2 +3 1+𝑧 2

2 𝑑𝑧

=

2𝑑𝑧

=

1+𝑧 2

+ 𝑐=

1 3


𝒙

𝒂𝒓𝒄𝒕𝒂𝒏

𝟐 𝒕𝒂𝒏 𝟐 + 𝟏 𝟑

2𝑧+1 3

+ 𝑐

2

= 2𝑎 𝑙𝑛

𝑎+𝑢 𝑎−𝑢

1 + 𝑧 2 2𝑑𝑧 . 1 − 𝑧2 1 + 𝑧2 𝑎2

𝑑𝑢 𝑤𝑕𝑒𝑟𝑒 𝑎 = 1, 𝑢 = 𝑧 − 𝑢2

+ 𝑐 = 𝑙𝑛

1+𝑧 1−𝑧

+𝐶 𝒙

𝟏 + 𝒕𝒂𝒏 𝟐 𝟐 𝒂+𝒖 = 𝒍𝒏 + 𝒄 = 𝒍𝒏 𝒙 + 𝒄 𝟐𝒂 𝒂−𝒖 𝟏 − 𝒕𝒂𝒏 𝟐

+𝒄


34

EXERCISE 10.10

INTEGRATION BY PARTS

1. 𝑥𝑐𝑜𝑠𝑥𝑑𝑥 =

𝑥𝑐𝑜𝑠𝑥𝑑𝑥

; 𝑙𝑒𝑡 𝑑𝑣 = 𝑐𝑜𝑥𝑑𝑥 𝑣 = 𝑠𝑖𝑛𝑥

= 𝑥𝑠𝑖𝑛𝑥 −

𝑢=𝑥 𝑑𝑢 = 𝑑𝑥

𝑠𝑖𝑛𝑥𝑑𝑥

= 𝒙𝒔𝒊𝒏𝒙 + 𝒄𝒐𝒔𝒙 + 𝑪

3. 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 ; 𝑢 = 𝑐𝑜𝑠2𝑥

𝑑𝑣 = 𝑒 −𝑥 𝑑𝑥 ; 𝑢 = 𝑠𝑖𝑛2𝑥

;

𝑣 = -𝑒 −𝑥

𝑑𝑢 = 𝑠𝑖𝑛2𝑥𝑑𝑥 ;

; 𝑑𝑣 = 𝑒 −𝑥 𝑑𝑥

; 𝑑𝑢 = 2𝑐𝑜𝑠2𝑥𝑑𝑥 ; 𝑣 = -𝑒 −𝑥

= -𝑒 𝑥 𝑐𝑜𝑠2𝑥 − 2𝑒 −𝑥 𝑠𝑖𝑛2𝑥𝑑𝑥 = -𝑒 𝑥 𝑐𝑜𝑠2𝑥 − 2 𝑒 −𝑥 𝑠𝑖𝑛2𝑥𝑑𝑥 = -𝑒 −𝑥 𝑐𝑜𝑠2𝑥 − 2[-𝑒 −𝑥 𝑠𝑖𝑛2𝑥 −

-𝑒 −𝑥 2𝑐𝑜𝑠2𝑥𝑑𝑥

= -𝑒 −𝑥 𝑐𝑜𝑠2𝑥 + 2𝑒 −𝑥 𝑠𝑖𝑛2𝑥 − 4 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 𝑎𝑑𝑑 4 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 𝑡𝑜 𝑏𝑜𝑡𝑕 𝑠𝑖𝑑𝑒𝑠 =

2𝑒 −𝑥 𝑠𝑖𝑛 2𝑥−𝑒 −𝑥 𝑐𝑜𝑠 2𝑥 5

=

𝒆−𝒙 𝟓

+𝐶

𝒔𝒊𝒏𝟐𝒙 − 𝒄𝒐𝒙𝟐𝒙 + 𝑪

5. 𝑎𝑟𝑐𝑡𝑎𝑛2𝑥𝑑𝑥 ; 𝑑𝑣 = 𝑑𝑥 ; 𝑢 = 𝑎𝑟𝑐𝑡𝑎𝑛2𝑥 2𝑑𝑥

𝑣 = 𝑥 ; 𝑑𝑢 = 1+𝑥 2 2𝑥𝑑𝑥 1+4𝑥 2

= 𝑥𝑎𝑟𝑐𝑡𝑎𝑛2𝑥 − = 𝑥𝑎𝑟𝑐𝑡𝑎𝑛2𝑥 − 2 1

𝑑𝑥 1+4𝑥 2 𝑑𝑢 𝑢

= 𝑥𝑎𝑟𝑐𝑡𝑎𝑛2𝑥 − 4 𝟏 𝟒

= 𝒙𝒂𝒓𝒄𝒕𝒂𝒏𝟐𝒙 − 𝒍𝒏 𝟏 + 𝟒𝒙𝟐 + 𝑪


35

EXERCISE 10.10


7. 𝑠𝑒𝑐 3 𝑥𝑑𝑥 ; 𝑑𝑣 = 𝑠𝑒𝑐 2 𝑥𝑑𝑥 ; 𝑢 = 𝑠𝑒𝑐𝑥 𝑣 = 𝑡𝑎𝑛𝑥

𝑑𝑢 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥

𝑠𝑒𝑐 2 𝑥𝑠𝑒𝑐𝑥𝑑𝑥

=

= 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 −

𝑠𝑒𝑐 2 𝑥 − 1 𝑠𝑒𝑐𝑥𝑑𝑥

= 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 + 𝑙𝑛 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 − 𝑠𝑒𝑐 3 𝑥𝑑𝑥 ; 𝑎𝑑𝑑

𝑠𝑒𝑐 3 𝑥𝑑𝑥 𝑜𝑛 𝑏𝑜𝑡𝑕 𝑠𝑖𝑑𝑒𝑠

2 𝑠𝑒𝑐 3 𝑥𝑑𝑥 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 + 𝑙𝑛 𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 𝟏 𝟐

𝑠𝑒𝑐 3 𝑥𝑑𝑥 =

𝒔𝒆𝒄𝒙𝒕𝒂𝒏𝒙 + 𝒍𝒏 𝒔𝒆𝒄𝒙 + 𝒕𝒂𝒏𝒙 + 𝑪

9. 𝑥𝑐𝑜𝑠 2 2𝑥𝑑𝑥 ; 𝑑𝑣 = 𝑑2𝑥𝑑𝑥 𝑣= 1 𝑥 2

=𝑥 =

𝑥2 2

1 𝑥 2

;

1

+ 8 𝑠𝑖𝑛4𝑥

1

1

𝑑𝑢 = 𝑥 1

+ 8 𝑠𝑖𝑛4𝑥 − (2 𝑥 + 8 𝑠𝑖𝑛4𝑥)𝑑𝑥

1 8

1 4

+ 𝑥𝑠𝑖𝑛4𝑥 − 𝑥 2 +

𝟏

𝑢 =𝑥

𝟏

1 𝑐𝑜𝑠4𝑥 32

+𝐶

𝟏

= 𝟒 𝒙𝟐 + 𝟖 𝒔𝒊𝒏𝟒𝒙 + 𝟑𝟐 𝒄𝒐𝒔𝟒𝒙 + 𝑪

11.

𝑥𝑎𝑟𝑐𝑠𝑖𝑛𝑥𝑑𝑥 1−𝑥 2

;

𝑑𝑣 =

𝑥𝑑𝑥

;

1−𝑥 2

𝑣 = - 1 − 𝑥2

;

𝑢 = 𝑎𝑟𝑐𝑠𝑖𝑛𝑥 𝑑𝑢 =

= - 1 − 𝑥 2 𝑎𝑟𝑐𝑠𝑖𝑛𝑥 − (- 1 − 𝑥 2 )(

𝑑𝑥 1−𝑥 2

𝑑𝑥 1−𝑥 2

)

= - 1 − 𝑥 2 𝑎𝑟𝑐𝑠𝑖𝑛𝑥 + 𝑑𝑥 = 𝒙 − 𝟏 − 𝒙𝟐 𝒂𝒓𝒄𝒔𝒊𝒏𝒙 + 𝑪


36

EXERCISE 10.10


13. 𝑠𝑖𝑛𝑥𝑙𝑛 1 + 𝑠𝑖𝑛𝑥 𝑑𝑥 ;

𝑢 = 𝑙𝑛 1 + 𝑠𝑖𝑛𝑥 𝑑𝑢 =

1 1+𝑠𝑖𝑛𝑥

;

𝑑𝑣 = 𝑠𝑖𝑛𝑥𝑑𝑥

𝑐𝑜𝑠𝑥𝑑𝑥 ;

𝑣 = 𝑐𝑜𝑠𝑥

𝑐𝑜𝑠 2 𝑥𝑑𝑥 1+𝑠𝑖𝑛𝑥

= -𝑐𝑜𝑠 𝑙𝑛 1 + 𝑠𝑖𝑛𝑥 +

1−𝑠𝑖𝑛 2 𝑥𝑑𝑥 1+𝑠𝑖𝑛𝑥

= -𝑐𝑜𝑠 𝑙𝑛⁡|1 + 𝑠𝑖𝑛𝑥 | + = -𝑐𝑜𝑠 𝑙𝑛 1 + 𝑠𝑖𝑛𝑥 +

1 − 𝑠𝑖𝑛𝑥 𝑑𝑥

= -𝑐𝑜𝑠 𝑙𝑛 1 + 𝑠𝑖𝑛𝑥 + 𝑥 + 𝑐𝑜𝑠𝑥 + 𝐶 = − 𝒄𝒐𝒔 𝒍𝒏 𝟏 + 𝒔𝒊𝒏𝒙 + 𝒙 + 𝒄𝒐𝒔𝒙 + 𝑪

15.

𝑒 𝑥 𝑥𝑑𝑥 (𝑥+1)2

;

𝑢 = 𝑒𝑥 𝑥 𝑑𝑢 = 𝑒 𝑥 𝑥 + 1 𝑑𝑥

=-

𝑒𝑥𝑥 𝑥+1

+ 𝑒 𝑥 𝑑𝑥 =

𝑑𝑣 = (𝑥+1)2 𝑑𝑥

;

𝑣 =-

𝑥3

=

1 𝑎𝑟𝑐𝑠𝑖𝑛 3

=

1 3 𝑥 𝑎𝑟𝑐𝑠𝑖𝑛𝑥 3

+ (3 𝑐𝑜𝑠 𝑒 −

=

1 3 𝑥 𝑎𝑟𝑐𝑠𝑖𝑛𝑥 3

+

=

1 3 𝑥 𝑎𝑟𝑐𝑠𝑖𝑛 3

𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟; 1

=3

1

1 3

1−𝑥 2

3− 1−𝑥 2 9 1−𝑥 2 2+𝑥 2

𝑥3 1−𝑥 2

𝑑𝑥 1−𝑥 2

;

𝑑𝑣 = 𝑥 2 𝑑𝑥 ; 𝑣 =

𝑥3 3

𝑑𝑥

1

+

1 𝑥+1

𝒆𝒙 +𝑪 𝒙+𝟏

17. 𝑥 2 𝑎𝑟𝑐𝑠𝑖𝑛𝑥𝑑𝑥; 𝑢 = 𝑎𝑟𝑐𝑠𝑖𝑛𝑥 ; 𝑑𝑢 = −3

1

;

9

𝑐𝑜𝑠 3 𝑒 )+ 9

𝐶

+𝐶 +𝐶

𝑑𝑥 ; 𝑎 = 1 ; 𝑣 = 𝑥 ; 𝑥 = 𝑠𝑖𝑛 𝑒 ; 𝑑𝑥 = 𝑐𝑜𝑠 𝑒 𝑑𝑒 ;

1 − 𝑥 2 = 𝑐𝑜𝑠𝑒

𝑠𝑖𝑛 3 𝑒 (𝑐𝑜𝑠𝑒𝑑𝑒) 𝑐𝑜𝑠𝑒

1

= 3 𝑠𝑖𝑛2 𝑒 𝑠𝑖𝑛𝑒𝑑𝑒 𝟏 𝟑

= (−𝒄𝒐𝒔𝒆 +

𝒄𝒐𝒔𝟑 𝒆 )+ 𝟑

𝑪 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy

37

EXERCISE 10.11

INTEGRATION OF RATIONAL FUNCTIONS

12𝑥+18 𝑥+2 𝑥+4 (𝑥−1)

1.

12𝑥 + 18 𝐴 𝐵 𝐶 = + + 𝑥 + 2 𝑥 + 4 (𝑥 − 1) (𝑥 + 2) (𝑥 + 4) (𝑥 − 1) 12𝑥 + 18 = 𝐴 𝑥 + 4 𝑥 − 1 + 𝐵 𝑥 + 2 𝑥 − 1 + 𝐶 𝑥 + 2 (𝑥 + 4) 12𝑥 + 18 = 𝐴(𝑥 2 + 3𝑥 − 4) + 𝐵 𝑥 2 + 𝑥 − 2 + 𝐶(𝑥 2 + 6𝑥 + 8) 12𝑥 + 18 = 𝐴𝑥 2 + 3𝐴𝑥 − 4𝐴 + 𝐵𝑥 2 + 𝐵 − 2𝐵 + 𝐶𝑥 2 + 6𝐶𝑥 + 8𝐶 𝐴𝑥 2 + 𝐵𝑥 2 + 𝐶𝑥 2 = 0 3𝐴𝑥 + 𝐵𝑥 + 6𝐶𝑥 = 12𝑥 4𝐴 + 𝐵 + 8𝐶 = 18 𝐴=1 𝐵 = −3 𝐶=2

=

𝑑𝑥 + (𝑥+2)

−3𝑑𝑥 + (𝑥+4)

2𝑑𝑥 (𝑥−1)

= 𝒍𝒏 𝒙 + 𝟐 − 𝟑 𝒍𝒏 𝒙 + 𝟒 + 𝟐𝒍𝒏 𝒙 − 𝟏 ⁡

3.

1=

𝑑𝑥 𝑥−1 (𝑥−4) 𝐴 𝐵 + (𝑥 − 1) (𝑥 − 4)

1 = 𝐴 𝑥 − 4 + 𝐵(𝑥 − 1) 1 = 𝐴𝑥 − 4𝐴 + 𝐵𝑥 − 𝐵

=

−1 3 (𝑥 −1)

1

𝑑𝑥 +

1 3 (𝑥 −4)

1

𝐴+𝐵 =0 −4𝐴 − 𝐵 = 1 𝐴 = −𝐵 1 𝐵= 3

𝑑𝑥

= − 3 𝑙𝑛 𝑥 − 1 + 3 𝑙𝑛 𝑥 − 4 + 𝐶

=

𝟏 𝒍𝒏 𝒙−𝟒 𝟑 𝒍𝒏 𝒙−𝟏

+C


38

EXERCISE 10.11


6𝑥 2 +23𝑥−9 𝑑𝑥 (𝑥 3 +2𝑥 2 −3𝑥)

5.

6𝑥 2 + 23𝑥 − 9 𝑑𝑥 𝑥 𝑥 + 3 (𝑥 − 2) 6𝑥 2 + 23𝑥 − 9 =

𝐴 𝐵 𝐶 + + 𝑥 (𝑥 + 3) (𝑥 − 1)

6𝑥 2 + 23𝑥 − 9 = 𝐴 𝑥 + 3 𝑥 − 1 + 𝐵 𝑥 𝑥 − 1 + 𝐶 𝑥 (𝑥 + 3) 6𝑥 2 + 23 − 9 = 𝐴 𝑥 2 + 2𝑋 − 3 + 𝐵 𝑥 2 − 𝑥 + 𝐶(𝑥 2 + 3𝑥) 𝐴+𝐵+𝐶 =6 2𝐴 − 𝐵 + 3𝐶 = 23 −3𝐴 + 0𝐵 + 0𝐶 = −9 𝐴=3 𝐵 = −2 𝐶=5 =3

𝑑𝑥 𝑥

−2

𝑑𝑥 + (𝑥+3)

5

𝑑𝑥 (𝑥−1)

= 𝟑𝒍𝒏 𝒙 − 𝟐𝒍𝒏 𝒙 + 𝟑 + 𝟓𝒍𝒏 𝒙 − 𝟏 + 𝑪 𝑥 3 +5𝑥 2 +9𝑥+7 𝑑𝑥 𝑥 2 +5𝑥+4

7.

𝑥 3 + 5𝑥 2 + 9𝑥 + 7 𝑑𝑥 𝑥 + 4 (𝑥 + 1)

By division of polynomials, 5𝑥 + 7 𝐴 𝐵 = + 𝑥 + 4 (𝑥 + 1) (𝑥 + 4) (𝑥 + 1) 5𝑥 + 7 = 𝐴 𝑥 + 1 + 𝐵(𝑥 + 4)

= =

𝑥𝑑𝑥 + 𝒙𝟐 𝟐

+

𝟏𝟑 𝒍𝒏 𝟑

13 𝑑𝑥 3

(𝑥 + 4) 𝒙+𝟒 +

+ 𝟐 𝒍𝒏 𝟑

𝑖𝑓 𝑥 = 4, 13 𝐴= 3 𝑖𝑓 𝑥 = −1 2 𝐵= 3

2 𝑑𝑥 3

(𝑥 + 1) 𝒙+𝟏 +𝑪


39

EXERCISE 10.11

9.


2𝑥+1 𝑥−2 (𝑥−3)2

2𝑥 + 1 =

𝐴 𝐵 𝐶 + + (𝑥 − 2) (𝑥 − 3) (𝑥 − 3)2

2𝑥 + 1 = 𝐴 𝑥 − 3

2

+𝐵 𝑥−3 𝑥−2 +𝐶 𝑥−2

2𝑥 + 1 = 𝐴 𝑥 2 − 6𝑥 + 9 + 𝐵 𝑥 2 − 5𝑥 + 6 + 𝐶 𝑥 − 2 𝐴+𝐵 =0 −6𝐴 − 5𝐵 + 𝐶 = 2 9𝐴 + 6𝐵 − 2𝐶 = 1 𝐴=5 𝐵 = −5 𝐶=7 =

5𝑑𝑥 + 𝑥−2

−5𝑑𝑥 + 𝑥−3

7𝑑𝑥 𝑥−3

2

7

= 5𝑙𝑛 𝑥 − 2 − 5𝑙𝑛 𝑥 − 3 + 𝑥−3 = 𝟓𝒍𝒏

𝒙−𝟐 𝟕 + 𝒙−𝟑 (𝒙 − 𝟑)


40

EXERCISE 10.11

11.


2𝑥−5 𝑑𝑥 𝑥(𝑥−1)

2𝑥 − 5 𝐴 𝐵 𝐶 𝐷 = + + + 𝑥(𝑥 − 1) 𝑥 (𝑥 − 1) (𝑥 − 1)2 (𝑥 − 13 2𝑥 − 5 = 𝐴 𝑥 − 1

3

+ 𝑏𝑥 𝑥 − 1

2

+ 𝑐𝑥 𝑥 − 1 + 𝐷𝑥

2𝑥 − 5 = 𝐴𝑥 3 − 3𝐴𝑥 2 + 3𝐴𝑥 − 𝐴 + 𝐵𝑥 3 − 2𝐵𝑥 2 + 𝐵𝑥 +C 𝑥 2 − 𝐶𝑥 + 𝐷𝑥 2𝑥 − 5 = 𝐴𝑥 3 = 3𝐴𝑥 2 − 2𝐵𝑥 2 + 𝐵𝑥 + 𝐶𝑥 2 − 𝐶𝑥 + 𝐷𝑥 2𝑥 − 5 = 𝐴𝑥 3 + 𝐵𝑥 2 − 3𝐴𝑥 2 − 2𝐵𝑥 2 + 𝐶𝑥 2 + 3𝐴𝑥 + 𝐵𝑥 − 𝐶𝑥 + 𝐷𝑥 − 𝐴 2𝑥 − 5 = 𝐴 + 𝐵 𝑥 3 + −3𝐴 − 2𝐵 + 𝑐 𝑥 2 + 3𝐴 + 𝐵 − 𝐶 + 𝐷 𝑥 − 𝐴 𝐴+𝐵 =0 −3𝐴 − 2𝐵 + 𝐶 = 0 3𝐴 + 𝐵 − 𝐶 + 𝐷 = 2 −𝐴 = −5 𝐴=5 𝐵 = −5 𝐶=5 𝐷 = −3 =

5𝑑𝑥 + 𝑥

=5

𝑑𝑥 −5 𝑥

−5𝑑𝑥 + (𝑥 − 1)

5𝑑𝑥 + (𝑥 − 1)2

𝑑𝑥 +5 (𝑥 − 1) 5

−3𝑑𝑥 (𝑥 − 1)3

𝑑𝑥 −3 (𝑥 − 1)2

𝑑𝑥 (𝑥 − 1)3

3

= 5𝑙𝑛 𝑥 − 5𝑙𝑛 𝑥 − 1 − (𝑥−1) + 2(𝑥−1)2 + 𝐶 = 𝟓𝒍𝒏

𝒙 𝟓 𝟑 − + +𝑪 𝒙 − 𝟏 (𝒙 − 𝟏) 𝟐(𝒙 − 𝟏)𝟐


41

EXERCISE 10.11


3𝑥 2 +17𝑥+32 𝑥 3 +8𝑥 2 +16𝑥

13.

3𝑥 2 + 17𝑥 + 32 𝑥(𝑥 + 4)2 3𝑥 2 + 17𝑥 + 32 𝐴 𝐵 𝐶 = + + 2 𝑥(𝑥 + 4) 𝑥 (𝑥 + 4) (𝑥 + 4)2 𝐴+𝐵 =3 8𝐴 + 4𝐵 + 𝐶 = 17 16𝐴 = 32 𝐴=2 𝐵=1 𝐶=3 2𝑑𝑥 + 𝑥

=

𝑑𝑥 + (𝑥 + 4)

−3𝑑𝑥 (𝑥 + 4)2 𝟑

= 𝟐𝒍𝒏 𝒙 + 𝒍𝒏 𝒙 + 𝟒 + 𝒙+𝟒 2𝑥+1 3𝑥−1 (𝑥 2 +2𝑥+2)

15.

2𝑥 + 1 𝐴 𝐵 2𝑥 + 2 + 𝐶 = + 2 2 3𝑥 − 1 (𝑥 + 2𝑥 + 2) (3𝑥 − 1) 𝑥 + 2𝑥 + 2 2𝑥 + 1 = 𝐴 𝑥 2 + 2𝑥 + 2 + 𝐵 2𝑥 + 2 3𝑥 − 1 + 𝐶 3𝑥 − 1 2𝑥 + 1 = 𝐴 𝑥 2 + 2𝑥 + 2 + 𝐵(6𝑥 2 + 4𝑥 + 2) + 𝐶 3𝑥 − 1 𝐴+𝐵 =0 2𝐴 + 4𝐵 + 3𝐶 = 2 2𝐴 + 2𝐵 − 𝐶 = 1 5 𝐴=− 2 5 𝐵= 2 𝐶 = −1 =−

5 2 5

𝑑𝑥 5 + (3𝑥 − 1) 2

(2𝑥 + 2) 𝑑𝑥 − 𝑥 2 + 2𝑥 + 2

𝑥2

𝑑𝑥 + 2𝑥 + 2

5

= − 2 𝑙𝑛 3𝑥 − 1 + 2 𝑙𝑛 𝑥 2 + 2𝑥 + 2 − 𝑙𝑛 𝑥 2 + 2𝑥 + 2 =

𝟓 𝒙𝟐 + 𝟐𝒙 + 𝟐 𝒍𝒏 − 𝒍𝒏 𝒙𝟐 + 𝟐𝒙 + 𝟐 𝟐 𝟑𝒙 − 𝟏


42

EXERCISE 10.11

17.


5𝑥 2 −𝑥+17 𝑑𝑥 𝑥+2 (𝑥 2 +9)

5𝑥 2 − 𝑥 + 17 𝐴 𝐵 2𝑥 + 𝐶 = + 2 𝑥 + 2 (𝑥 + 9) 𝑥 + 2 𝑥2 + 9 5𝑥 2 − 𝑥 + 17 = 𝐴 𝑥 2 + 9 + 2𝐵𝑥 + 𝐶 (𝑥 + 2) 5𝑥 2 − 𝑥 + 17 = 𝐴𝑥 2 + 9𝐴 + 2𝐵𝑥 2 + 4𝐵𝑥 + 𝐶𝑥 + 2𝐶 5𝑥 2 − 𝑥 + 17 = 𝐴𝑥 2 + 2𝐵𝑥 2 + 4𝐵𝑥 + 𝐶𝑥 + 9𝐴 + 2𝐶 5𝑥 2 − 𝑥 + 17 = 𝐴 + 2𝐵 𝑥 2 + 4𝐵 + 𝐶 𝑥 + 9𝐴 + 2𝐶 𝑥 2 = 𝐴 + 2𝐵 = 5 𝑥 = 4𝐵 + 𝐶 = −1 𝑐 = 9𝐴 + 2𝐶 = 17 𝐴 + 2𝐵 = 5 − 2 4𝐵 + 𝐶 = −1

= −2𝐴 − 4𝐵 = −10 =

4𝐵 + 𝐶 = −1 −2𝐴 + 𝐶 = −11

−2𝐴 + 𝐶 = −1 − 2 9𝐴 + 2𝐶 = 17

= 4𝐴 − 2𝐶 = 22 =

9𝐴 + 2𝐶 = 17 13𝐴 = 39

A=3 9(3)+2C=17

4B-5=-1

27+2C=17

4B=-1+5

2C=17-27

4B=4

2C=-10

B=1

C=-5 =

=3

3 1 2𝑥 − 5 + 𝑑𝑥 𝑥+2 𝑥2 + 9 𝑑𝑥 𝑥+2

+2

𝑥𝑑𝑥 − 𝑥 2 +9

5

𝑑𝑥 𝑥 2 +9 𝟓

= 𝟑𝒍𝒏 𝒙 + 𝟐 + 𝒍𝒏 𝒙𝟐 + 𝟗 − 𝟑 𝑨𝒓𝒄𝒕𝒂𝒏

𝒙 + 𝟑

𝑪


43

EXERCISE 10.11

19.


4𝑥 2 +21𝑥+54 𝑥 2 +6𝑥+13

4 − 3𝑥 − 2 𝑥 2 + 6𝑥 + 13 𝐴 2𝑥 + 6 + 𝐵 𝑥 2 + 6𝑥 + 13 𝐴 2𝑥 + 6 + 𝐵 = 3𝑥 − 2 2𝐴 + 𝐵 = 3 𝐵 = −11 𝐴=

3 2

3 2𝑥 + 6 𝑑𝑥 𝑑𝑥 =ʃ4−[ ʃ 2 + (−11ʃ 2 + 6𝑥 + 13)] 2 𝑥 + 6𝑥 + 13 𝑥 = −11ʃ

𝑥2

𝑑𝑥 + 6𝑥 + 9 + 13 − 9

𝑑𝑥 𝑥 + 3 2 + 13 − 9

= −11ʃ 1 2

= −11( 𝑎𝑟𝑐𝑡𝑎𝑛 = 4𝑥 −

𝑥+3 ) 2

3 𝑙𝑛⁡| 𝑥 2 2 𝟑

2

+ 6𝑥 + 13| −

= 𝟒𝒙 − 𝟐 𝒍𝒏⁡| 𝒙𝟐 + 𝟔𝒙 + 𝟏𝟑| +

11 𝑥+3 𝑎𝑟𝑐𝑡𝑎𝑛 2 2 𝟏𝟏 𝒙+𝟑 𝒂𝒓𝒄𝒕𝒂𝒏 𝟐 + 𝟐

𝑪


44

EXERCISE 10.11

21.


𝑥 3 +7𝑥 2 +25𝑥+35 𝑥 2 +5𝑥+6

𝑥+2+

9𝑥 + 23 𝑑𝑥 + 5𝑥 + 6

𝑥2

9𝑥 + 23 𝐴 𝐵 = + 𝑥 + 3 (𝑥 + 2) 𝑥 + 3 𝑥 + 2 9𝑥 + 23 = 𝐴 𝑥 + 2 + 𝐵(𝑥 + 3)

x=-3 9(-3)+23= A(-3+2)+B(-3+3) -27+23=A(-1)+B(0) -4=-A A=4 If x=-2 9(-2)+23= A(-2+2)+B(-2+3) -18+23=A(0)+B 5=B B=5 −2 5 + 𝑑𝑥 𝑥+3 𝑥+2

=

𝑥+2+

=

𝑥𝑑𝑥 + 2 𝑑𝑥 − 4

𝑑𝑥 𝑥+3

+5

𝑑𝑥 𝑥+2

𝒙𝟐 = + 𝟐𝒙 − 𝟒𝒍𝒏 𝒙 + 𝟑 + 𝟓𝒍𝒏 𝒙 + 𝟐 + 𝒄 𝟐


45

EXERCISE 10.11

23.


𝑥 2 −𝑥−8 (2𝑥−3)(𝑥 2 +2𝑥+2)

𝐴 𝐵 2𝑥 − 2 + 𝐶 + 2 2𝑥 − 3 𝑥 + 2𝑥 + 2

A(𝑥 2 + 2𝑥 + 2) + 𝐵 2𝑥 + 2 2𝑥 − 3 + 𝐶(2𝑥 − 3) A(𝑥 2 + 2𝑥 + 2) + 𝐵 4𝑥 2 − 2𝑥 − 6 + 𝐶(2𝑥 − 3) A+4B=1 2A-2B+2C=-1 2A-6B-3C=-8 1

A=-2 1

A=2 C=1 𝑑𝑥 1 + (2𝑥 − 3) 2

−1

2𝑥 + 2 𝑑𝑥 + 𝑥 2 + 2𝑥 + 2

1 −𝑙𝑛(2𝑥 − 3) + 𝑙𝑛│𝑥 2 + 2𝑥 + 2│ + 2

𝑑𝑥 𝑥 2 + 2𝑥 + 2 𝑑𝑥 𝑥 + 1 2 + 12

1

= − 2 − 𝑙𝑛 2𝑥 − 3 + 𝑙𝑛│𝑥 2 + 2𝑥 + 2│ + 𝑎𝑟𝑐𝑡𝑎𝑛 𝑥 + 1 + 𝐶 =

𝟏 𝒙𝟐 + 𝟐𝒙 + 𝟐 𝒍𝒏│ │ + 𝒂𝒓𝒄𝒕𝒂𝒏 𝒙 + 𝟏 + 𝒄 𝟐 𝟐𝒙 − 𝟑


46

EXERCISE 10.11

25. ʃ


𝑥 5 +2𝑥 3 −3𝑥 𝑥 2 +1 3

=ʃ

𝑥 5 + 2𝑥 3 − 3𝑥 𝑥 6 + 3𝑥 4 + 2𝑥 2 + 1

=

𝐴 2𝑥 + 𝐵 𝐶 2𝑥 + 𝐷 𝐸 2𝑥 + 𝐹 + 2 + 2 𝑥2 + 1 𝑥2 + 1 𝑥 +1 2 𝑥 +1 3

= 𝐴 2𝑥 𝑥 2 + 1

2

+ 𝐵 𝑥2 + 1

2

3

+ 𝐶 2𝑥 𝑥 2 + 1 + 𝐷 𝑥 2 + 1 + 𝐸 2𝑥 + 𝐹

= 𝐴 2𝑥 𝑥 4 + 2𝑥 2 + 1 + 𝐵 𝑥 4 + 2𝑥 2 + 1 + 𝐶 2𝑥 3 + 2𝑥 + 𝐷 𝑥 2 + 1 + 𝐸 2𝑥 + 𝐹 = 𝐴 2𝑥 5 + 4𝑥 3 + 2𝑥 + 𝐵 𝑥 4 + 2𝑥 2 + 1 + 𝐶 2𝑥 3 + 2𝑥 + 𝐷 𝑥 2 + 1 + 𝐸 2𝑥 + 𝐹 1 2

𝑥 5 : 2𝐴 = 1

; 𝐴=

𝑥4: 𝐵 = 0

; 𝐵=0

𝑥 3 : 4𝐴 + 2𝐶 = 2

; 𝐶=0

𝑥 2 : 2𝐵 + 𝐷 = 0

; 𝐷=0

𝑥: 2𝐴 + 2𝐶 + 2𝐸 = −3 ; 𝐸 = 0 𝑐: 𝐵 + 𝐷 + 𝐹 = 0 =

; 𝐹=0

𝟏 𝟏 𝒍𝒏 𝒙𝟐 + 𝟏 + 𝟐 𝟐 𝒙 +𝟏

𝟐

+𝑪


47

EXERCISE 10.11


27. 𝑥 4 + 2𝑥 3 + 11𝑥 2 + 8𝑥 + 16 𝑥(𝑥 2 + 4)2 𝐴 𝐵 2𝑥 + 𝐶 𝐷 2𝑥 + 𝐸 [ + + 2 ][(𝑥 2 + 4)2 ] 𝑋 (𝑥 2 + 4) (𝑥 + 4)2

A 𝑥2 + 4

2

+ 𝐵 2𝑥 𝑥 (𝑥 2 + 4) + 𝐶 𝑥 2 + 4 (𝑥) + 𝐷(2𝑥)(𝑥) + 𝐸(𝑥)

A(𝑥 4 + 8𝑥 2 + 16) + 𝐵 2𝑥 4 + 8𝑥 2 + 𝐶 𝑥 3 + 4𝑥 + 𝐷2𝑥 2 + 𝐸𝑥 𝑥 4 : 𝐴 + 2𝐵 = 1

A=1

𝑥3: 𝐶 = 2

B=0

𝑥 2 : 8A+8B+2D=11

C=2

X: 4C + E=8

D = 3/2

C : 16A = 16

E=0

=

𝑑𝑥 2𝑑𝑥 3 + 2 + 𝑥 𝑥 +4 2

= 𝑙𝑛𝑥 + 2

1 2


2𝑥𝑑𝑥 (𝑥 2 + 4)2 𝑥 2

−

3 2 𝑥 2 +4

+𝐶

𝒙 𝟑 = 𝒍𝒏𝒙 + 𝒂𝒓𝒄𝒕𝒂𝒏 − +𝑪 𝟐 𝟐 𝒙𝟐 + 𝟒


48

EXERCISE 11.1

SUMMATION NOTATION

𝑛=10

∗ 𝑛 = 10

𝟓.

𝑛

12𝑖 3

𝟏.

𝑛

𝑖=1

𝑖=1

𝑖3

𝑛=10

𝑖=1 2

10 10 + 1 4

= 12

𝑖3 − 𝑖

=

𝑛=10

= 12

𝑖(𝑖 − 1)(𝑖 + 1) 𝑖=1

𝑖3 − 𝑖

=

2

𝑖=1 𝑛=10

=

= 3(100 121 )

𝑛=10 3

𝑖 + 𝑖=1

= 𝟑𝟔𝟑𝟎𝟎

=

𝑖 𝑖=1

10 2 10+1 2 4

−

10 10+1 2

= 𝟐𝟗𝟕𝟎 𝑛=10

(12𝑖 2 + 4𝑖 )

𝟑.

𝒏=𝟏𝟎

𝟕.

𝑖=1 𝑛=10

𝑖=1

𝑖

𝑛=10

9𝑖 2 + 6𝑖 + 1

=

𝑖=1

10(10 + 1)(2 10 + 1) 10(10 + 1) = 12 +4 6 2

𝑖=1

𝑖2 + 6

=9

= 2 110 21 + 2 110 = 𝟒𝟖𝟒𝟎

𝟐

𝒊=𝟏

𝑛=10

𝑖2 + 4

= 12

𝟑𝒊 + 𝟏

=9

𝑖+

10(10+1)(2 10 +1) 6

+6

1 10(10+1) 2

+ 10

= 𝟑𝟖𝟎𝟓

𝟗. 𝒂𝟏 − 𝒃𝟏 + 𝒂𝟐 − 𝒃𝟐 + ⋯ + (𝒂𝒏 − 𝒃𝒏 ) 𝒏

=

𝒂𝒊 − 𝒃𝒊 𝒊=𝟏


49

EXERCISE 11.1

SUMMATION NOTATION

𝟏𝟏. 𝑓 𝑥1 ∆𝑥1 + 𝑓 𝑥2 ∆𝑥2 + ⋯ + 𝑓 𝑥𝑛 ∆𝑥𝑛 𝒏

=

𝒇(𝒙𝒊 ) ∆𝒙𝒊 𝒊=𝟏

𝟏𝟑. 14 + 24 + 34 + ⋯ + 𝑛4 𝒏

𝒊𝟒

= 𝒊=𝟏

𝟏𝟓. 𝑎1 𝑏1+𝑎2 𝑏2+𝑎3 𝑏3 + ⋯ + 𝑎𝑛 𝑏𝑛 𝒏

=

𝒂𝒊 𝒃𝒊 𝒊=𝟏

𝟏𝟕. 𝑢13 + 𝑢23 + 𝑢33 + ⋯ + 𝑢𝑛3 𝒏

𝒖𝟑𝒊

= 𝒊=𝟏


50

EXERCISE 11.2

𝟏.

THE DEFINITE INTEGRAL

2 3𝑥 2 𝑑𝑥 1

𝑎 = 0 ;𝑏 = 2 ∆𝑥 = =

(𝑥 − 1)𝑑𝑥

𝑎=0

;

2−0 𝑛

∆𝑥 =

1−0 𝑛

2 𝑛

= 𝑙𝑖𝑚 2

= 𝑙𝑖𝑚 3

4𝑖 2 2 ( ) 𝑛2 𝑛

= 𝑙𝑖𝑚⁡3

8𝑖 2 𝑛3

𝑛→∞

= 𝑙𝑖𝑚 24 𝑛→∞

= 𝑙𝑖𝑚𝑛→∞ 24

𝑛 2 +1 2𝑛+1 6 𝑛3 1 0 0 0 2𝑛 3 +𝑛 2 +2𝑛 2 +𝑛 6𝑛 3

} 1 𝑛

−

1 2𝑛3 + 𝑛2 + 2𝑛2 + 𝑛 𝑛3 6

1 0 𝑛 2 −𝑛 𝑛2

−1

2 3

= −1 =−

𝟏 𝟑

5 2𝑥 1

5.

𝑛(𝑛 + 1)(2𝑛 + 1) 1 6 𝑛3

𝑖 𝑛3

]−

1 𝑛 𝑛+1 2𝑛+ 𝑛2 6

= 𝑙𝑖𝑚𝑛→∞ 2

𝑛→∞

𝑛→∞

1 𝑛

𝑛→∞

2𝑖 2 ( ) 𝑛 𝑛

𝑖 𝑛

𝑥 2 − 𝑥𝑑𝑥

= 𝑙𝑖𝑚 2

𝑖=1

𝑍𝑖 =

𝑖2

2𝑖 𝑛 3

;

= { 𝑙𝑖𝑚 2 [ ( 2 )

2 =0+𝑖 𝑛

= 𝑙𝑖𝑚 𝑛=∞

𝑏=1

𝑛→∞

𝑍𝑖 = 𝑎 + 𝑖∆𝑥

=

1 2𝑥 0

3.

∆𝑥 =

+ 3𝑑𝑥

5−1 𝑛

;

𝑍𝑖 = 1 + 𝑖

4 𝑛

4

=𝑛

1

= 𝑙𝑖𝑚𝑛→∞ 24 =𝟖

4𝑖

4

=𝑙𝑖𝑚𝑛=∞ (1 + 𝑛 ) ∙ 𝑛 + 3𝑛 4 𝑛

=𝑙𝑖𝑚𝑛=∞ =𝑙𝑖𝑚𝑛=∞

4𝑛 𝑛

+ 16

16𝑖 𝑛2

+ 𝑛2 +

+

3𝑛

𝑛(𝑛+1) 2

+ 3𝑛

= 𝟑𝟔


51

EXERCISE 11.2

2

𝟕.

THE DEFINITE INTEGRAL

𝑥 3 𝑑𝑥

0

∆𝑥 =

2 2𝑖 ; 𝑍𝑖 = 𝑛 𝑛 3

= 𝑙𝑖𝑚

2𝑖 𝑛

= 𝑙𝑖𝑚

8𝑖 3 𝑛3

𝑛→∞

𝑛→∞

2 𝑛 2 𝑛

16 𝑛2 𝑛 + 1 𝑛→∞ 𝑛 4 4

2

= 𝑙𝑖𝑚

𝑖3

4 2 2 (𝑛 (𝑛 + 2𝑛 + 1) 𝑛→∞ 𝑛 4

= 𝑙𝑖𝑚

= 𝑙𝑖𝑚𝑛→∞

4𝑛 4 𝑛4

8

+ 𝑛3 +

4𝑛 2 𝑛3

=4+0+0 = 4


52

EXERCISE 11.3

2

𝟏.

SOME PROPERTIES OF THE DEFINITE INTEGRAL

3𝑥 2 − 2𝑥 + 1 𝑑𝑥

3

𝟕.

1

2

3𝑥 3 2𝑥 2 + +𝑥 3 2

=

𝑢 = 𝑥2 + 1 𝑑𝑢 = 2𝑥𝑑𝑥

=8−4+2−1+1−1 =

=5

𝑥𝑑𝑥 𝑥2 + 1

=

1 2 1 2

3 2

𝑑𝑢 𝑢

𝑙𝑛 10 − 𝑙𝑛 5

= 𝟎. 𝟑𝟒𝟕 3

3𝑥 2 +

𝟑. 1

4 𝑑𝑥 𝑥2 9.

3

3𝑥 4 + 3 𝑥

=

0

=

4

= 27 − 3 − 1 + 4 =

0 𝑑𝑦 −1 −(𝑥 2 +2𝑥−1)

−1

𝑑𝑦 −(𝑥 + 2𝑥 + 1 − 1 − 1)

0

𝟖𝟔 𝟑

= −1

𝑑𝑦 −[ 𝑥 + 1

0

= −1 7

𝟓.

3

1+

𝑥2

𝑑𝑥

𝑢 = 1 + 𝑥2 𝑑𝑢 = 2𝑥𝑑𝑥 1

=2 =

3

4 1+𝑥 2 3

4

+ 2]

𝑑𝑦 − 𝑥+1

2

0

𝑑𝑦

−1

2− 𝑥+1

=

0

2

+2

2

𝑙𝑒𝑡 𝑎 = 2 ; 𝑢 = (𝑥 + 1) = 𝐴𝑟𝑐𝑠𝑖𝑛 =

𝑥+1 2

+𝑐

𝝅 𝟒

𝟒𝟓 𝟖


53

EXERCISE 11.3 𝑒

𝟏𝟏. 0

SOME PROPERTIES OF THE DEFINITE INTEGRAL 1

𝑥𝑑𝑥 𝑥2 + 𝑒

0

𝑙𝑒𝑡 𝑢 = 𝑥 2 + 𝑒 ; 𝑑𝑢 = 2𝑥𝑑𝑥 ; 𝑑𝑢 𝑒 2

= 1 2

𝑑𝑢 = 𝑥𝑑𝑥 2

1

𝑥 ∙ 2 − 𝑥 𝑑𝑥 0

𝑢

𝑜

=

2𝑥 − 𝑥 2 𝑑𝑥

𝟏𝟓.

𝑒 𝑜

𝑑𝑢 𝑢

=

1 𝑒 𝑙𝑛𝑢 0 2

=

1 𝑒 ln 𝑥 2 + 𝑒 0 2

1 = ln 𝑒 2 + 𝑒 − 𝑙𝑛 0 + 𝑒 2

𝑎 ; 𝑙𝑛𝑎 − 𝑙𝑛𝑏 = 𝑙𝑛 𝑏

1 𝑒2 + 𝑒 1 𝑒 𝑒+1 = 𝑙𝑛 = 𝑙𝑛 2 𝑒 2 𝑒 1

= 2 𝑙𝑛 𝑒 + 1 = 𝑙𝑛 𝑒 + 1

𝑠𝑖𝑛𝜃 =

𝜋

4

=

𝜋

=8

1 𝑎𝑟𝑐𝑡𝑎𝑛 1 2

=

𝝅 𝟖

4

4

0

u= x; du=dx; a=2

=

2𝑐𝑜𝑠𝜃 ∙ 2𝑠𝑖𝑛𝜃 ∙ 4𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃𝑑𝜃

𝑠𝑖𝑛2 𝜃𝑐𝑜𝑠 2 𝜃 𝑑𝜃

0 𝜋

1 𝑥 = 𝑎𝑟𝑐𝑡𝑎𝑛 2 2

2𝑠𝑖𝑛𝜃 = 𝑥

𝑜

=8 2 𝑑𝑥 0 𝑥 2 +4

; 2𝑐𝑜𝑠𝜃 = 2 − 𝑥

𝑥 = 2𝑠𝑖𝑛 𝜃 ; 𝑑𝑥 = 4𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃𝑑𝜃 𝐴𝑡 𝑥 = 1, 𝜃 = 𝜋 4 ; 𝑥 = 0, 𝜃 = 0

1 2

= 𝒍𝒏 𝒆 + 𝟏

𝟏𝟑.

2−𝑥 2 𝑥 ; 2 2

cos 𝜃 =

=2

𝜋 0

4

1 − 𝑐𝑜𝑠2𝜃 2

1 + 𝑐𝑜𝑠2𝜃 𝑑𝜃 2

1 − 𝑐𝑜𝑠 2 2𝜃 𝑑𝜃

=𝝅 𝟒

1

𝟏𝟕.

𝑥𝑒 𝑥 𝑑𝑥

0

𝑢=𝑥 ; 𝑑𝑢 = 𝑑𝑥 ; = 𝑥𝑒 𝑥 −

𝑑𝑣 = 𝑒 𝑥 𝑑𝑥 𝑣 = 𝑒𝑥

1 𝑥 𝑒 𝑑𝑥 0

= 𝑥𝑒 𝑥 − 𝑒 𝑥

= 1−1+0−1 = 1


54

EXERCISE 11.3

𝜋 2

𝟏𝟗.

SOME PROPERTIES OF THE DEFINITE INTEGRAL

𝜋

𝑠𝑖𝑛 𝑥𝑐𝑜𝑥𝑑𝑥

0

𝑙𝑒𝑡 𝑢 = 𝑠𝑖𝑛𝑥 ; 𝑑𝑢 = 𝑐𝑜𝑠𝑥𝑑𝑥 𝑢2 𝑑𝑢

=

6−1 6−3 6−5 2−1 6+2 6+2−2 6+2−4 6+2−6

=2

𝑢3 3

=

𝑠𝑖𝑛 3 𝑥 3

=

𝑠𝑖𝑛6 𝑢 𝑐𝑜𝑠 2 𝑢 𝑑𝑢

𝑜

𝑜

=

𝜋

=2

𝜋 2

=

𝑥 𝑥 𝑠𝑖𝑛6 𝑐𝑜𝑠 2 𝑑𝑥 2 2 0 𝑥 𝑑𝑥 𝑢 = ; 𝑑𝑢 = 2 2 𝟐𝟓.

2

𝟏 𝟑

𝜋 2

𝟓𝝅 𝟏𝟐𝟖

𝜋 4

𝟐𝟕.

𝑠𝑖𝑛2 4𝑥 𝑐𝑜𝑠 2 2𝑥 𝑑𝑥

8

𝜋 2

𝟐𝟏.

𝑠𝑖𝑛6 𝑥𝑐𝑜𝑠 4 𝑥 𝑑𝑥

𝑜

=

2−1 2−1 2+2 2+2−2

=4

1 2

𝜋

=

6−1 6−3 6−5 (4−1)(4−3)( 2 ) (6+4)(6+4−2)(6+4−4)(6+4−6)(6+4−8)

=

𝟑𝝅 𝟓𝟏𝟐

=

𝜋 2

𝝅 𝟏𝟔

2

𝟐𝟗.

4 − 𝑥2

3 2

𝑑𝑥 ; 𝑙𝑒𝑡 𝑥 = 2𝑠𝑖𝑛∅

0

𝑑𝑥 = 2𝑐𝑜𝑠∅𝑠𝑖𝑛∅ 𝜋 2

𝟐𝟑.

2

𝑠𝑖𝑛7 𝑥

=

4 − 2𝑠𝑖𝑛∅

=

𝟏𝟔 = 𝟑𝟓

(2𝑐𝑜𝑠∅𝑑∅)

0

𝑜 (4−1)(7−3)(7−5) 7(7−2)(7−4)(7−6)

3

2 2

2

=

3

(4 𝑐𝑜𝑠 2 ∅)2 2𝑐𝑜𝑠∅𝑐𝑜𝑠∅𝑑∅

0 2

=

8 𝑐𝑜𝑠 3 ∅ 2𝑐𝑜𝑠∅ 𝑑∅

0

=( =

4−1

4−3

4 4−2

𝜋 2

𝟑𝝅 𝟏𝟔


55

EXERCISE 12.1

1. 𝑦 = 3𝑥 2 ;

AREA UNDER A CURVE

𝑓𝑟𝑜𝑚 𝑥 = 1 𝑡𝑜 𝑥 = 2

3. 𝑥𝑦 = −1 ; 𝑓𝑟𝑜𝑚 𝑥 = 1 𝑡𝑜 𝑥 = 2 𝑦=−

1 𝑥

2

𝐴=

2

𝑦𝑑𝑥

𝐴=

1

𝐴=

2 3𝑥 2 𝑑𝑥 1 2 3

𝐴= 𝑥

𝐴= 2

𝑦𝑑𝑥 1

1 3

− 1

𝐴=

2 1 − 𝑑𝑥 1 𝑥 2

𝐴 = [− 𝑙𝑛 𝑥] 3

𝑨 = 𝟕 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔

1

𝐴 = {[− 𝑙𝑛 2] − [− 𝑙𝑛 1]} 𝐴 = − 𝑙𝑛 2; 𝑏𝑢𝑡 𝑡𝑕𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑎𝑟𝑒𝑎, 𝑕𝑒𝑛𝑐𝑒, 𝑨 = 𝒍𝒏𝟐 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔


56

EXERCISE 12.1

AREA UNDER A CURVE

5. 𝑦 = 3𝑙𝑛𝑥, 𝑥 = 2 𝑡𝑜 𝑦 = 4

9. 𝑥 + 𝑦 = 3 & 𝑡𝑕𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑎𝑥𝑒𝑠

𝑎

𝑑𝐴 =

𝑦𝑑𝑥

0 4

𝐴= 3

𝑙𝑛𝑥𝑑𝑥 2

= 3[𝑥𝑙𝑛𝑥 – 𝑥] = 3[4 𝑙𝑛 4 − 4] − 3[2 𝑙𝑛 2 − 2] = 3[4 𝑙𝑛 4 − 4 − 2𝑙𝑛 2 + 2] = 3[8𝑙𝑛2 − 2𝑙𝑛2 − 2] 𝐴=

= 3[6𝑙𝑛2 − 2] = 6[3𝑙𝑛2 − 1]

3 0

3 − 𝑥 𝑑𝑥

𝐴 = 3𝑥 −

𝑨 = 𝟔[𝒍𝒏𝟖 − 𝟏] 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔

𝑥3 2

𝐴= 3 3 − 𝑨= 7. 𝑦 = 9 − 𝑥 2 𝐴=

3 −3

; 𝑥 = −3 𝑡𝑜 𝑥 = 3

3 0 3 2 2

𝟗 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 𝟐

4 − 𝑥 2 𝑑𝑥

𝑨 = 𝟔 𝒔𝒒𝒖𝒂𝒓𝒆 𝒖𝒏𝒊𝒕𝒔


57

EXERCISE 12.1

AREA UNDER A CURVE

11. 𝑦 2 = 4𝑥, 𝑥 = 1 𝑎𝑛𝑑 𝑥 = 4

𝟏𝟑. 𝑥𝑦 = 1, 𝑦 = 𝑥, 𝑥 = 2, 𝑦 = 0

4

𝐴=

4𝑥𝑑𝑥 1 4

𝐴= 𝐴=

1

4𝑥 2 𝑑𝑥 1

𝑥𝑦 = 1; 𝑦 = 𝑥

8 3 𝑥4 3

𝑥(𝑥) = 1

8(4)3/2 8(1)3/2 𝐴= − 3 3 𝐴=

64 3

−

8 3

𝟓𝟔 𝑨= 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 𝟑

𝑥=1 ; 𝑦=1 2

𝐴1 = 1

; (1,1)

1 𝑑𝑥 𝑥

= (𝑙𝑛 𝑥) = 𝑙𝑛 2 − 𝑙𝑛 1 𝐴1 = 𝑙𝑛 2 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠 𝐴2 = =

1 𝑏𝑕 2

1 1 1 2

𝐴2 =

1 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠 2

𝐴𝑡 = 𝐴1 + 𝐴2 𝟏 𝑨 = (𝒍𝒏 𝟐 + )𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 𝟐


58

EXERCISE 12.2

AREA BETWEEN TWO CURVES

5. y = x 2 ; y = 2 − x 2

1. 𝑦 = 𝑥 2 ; 𝑦 = 2𝑥 + 3 𝑦 = 2𝑥 + 3 𝑥 2 = 2𝑥 + 3 𝑥 2 − 2𝑥 − 3 = 0 𝑥−3 𝑥+1 = 0 𝑥 = 3, 𝑥 = −1 𝐴=

3 −1

𝑑𝑦 = 2𝑥 ; (0,0) 𝑑𝑥 𝑥 = 0 ,𝑦 = 0 𝑑2 𝑦 = 2 (𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑢𝑝𝑤𝑎𝑟𝑑) 𝑑𝑥 2

2𝑥 + 3 − 𝑥 2 𝑑𝑥

= [𝑥 2 + 3𝑥 −

𝑥3 ] 3 3 -1

= 32 + 3(3) −

(3)3

𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛: − (−1)2 + 3(−1) −

3

(−1)3 3

5

= 9+3 𝑨=

3. 𝑥 2 = 𝑦 − 1

(2𝑥 + 2)(𝑥 − 1) 2𝑥 + 2 = 0𝑥 − 1 = 0 2 2𝑥 = − 𝑥 = 1 2

; 𝑥 =𝑦−3

𝑥 = −1 𝑦 = 1

Y1=Y2 𝑦−3 2 =𝑦−1 𝑦 2 − 6𝑦 + 9 = 𝑦 − 1 𝑦−5 𝑦−2 =0 𝑦 = 5 ,𝑦 = 2 𝑥 =5−3=2 2

𝑥2 = 2 − 𝑥2 𝑥2 − 2 + 𝑥2 = 0

𝟑𝟐 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 𝟑

𝐴=

y1= y2

𝑑𝐴 = [𝑌1 − 𝑌2]𝑑𝑥 1 −1 1

=

𝑥 + 3 − (𝑥 2 + 1) 𝑑𝑥

2 −1

=

𝑥2 𝑥3 + 2𝑥 − 2 3

𝑥 + 2 − 𝑥 2 𝑑𝑥

=

22 2

+ 2(2) −

=

10 3

+6 =A=

7

23 3

= 2𝑥 − 2 3

2

2𝑥 3 3

− =

−1 2 2 𝟗 𝟐

+ 2(−1) −

(−1)3 3

=

2

2

= 2 − 3 − [−2 + 3]

=2− +2−

-1

27 6

(2 − 2𝑥 2 ) 𝑑𝑥

−1

−1

=

(2 − 𝑥 2 − 𝑥 2 ) 𝑑𝑥

𝑑𝐴 =

2 3

=

12−4 3

𝟖 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 𝟑

𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔


59

EXERCISE 12.2


7. 𝑦 = 𝑠𝑖𝑛𝑥 ; 𝑥 = 𝑐𝑜𝑠𝑥 ; 𝑥 = x 0 90 180 270 360

y 0 1 0 -1 0

𝐴2 =

x 0 90 180 270 360

𝜋 2 𝜋 4

𝑎𝑛𝑑 𝑥 =

𝜋 2

11. 𝑦 = 𝑥 3 , 𝑦 = 8, 𝑥 = 0 𝑑𝑦 = 3𝑥 2 𝑑𝑥

y 1 0 -1 0 1

, 0 = 3𝑥 2

𝑦 = 0 ,𝑥 = 0 𝑑2 𝑦 = 6𝑥(𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑢𝑝𝑤𝑎𝑟𝑑) 𝑑𝑥 2 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛:

𝑠𝑖𝑛𝑥𝑑𝑥 = [-𝑐𝑜𝑠𝑥]

y1= y2

𝜋 4

𝜋 2

= [-𝑐𝑜𝑠 ] − [-𝑐𝑜𝑠 ] = 𝐴1 =

𝜋 4

𝜋 2 𝜋 4

2 2

𝑥3 = 8

𝜋 𝜋 𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑠𝑖𝑛𝑥 = 𝑠𝑖𝑛𝑥 − 𝑠𝑖𝑛 2 4

𝑥3 − 8 = 0 𝑥3 = 8 𝑥=

2 = 1− 2

3

8

𝑥=2

𝑨𝟐 − 𝑨𝟏 = 𝟐 − 𝟏 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔

𝑤𝑕𝑒𝑛 𝑥 = 2 𝑦 = 8 , (2,8) 𝑤𝑕𝑒𝑛 𝑥 = −2

9. 𝑥 2 = 4𝑦 , 𝑦 = 𝑦=

𝑥2

8 𝑥 2 +4

4

𝑥 2 𝑥 2 + 4 = 32 2 8 𝑥2 𝐴= − 𝑑𝑥 2 4 −2 𝑥 + 4 𝐴 = 4.95

𝑦 = −2

3

, 𝑦 = −8

(-2,-8) 𝑑𝐴 = [𝑌1 − 𝑌2]𝑑𝑥 2

𝑑𝐴 =

(8 − 𝑥 3 ) 𝑑𝑥

0

= 𝟏𝟔 − 𝟒 = 𝟏𝟐 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔


60

EXERCISE 12.2


13. 𝑦 = 2𝑥 + 1 , 𝑦 = 7 − 𝑥 , 𝑥 = 8

𝑒

8

𝐴=

15. 𝑦 = 𝑙𝑛𝑥 3 , 𝑦 = 𝑙𝑛𝑥; 𝑥 = 𝑒

2𝑥 + 1 − 7 − 𝑥 𝑑𝑥

𝐴= 1

2

𝑒

8

=

2𝑥 + 1 − 7 + 𝑥 𝑑𝑥

=

8 2

3𝑥 − 6 𝑑𝑥

=

3𝑥 2 = − 6𝑥 2 =

3(8)2 2

𝑒 𝑙𝑛𝑥 3 1

𝑢 = 𝑙𝑛𝑥 3

8 2

− 6(8) −

[(𝑙𝑛𝑥 3 ) − (𝑙𝑛𝑥)]𝑑𝑥

1

2

=

(𝑦2 − 𝑦1 )𝑑𝑥

𝑑𝑢 = 3(2)2 2

− 6(2)

= 𝑑𝑥 ; 𝑑𝑢 = 𝑒

1

= 𝟓𝟒 𝒔𝒒 𝒖𝒏𝒊𝒕𝒔

𝑙𝑛𝑥

; 𝑣 = 𝑥 ; 𝑢 = 𝑙𝑛𝑥 ; 𝑑𝑣 = 𝑑𝑥

3𝑥 2 𝑑𝑣 𝑥3

= 𝑥𝑙𝑛𝑥 3 −

𝑒 1

−

= 𝑥𝑙𝑛𝑥 3 −

3𝑥 2 𝑥( 3 ) 𝑥 3𝑥

= 𝑥𝑙𝑛𝑥 3 − 3𝑥

𝑒 1

𝑑𝑥 𝑥

𝑒 – [𝑥𝑙𝑛𝑥 − 1

;

𝑣=𝑥

𝑑𝑥 𝑒 𝑥( )] 𝑥 1

− [𝑥𝑙𝑛𝑥 − 𝑥] 𝑒1

= 𝟐 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔


61

EXERCISE 12.2


17. 𝑦 2 = 2𝑎𝑥 , 𝑦 2 = 4𝑎𝑥 − 𝑎2

𝑤𝑕𝑒𝑛 𝑥 = −4𝑎

𝑦 2 = 2𝑎𝑥𝑦 2 = 4𝑎𝑥 − 𝑎2

𝑥 = (−4𝑎)2

𝑥= 𝑑𝑥 𝑑𝑦

𝑦2 2𝑎

;x=

𝑦 2 +𝑎 2 4𝑎

2𝑦

=

16𝑎2 2𝑎

= 8𝑎

= 2𝑎

𝑑𝑥 𝑦 = 𝑑𝑦 𝑎

𝐴

𝑎

𝑑𝐴 =

0 = 0 ; (0,0)

𝑜

[ −𝑎

𝑦 2 + 𝑎2 𝑦 2 − ]𝑑𝑦 4𝑎 2𝑎

𝑑2 𝑥 1 = 𝑜𝑝𝑒𝑛 𝑡𝑜 𝑡𝑕𝑒 𝑟𝑖𝑔𝑕𝑡 𝑑𝑦 2 𝑎

=

𝑎 𝑦 2 +𝑎 2 −𝑦 2 ( 4𝑎 )𝑑𝑧 −𝑎

𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛:

=

𝑦3 𝑎2 𝑦 2𝑦 3 + − 12𝑎 4𝑎 12𝑎

=

𝑎3 𝑎2 𝑎 2𝑎3 (−𝑎)3 𝑎2 (−𝑎) 2(−𝑎)3 + − − + − 12𝑎 4𝑎 12𝑎 12𝑎 4𝑎 12𝑎

=

𝑎3 − 2𝑎3 + 𝑎3 − 2𝑎3 𝑎3 + 𝑎3 + 12𝑎 4𝑎

=

−2𝑎 3 12𝑎

X1 = X2 𝑦 2 𝑦 2 + 𝑎2 = 2𝑎 4𝑎 4𝑎𝑦 2 = 2𝑎𝑦 2 + 2𝑎3 4𝑎𝑦 2 − 2𝑎𝑦 2 − 2𝑎3 = 0 2𝑎𝑦 2 − 2𝑎3 = 0 2𝑎𝑦 2 = 2𝑎3 2𝑎3 𝑦2 = 2𝑎 2 𝑦 = 𝑎2 𝑦 = 𝑎2 𝑦 = ±𝑎 𝑎2

+

2𝑎 3 4𝑎

=

a -a

−2𝑎 3 +6𝑎 3 12𝑎

4𝑎 3

= 12𝑎 A=

a2 sq. 3

units

𝑎

X1 = X2=2𝑎 = 2 𝑤𝑕𝑒𝑛 𝑥 = 4𝑎 𝑥= =

(4𝑎)2 2𝑎

16𝑎2 2𝑎

= 8𝑎 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy

62

EXERCISE 12.2


𝟏𝟗. 𝑦 2 = 𝑥 + 1 ; 𝑦 = 1 − 𝑥

𝟐𝟏. 𝑦 2 = 4𝑥 ; 𝑦 = 4𝑥 − 4

𝑣1= 𝑦 2 − 1; 𝑦𝑥 = 1 𝑑𝑥 = 2𝑦 ; 𝑥2 = 1 − 𝑦 𝑑𝑦 𝑥 = 0; 𝑦 = 0 𝑑2 𝑥 = 2 (𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑡𝑜 𝑡𝑕𝑒 𝑟𝑖𝑔𝑕𝑡) 𝑑𝑦 2 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 𝑥1= 𝑦2 ; 𝑦 2 − 1 = 𝑦 𝑦2 + 𝑦 − 2 = 0 (𝑦 − 1)(𝑦 + 2) 𝑦−1=0 𝑦+2=0

4𝑥 = 𝑦 2 2𝑥 = 𝑦 + 4

y=1

𝑥=

𝑑𝑥 1 = 2𝑦 𝑑𝑦 4 1 0 = 2𝑦 4 0=0 0,0

y=-2

𝑣=0

𝑦2 𝑦+4 𝑥= 4 2

𝑑2𝑥 𝑑𝑦 2

𝑦=3

𝑤𝑕𝑒𝑛 𝑥 = 1, 𝑦 = 2

= (concave to the right)

𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑦2 𝑦 + 4 = 4 2

𝑤𝑕𝑒𝑛 𝑥 = 2, 𝑦 = 5 𝑤𝑕𝑒𝑛 𝑦 = 1, 𝑥 = 0 𝑤𝑕𝑒𝑛 𝑦 = 2, 𝑥 = 3 𝑤𝑕𝑒𝑛 𝑦 = 3, 𝑥 = 8 𝑡𝑕𝑒𝑛;

2𝑦 2 − 4𝑦 + 4(4) 2𝑦 2 − 4𝑦 − 16 = 0

𝑑𝐴 = 𝑋2 − 𝑋1 𝑑𝑦 1

1

1 − 𝑦 − 𝑦 2 − 1 𝑑𝑦

𝑑𝐴 = −2

2𝑦 − 8 𝑦 + 2 2𝑦 − 8 = 0𝑦 + 2 = 0 𝑦 = 4; 𝑥 = 4(1, 2)

−2

𝐴 = 1 − 𝑦 − 𝑦2 + 1 𝐴 = 2−𝑦−

1 −2

(4, 4)

𝑦 2 1−2

𝑑𝐴 = (𝑥2 − 𝑥1 )𝑑𝑦

3 1

𝐴 = 2𝑦 −

𝐴 = 2(1) −

𝑦2 𝑦 − 2 3

(1)2 (1)3 − 2 3

1

1

𝐴= −2 − 𝐴 = 2(−2) −

(−2)2 (−2)3 − 2 3

4 𝑦+4 −2 2

−

𝑦2 4

𝑑𝑦

𝑨 = 𝟗 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔

8

𝐴=2−2−3+4+2−3 𝑨=

𝟗 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 𝟐


63

EXERCISE 12.2


23. 𝑦 2 = 𝑥 + 4 , 𝑥 − 2𝑦 + 1 = 0

3

𝐴=

2𝑦 − 1 − 𝑦 2 − 4 𝑑𝑦

−1 3

3 + 2𝑦 − 𝑦 2 𝑑𝑦

= −1

= 3𝑦 + 𝑦 2 − =

3 𝑦3 3 −1

𝟑𝟐 𝟑

25. 𝑦 = 𝑒 2𝑥 , 𝑦 = 𝑒 , 𝑥 = 2

2

𝐴=

𝑒 2𝑥 − 𝑒 𝑥 𝑑𝑥

0

= =

𝑒 2𝑥 − 𝑒𝑥 2 𝑒4 2

2 0 1

− 𝑒2 − 2 + 1

= 𝟏 𝟐 𝒆𝟐 − 𝟏

𝟐


64

EXERCISE 12.4

VOLUME OF A SOLID OF REVOLUTION

1. 𝑦 = 𝑥 2 − 2𝑥 , 𝑥 − 𝑎𝑥𝑖𝑠, 𝑎𝑏𝑜𝑢𝑡 𝑡𝑕𝑒 𝑥 − 𝑎𝑥𝑖𝑠 𝑑𝑦 = 2𝑥 − 2 , 𝑒𝑞𝑢𝑎𝑡𝑒 𝑡𝑜 𝑧𝑒𝑟𝑜 𝑑𝑥 0 = 2𝑥 − 2 ; 𝑦 = 12 − 2(1) 𝑥=1

;

𝑦 = −1

𝑑2𝑦 =2 𝑑𝑥 2

𝑦 = 𝑥 2 − 2𝑥

𝑣 1, −1 x y

0 0

1 -1

2 0

3 3

1 -1

𝑑𝑣 = 𝜋𝑦 2 𝑑𝑥

dx

𝑑𝑣 = 𝜋 𝑥 2 − 2𝑥 2 𝑑𝑥

2

y (1,-1)

-2

ʃ𝑑𝑣 = 𝜋ʃ 𝑥 4 − 4𝑥 3 + 4𝑥 2 𝑑𝑥 𝑣=𝜋 =𝜋

𝑥5 5

−

1 2 5

4𝑥 4 4

5

+

4𝑥 3 3

− 24 +

4 2 3

=𝜋

32 32 − 16 + 5 3

=𝜋

96 − 240 + 160 15

=𝜋

16 15

𝑽=

3

− 0

𝟏𝟔𝝅 𝒖𝒏𝒊𝒕𝒔𝟑 𝟏𝟓


65

EXERCISE 12.4


𝟑. 𝑥 + 𝑦 = 5 ; 𝑦 = 0 ; 𝑥 = 0 ; 𝑎𝑏𝑜𝑢𝑡 𝑦 = 0

𝟓. 𝑥 + 𝑦 = 6 ; 𝑦 = 3 ; 𝑥 = 0 ; 𝑎𝑏𝑜𝑢𝑡 𝑦 − 𝑎𝑥𝑖𝑠

𝑤𝑕𝑒𝑛 𝑥 = 0 ; 𝑦 = 5

𝑥 = (6 − 𝑦)

𝑤𝑕𝑒𝑛 𝑦 = 0 ;

𝑥=5

𝑑𝑣 = 𝜋𝑥 2 𝑑𝑦

𝑑𝑣 = 𝜋𝑦 2 𝑑𝑥

; 𝑏𝑢𝑡 𝑦 = 5 − 𝑥

𝑑𝑣 = 𝜋 6 − 𝑦 𝑑𝑦

𝑦2 = 5 − 𝑥

2

𝑑𝑣 = 𝜋 36 − 12𝑦 + 𝑦 2 𝑑𝑦

𝑑𝑣 = 𝜋 5 − 𝑥 2 𝑑𝑥 𝑣

5

𝑑𝑣 = 𝜋 0

𝑣 0

25 − 10𝑥 + 𝑥 2 𝑑𝑥

𝑉=𝜋

10𝑥 2 𝑥 3 + 2 3

25 5 − 5 5

𝑉 = 𝜋 125 − 125 + 𝑽=

2

125 3

0

𝑉=

+

1 5 3

3

36 − 12𝑦 + 𝑦 2 𝑑𝑦

𝑦2 𝑦3 𝑉 = 𝜋 36𝑦 − 12 + 2 3

0

𝑉 = 𝜋 25𝑥 −

3

𝑑𝑣 =

− 0

36 3 − 6 3

2

+

𝑉 = 𝜋 36 3 − 6 9 +

−0

1 3 3

2

− 0

1 3 27

𝑉 = 𝜋[ 36 3 − 6 9 + 9]

𝟏𝟐𝟓𝝅 𝒖𝒏𝒊𝒕𝒔𝟑 𝟑

𝑉 = 𝜋(9)(12 − 6 + 1) 𝑉 = 𝜋(9)(7)

y

𝑽 = 𝟔𝟑𝝅 𝒖𝒏𝒊𝒕𝒔𝟑

𝑥=0

𝑦𝑥 = 0 5

(0,6)

3 2 1 𝑑𝑥

𝑥+𝑦 =6

5 y

𝑦=3

3 3

5

x 𝑦=0

𝑑𝑦 1 0

(6,0) 𝑥 1

3

5

𝑥


66

EXERCISE 12.4


𝟕. 𝑥𝑦 = 4, 𝑥 = 2, 𝑦 = 4; 𝑎𝑏𝑜𝑢𝑡 𝑦 = 4

9. 𝑦 2 = 4𝑎𝑥, 𝑥 = 𝑎; 𝑎𝑏𝑜𝑢𝑡 𝑥 = 𝑎

𝑉 = 𝜋𝑟 2 𝑕

2

𝑉 = 𝜋𝑟 𝑕 𝑉 = 𝜋(4 − 𝑦)2 𝑑𝑥

𝑉 = 𝜋(𝑎 − 𝑥)2 𝑕 𝑣

𝑉 = 𝜋 4− 𝑣

4 𝑥 2

𝑑𝑣 = 𝜋 0

0

𝑑𝑥 32 16 (16 − + 2 ) 𝑑𝑥 𝑥 𝑥

2𝑎

𝑑𝑣 =

2 0

−2𝑎 2𝑎

𝑉= 𝜋

𝜋(𝑎 −

(𝑎2 −

−2𝑎

16 𝑉 = 𝜋 16 2 − 32𝑙𝑛2 − −0 2

𝑉 = 𝜋 𝑎2 𝑦 −

𝑉 = 8𝜋 4 − 4 𝑙𝑛 2 − 1

𝑉 = 𝑎2 2𝑎 −

𝑽 = 𝟖𝝅 𝟑 − 𝟒 𝒍𝒏 𝟐 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔

𝑦2 2 ) 𝑑𝑦 4𝑎

𝑦2 𝑦4 + ) 𝑑𝑦 2 16𝑎2

𝑦3 𝑦5 + 6 16 5 𝑎2

2𝑎3 2𝑎5 −2𝑎3 −2𝑎5 + − 𝑎2 −2𝑎 − + 6 16(5)𝑎2 16 16(5)𝑎2 2

1

𝑉 = 4𝑎3 𝜋 1 − 3 + 5 𝑽=

𝟑𝟐𝒂𝟑 𝝅 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔 𝟏𝟓


67

EXERCISE 12.4


𝟏𝟏. 𝑦 = 𝑠𝑖𝑛 𝑥, 𝑥 = 0, 𝑦 = 1; 𝑎𝑏𝑜𝑢𝑡 𝑦 = 1

𝑉 = 𝜋𝑟 2 𝑕 𝑉 = 𝜋(1 − 𝑦)2 𝑑𝑥 𝜋 2

𝑣

𝑣= 0

𝜋 (1 − 𝑠𝑖𝑛 𝑥 )2 𝑑𝑥

0 𝜋 2

𝑉= 𝜋

0

1 − 2𝑠𝑖𝑛𝑥 + 𝑠𝑖𝑛2 𝑥 𝑑𝑥

𝑉 = 𝜋[𝑥 + 2 𝑐𝑜𝑠 𝑥 +

𝑥 𝑠𝑖𝑛2𝑥 − ] 2 4

𝑉= 𝜋

3𝑥 𝑥 𝑠𝑖𝑛 2𝑥 + 2 𝑐𝑜𝑠 𝑥 + − 2 2 4

𝑉= 𝜋

3𝑥 𝑠𝑖𝑛 2𝑥 + 2 𝑐𝑜𝑠 𝑥 − 2 4

𝑉= 𝜋

3𝜋 + 0 − 4(0) − 0 + 2 + 0 4

𝑉= 𝑽=

3𝜋 2 4

− 2𝜋

𝝅 𝟑𝝅 − 𝟖 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔 𝟒


68

EXERCISE 12.5

THE WASHER METHOD 9

1. 𝑦 = 𝑥 2 , 𝑥 = 3, 𝑦 = 0; 𝑎𝑏𝑜𝑢𝑡 𝑡𝑕𝑒 𝑦 − 𝑎𝑥𝑖𝑠

𝑉=𝜋

32 − 𝑥 2 𝑑𝑦

0 9

𝑉=𝜋

9 − 𝑦 𝑑𝑦 0

𝑉 = 𝜋 9𝑦 −

𝑦2 9 2 0

𝑉 = 𝜋 9(9) − 𝑽=

(9)2 9 2 0

𝟖𝟏𝝅 𝑪𝑼𝑩𝑰𝑪 𝑼𝑵𝑰𝑻𝑺 𝟐

3. 𝑦 2 = 4𝑎𝑥, 𝑥 = 𝑎; 𝑎𝑏𝑜𝑢𝑡 𝑡𝑕𝑒 𝑦 − 𝑎𝑥𝑖𝑠 x 0 a

𝑦 2 = 4𝑎𝑥

y 0 2a

dy x 2𝑎

X=a

𝑎2 − 𝑥 2 𝑑𝑦

𝑉= 𝜋 −2𝑎 2𝑎

𝑦2 = 𝜋 (𝑎 − 4𝑎 −2𝑎 2𝑎

= 𝜋

2

𝑎2 −

−2𝑎

2

)𝑑𝑦

𝑦4 𝑑𝑦 16𝑎2

𝑦5 2𝑎 = 𝜋 𝑎 𝑦− 80𝑎2 −2𝑎 2

= 𝜋 (2𝑎3 −

32𝑎5 32𝑎5 3 ) − (−2𝑎 + ) 80𝑎2 80𝑎2

= 𝜋 (2𝑎3 −

2𝑎 3 )− 5

𝑽=

(2𝑎3 +

2𝑎 3 ) 5

𝟏𝟔𝝅𝒂𝟑 𝟓


69

EXERCISE 12.5

THE WASHER METHOD

5. 𝑥 2 +𝑦 2 = 𝑎2 , 𝑥 = 𝑏 𝑎 𝑉 = 4𝜋 0 𝑎2 − 𝑦 2 + 𝑏 𝑑𝑦 𝑉 = 4𝜋 𝑎2 𝑦 − 𝑉 = 4𝜋 𝑉 = 4𝜋 𝑽=

𝑦3 3

+ 𝑏𝑦

𝑎3 𝑎3 − 3 − 2𝑎 3 − 𝑎𝑏 3

𝑎𝑏

a

(-a,0)

o

(a,0)

x=b

𝟖𝝅𝒂𝟑 𝟑

7. 𝑥 2 + 𝑦 2 = 25 , 𝑥 + 𝑦 = 5 ; 𝑦 = 0 𝑉= 𝜋 𝑽=

5 0

25 − 𝑥 2 − 5 − 𝑥

2

𝑑𝑥

𝟏𝟐𝟓𝝅 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔 𝟑


70

EXERCISE 12.5

THE WASHER METHOD

9. 𝑦 2 = 4𝑥, 𝑥 2 = 4𝑦; 𝑎𝑏𝑜𝑢𝑡 𝑡𝑕𝑒 𝑥 − 𝑎𝑥𝑖𝑠

𝑦2 = 𝑦1

2

𝑥2 4𝑥 = 4 3 64𝑥 = 𝑥 𝑥 = 4, 𝑦 = 4: 𝑃𝑂𝐼 (4,4) 4

𝑉=𝜋

4𝑥

2

−

0 4

𝑥2 4

2

𝑑𝑥

𝑥4 𝑑𝑥 16 0 𝑥5 4 𝑉 = 𝜋 2𝑥 2 − 80 0 (4)5 𝑉 = 𝜋 2(4)2 + 80 𝟗𝟔𝝅 𝑽= 𝑪𝑼𝑩𝑰𝑪 𝑼𝑵𝑰𝑻𝑺 𝟓 𝑉=𝜋

11. 𝑦 2 = 8𝑥, 𝑌 = 2𝑥; 𝑎𝑏𝑜𝑢𝑡 𝑦 = 4

4𝑥 −

𝑦2 = 𝑦1 8𝑥 = 2𝑥

2

8𝑥 = 4𝑥 2 𝑥 = 2, 𝑦 = 4: 𝑃𝑂𝐼 (2,4) 4𝑥 3 2 𝑉 = 𝜋 4𝑥 2 − 3 0 4(2)3 𝑉 = 𝜋 4(2) + 3 2

𝑽=

𝟏𝟔𝝅 𝑪𝑼𝑩𝑰𝑪 𝑼𝑵𝑰𝑻𝑺 𝟑


71

EXERCISE 12.6

THE CYLINDRICAL SHELL METHOD

𝟏. 4𝑦 = 𝑥 3 , 𝑦 = 0, 𝑥 = 2, ; 𝑎𝑏𝑜𝑢𝑡 𝑥 = 2

V = 2π

2 𝑥𝑦𝑑𝑥 0

V = 2π

2 0

V = 2π

2 𝑥2 [ 0 2

V = 2π V = 2π V = 2π V=

2−𝑥

𝑥4 4

−

−

(2)4 4

−

𝑥3 4

dx

𝑥4 ]𝑑𝑥 4

𝑥5 20

2 0

(2)5 20

3 5

𝟒𝝅 cubic units 𝟓

2 0

3. 𝑥 = 4𝑦 – 𝑦 2 , 𝑦 = 𝑥 , 𝑎𝑏𝑜𝑢𝑡 𝑦 = 0

V = 2π

3 𝑥𝑦𝑑𝑦 0

V = 2π

3 0

V = 2π

3 0

V = 2π

4𝑦 − 𝑦 2 − 𝑦 𝑦𝑑𝑦 4𝑦 2 − 𝑦 3 − 𝑦 2 𝑑𝑦

4 2 𝑦 3

V = 2π 𝑦 3 −

𝑦4 4

V = 2π (3)3 − V=

𝟐𝟕𝝅 𝒄𝒖𝒃𝒊𝒄 𝟐

1 4 𝑦 4

−

1

− 3 𝑦3

3 0

(3)4 4

3 0

𝒖𝒏𝒊𝒕𝒔


72

3 0

EXERCISE 12.6


𝜋

5. 𝑦 = 𝑠𝑖𝑛𝑥, 𝑦 = 𝑐𝑜𝑠𝑥, 𝑥 = 2 𝑉 = 2𝜋 𝑽=

𝜋 2 𝜋 4

𝑥 𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥 𝑑𝑥

𝝅 𝟒 + 𝟐𝝅 − 𝟐𝝅 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔 𝟐

7. 𝑥 = 2 𝑦 𝑉 = 2𝜋 𝑽=

9 0

𝜋 2

,𝑥 = 0 ,𝑦 = 0

Y=9

9 − 𝑦 2 𝑦 𝑑𝑦

𝟏𝟐𝟗𝟔𝝅 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔 𝟓

9.𝑦 = 𝑙𝑛𝑥 , 𝑥 = 𝑒 , 𝑦 = 0 𝑉 = 2𝜋

𝑒 𝑥 1

𝑙𝑛𝑥 𝑑𝑥 (e,1)

𝑽 = 𝟏𝟑. 𝟕𝟕 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔

(1,0) X=e


73

EXERCISE 12.6


11. 𝑦 2 = 8𝑥 , 𝑥 = 0 , 𝑦 = 4 ; about 𝑦 = 4

4

𝑉 = 2𝜋

4−𝑦 0

=

𝜋 4

4

𝑦2 𝑑𝑦 8

4𝑦 2 − 𝑦 3 𝑑𝑦

0

=

𝜋 4𝑦 3 4 3

=

𝟏𝟔𝝅 𝟑

−

4 𝑦4 4 0

13. ( 𝑥 – 3 ) 2 + 𝑦 2 = 9; 𝑎𝑏𝑜𝑢𝑡 𝑡𝑕𝑒 𝑦 – 𝑎𝑥𝑖𝑠. 3

𝑉 = 8𝜋

2 x 9  ( x  3) dx

0 𝑉 = 8𝜋(

− ( 9 – ( x − 3 ) 2 3 27 𝑠𝑖𝑛𝑥 − 3 9 )2 + + (𝑥 − 3)( 9 − 𝑥 − 3 3 2 3 2

𝑉 = 8𝜋(27𝑠𝑖𝑛𝜃 −

2

3 0

27 𝑠𝑖𝑛 − 1) 2

27 𝑉 = 8𝜋( )(−𝑠𝑖𝑛 − 1 + 𝑠𝑖𝑛𝜃) 2 𝜋 𝑉 = 108𝜋( ) 2 𝑽 = 𝟓𝟒𝝅𝟐


74

EXERCISE 12.6


15. 𝑥 2 + 𝑦 2 = 𝑎2 ; 𝑎𝑏𝑜𝑢𝑡 𝑥 = 𝑏 𝑏 > 𝑎 𝑎

𝑉= 𝜋

𝑏−𝑥 −𝑎 𝑎

𝑉= 𝜋

2

− 𝑏−𝑥

2

𝑑𝑦

𝑏 2 − 𝑏𝑥 + 𝑥 2 − 𝑏 2 − 𝑏𝑥 + 𝑥 2 𝑑𝑦

−𝑎 𝑎

𝑉= 𝜋

4𝑏𝑥𝑑𝑦 −𝑎

𝑛𝑜𝑡𝑒: 𝑥 2 + 𝑦 2 = 𝑎2 𝑉 = 4𝑏𝜋 𝑉 = 4𝑏𝜋

𝑎 −𝑎

𝑦 2 − 𝑎2

=𝑥=

𝑦 2 − 𝑎2 𝑑𝑦

𝑦 𝑎2 𝑎 𝑦 2 − 𝑎2 − ln 𝑦 + 𝑦 2 − 𝑎2 + 𝑐 −𝑎 2 2

𝑽 = 𝟐𝝅𝟐 𝒂𝟐 𝒃 𝑥 2 + 𝑦 2 = 𝑎2

a

a

a

a


75

EXERCISE 12.7

𝟏.

VOLUME OF SOLIDS WITH KNOWN CROSS SECTIONS

𝑥 2 + 𝑦 2 = 36

𝟑. 9𝑥 2 + 16𝑦 2 = 144

𝑆2 𝐴 𝑥 = , 2

𝑆 = 2𝑦

𝐴 𝑥 = 2𝑦 2 ,

𝑦=

36 − 𝑥 2

6

𝑣=

𝐴 𝑥 𝑑𝑥 −6

1 𝐴 𝑥 = (2𝑦)(𝑦) 2 𝐴 𝑥 = 𝑦2 8

0

2𝑥 2 𝑑𝑥

𝑣=

𝑦 2 𝑑𝑥

𝑉=2

6

𝑉=2

−6

𝑣=

1 𝐴 𝑥 = 𝑏𝑕 2

6 2(3𝑥 −6

− 𝑥 2 )𝑑𝑥

8 144−9𝑥 2 0 16

𝑑𝑥

𝑽 = 𝟒𝟖 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔

𝒗 = 𝟓𝟕𝟔 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔


76

EXERCISE 12.7

VOLUME OF SOLIDS WITH KNOWN CROSS SECTIONS

𝟓.

𝐴 𝑦 = (1 − 𝑥)(2𝑦 2 ) 2

𝑉 = 2 (1 − 𝑥 )2𝑦 2 𝑑𝑦 0 2

𝑉 = 2 (1 − 0

𝑦2 2 )𝑦 𝑑𝑦 4

64

𝑉 = 15 𝑽 = 𝟒. 𝟐𝟔𝟔𝟕 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔


77

EXERCISE 12.8

LENGTH OF AN ARC

3

𝟏. 𝑦 = 𝑥 2 𝑓𝑟𝑜𝑚 𝑥 = 0 𝑡𝑜 𝑥 = 5

2

2

3

𝑦 = 𝑥2 3 1 𝑑𝑦 = 𝑥 2 𝑑𝑥 2 𝑑𝑦 3 1 = 𝑥2 𝑑𝑥 2 𝑠= 0

𝑆=

=

𝑑𝑦 1 + ( )2 𝑑𝑥 𝑑𝑥

=

5 0

1+

1+ − 0

2

𝑆=

𝑥

0

𝑦3 1

𝑥3

2

𝑑𝑥

1

𝑥3 + 𝑦3

2

5 0

1

9

9 5

2

3. 𝑡𝑕𝑒 𝑒𝑛𝑡𝑖𝑟𝑒 𝑕𝑦𝑝𝑜𝑐𝑦𝑐𝑙𝑜𝑖𝑑 𝑥 3 + 𝑦 3 + 𝑎3

2 3

2

𝑑𝑥

2

𝑁𝑜𝑡𝑒: 𝑎3 = 𝑥3 + 𝑦3

3 1 ( 𝑥 2 )2 dx 2

2

9

9 4

𝑎3

𝑆=

1 + 𝑥 𝑑𝑥

2

𝑥3

0

𝒔 = 𝟏𝟐. 𝟒𝟎𝟕 𝒖𝒏𝒊𝒕𝒔

9

𝑆= 0

𝑑𝑥

1

𝑎3 1

𝑥3

𝑑𝑥 2

3𝑥 3 9 𝑆= 𝑎 2 0 1 3

𝑆=

3𝑎 2

𝑆=4

3𝑎 2

𝑺 = 𝟔𝒂

X=0

x=5


78

EXERCISE 12.8

LENGTH OF AN ARC

5. 𝑦 = 𝐴𝑟𝑐𝑠𝑖𝑛𝑒 𝑥 , 𝑓𝑟𝑜𝑚 𝑦 =

𝜋 6

𝑡𝑜 𝑦 =

𝜋 2

𝑦 = 𝐴𝑟𝑐𝑠𝑖𝑛𝑒 𝑥 ; 𝑙𝑛 𝑠𝑖𝑛𝑦 = 𝑥 1 𝑐𝑜𝑠𝑦𝑑𝑦 = 𝑑𝑥 𝑠𝑖𝑛𝑦

7. 𝑜𝑛𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑕𝑒 𝑐𝑦𝑐𝑙𝑜𝑖𝑑 𝑥 = 𝑎 𝜃 − 𝑠𝑖𝑛𝜃 , 𝑦 = 𝑎(1 − 𝑐𝑜𝑠𝜃) 𝑥 = 𝑎(𝜃 − 𝑠𝑖𝑛𝜃) 𝑑𝑥 = 𝑎(𝑑𝜃 − 𝑐𝑜𝑠𝜃𝑑𝜃)

𝑦 = 𝑎(1 − 𝑐𝑜𝑠𝜃) 𝑑𝑦 = 𝑎(𝑠𝑖𝑛𝜃𝑑𝜃)

𝑑𝑥 𝑑𝜃

𝑑𝑦 𝑑𝜃

= 𝑎(1 − 𝑐𝑜𝑠𝜃)

𝑑𝑥 𝑐𝑜𝑠𝑦 = 𝑑𝑦 𝑠𝑖𝑛𝑦

2𝜋

𝑠=

𝑑𝑥 = 𝑐𝑜𝑡𝑦 𝑑𝑦 𝑆= =

𝜋 2 𝜋 6

𝜋 2 𝜋 6

2

+ 𝑎2 sin2 𝜃

0

𝑠=𝑎 1+

𝑎2 1 − 𝑐𝑜𝑠𝜃

= 𝑎𝑠𝑖𝑛𝜃

𝑑𝑥 2 𝑑𝑦

𝑑𝑦

2𝜋 0

1 − 𝑐𝑜𝑠𝜃

2

+ sin2 𝜃

𝒔 = 𝟖𝒂

1 + 𝑐𝑜𝑡 2 𝑦 𝑑𝑦

𝑺 = 𝟏. 𝟑𝟏𝟔𝟗𝟔 𝒖𝒏𝒊𝒕𝒔

9. 𝑇𝑕𝑒 𝐶𝑎𝑟𝑑𝑖𝑜𝑖𝑑 𝑟 = 2 1 − 𝑐𝑜𝑠𝜃

𝜋 2

𝑟 = 2 1 − 𝑐𝑜𝑠𝜃 𝑑𝑟 = 2 𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝑟 = 2𝑠𝑖𝑛𝜃 𝑑𝜃 𝑟 2 = 4(1 − 𝑐𝑜𝑠𝜃)2 2𝜋

𝑆=

4(1 − 𝑐𝑜𝑠𝜃)2 + 4𝑠𝑖𝑛2 𝜃 𝑑𝜃

0

𝑆=2 𝜋 6

2𝜋 0

(1 − 𝑐𝑜𝑠𝜃)2 + 𝑠𝑖𝑛2 𝜃 𝑑𝜃

𝑺 = 𝟏𝟔 𝒖𝒏𝒊𝒕𝒔


79

EXERCISE 12.9

AREA OF A SURFACE OF REVOLUTION

1. 𝑥 2 + 𝑦 2 = 16 ; 𝑓𝑟𝑜𝑚 𝑥 = 2 𝑡𝑜 𝑥 = 4

3. 𝑦 2 = 12𝑥 ; 𝑓𝑟𝑜𝑚 𝑥 = 0 𝑡𝑜 𝑥 = 3

4

𝑆 = 2𝜋

3

𝑦𝑑𝑠

𝑆 = 2𝜋

2

𝑦=

𝑦 = 12𝑥

16 − 𝑥 2

𝑑𝑦 1 = 16 − 𝑥 2 𝑑𝑥 2

1 2

−

𝑑𝑦 1 = 12𝑥 𝑑𝑥 2

(−2𝑥)

2

𝑑𝑠 =

𝑑𝑦 1+ 𝑑𝑥

𝑑𝑠 =

𝑥2 1+ 𝑑𝑥 16 − 𝑥 2

𝑑𝑠 =

16 − + 16 − 𝑥 2

16 − 𝑥 2

𝑆 = 2𝜋

𝑥2

(12)

𝑑𝑠 =

16 − 𝑥 2

36 𝑑𝑥 12𝑥

12𝑥 + 36

𝑑𝑠 =

12𝑥 2 3𝑥 + 9

𝑑𝑥

12𝑥

𝑑𝑥

𝑑𝑥

3

12𝑥

2 3𝑥 + 9

0

𝑑𝑥

2

𝑆 = 2𝜋

1+

𝑆 = 2𝜋

4 4

𝑑𝑠 =

𝑑𝑥

𝑥2

1 2

𝑑𝑦 6 = 𝑑𝑥 12𝑥

𝑑𝑦 𝑥 =− 𝑑𝑥 16 − 𝑥 2

𝑑𝑠 =

𝑦𝑑𝑠 0

4 16 − 𝑥 2

𝑆 = 4𝜋 𝑑𝑥

3 0

12𝑥

𝑑𝑥

3𝑥 + 9 𝑑𝑥

𝑺 = 𝟏𝟑𝟕. 𝟖𝟔𝟎 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔

4 4𝑑𝑥 2

𝑺 = 𝟏𝟔𝝅 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔

5. 𝑦 = 𝑥 3 ; 𝑓𝑟𝑜𝑚 𝑥 = 0 𝑡𝑜 𝑥 = 1 𝑑𝑦 = 3𝑥 2 𝑑𝑥 𝑑𝑠 = 𝑆 = 2𝜋

1 + 9𝑥 4 𝑑𝑥 1 3 𝑥 0

1 + 9𝑥 4 𝑑𝑥

𝑺 = 𝟑. 𝟓𝟔𝟑𝟏 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔


80

EXERCISE 12.9


7. 𝑥 = 𝑐𝑜𝑠2𝑦 ; 𝑓𝑟𝑜𝑚 𝑦 = 0 𝑡𝑜 𝑦 = 𝜋 4

𝑆 = 2𝜋

𝜋 4

𝑥𝑑𝑠

0

𝑑𝑥 = − 𝑠𝑖𝑛 2𝑦(2) 𝑑𝑦 𝑑𝑠 = 𝑆 = 2𝜋

1 + 4 𝑠𝑖𝑛2 2𝑦 𝜋 4

0

𝑐𝑜𝑠 2𝑦 1 + 4 𝑠𝑖𝑛2 2𝑦 𝑑𝑦

𝑺 = 𝟒. 𝟗𝟑𝟔𝟔𝟓 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔

9. 4 − 𝑥 2 𝑓𝑟𝑜𝑚 𝑥 = 0 𝑡𝑜 𝑥 = 2 2

𝑆 = 2𝜋

𝑥𝑑𝑠 0

𝑑𝑦 = −2𝑥 𝑑𝑥 𝑑𝑠 = 1 + 4𝑥^2 𝑑𝑥 𝑆 = 2𝜋

2 𝑥 0

1 + 4𝑥 2 𝑑𝑥

𝑺 = 𝟑𝟔. 𝟏𝟕𝟔𝟗 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔


81

EXERCISE 12.9


13. 𝑦 = 𝑚𝑥 ; 𝑥 = 0 ; 𝑥 = 1 ; 𝑎𝑏𝑜𝑢𝑡 𝑡𝑕𝑒 𝑥 − 𝑎𝑥𝑖𝑠 1

𝑆 = 2𝜋

𝑚𝑥

1 + 𝑚2

𝑑𝑥

0 1

𝑆 = 2𝜋 𝑚

1 + 𝑚2 =

𝑥 𝑑𝑥 0

𝑆 = 2𝜋 𝑚 𝑆 = 2𝜋 𝑚

1 + 𝑚2( 𝑥2/2 )

1 0

1 + 𝑚2( ½ )

𝑺 = 𝝅 𝒎 𝟏 + 𝒎𝟐


82

EXERCISE 13.1

1.

FORCE OF FLUID PRESSURE

𝐹 = 𝑤𝐴𝑥 = (62.5𝑙𝑏/𝑓𝑡 3 )(96𝑓𝑡 2 )(4𝑓𝑡) = 24000𝑙𝑏

𝑃=

𝐹 𝐴

𝑃=

𝑤𝐴𝑥 𝐴

3. 𝐹 = 𝑤𝐴𝑥 1 𝐹=𝑤 5 3 2

2 2 + ( )(3) 3

𝑭 = 𝟑𝟎𝒘 𝒍𝒃

2 3

𝑃 = 𝑤𝑥

5

62.5𝑙𝑏3 1𝑓𝑡 2 𝑃=( )(4𝑓𝑡)( ) 𝑓𝑡 144𝑖𝑛 2 𝑃=

3 5

(625)(4) 144

5 3

𝑷 = 𝟏. 𝟕𝟒 𝒑𝒔𝒊

5

5. 𝐹 = 50𝑤

12ft

𝑏𝑎𝑠𝑒 = 3𝑓𝑡 8ft

𝑥

𝐹 = 𝑤𝐴𝑥 50 =

1 𝑕 3 2

50 =

𝑕2 2

1 𝑕 3

100 = 𝑕2 𝒉 = 𝟏𝟎𝒇𝒕

3 5

h 5


83

EXERCISE 13.1

FORCE OF FLUID PRESSURE

7. 𝐹 = 𝑤𝐴𝑥 = 𝑤[(𝜋)(3)(2)](2) 𝑭 = 𝟏𝟐𝝅 𝒘 𝑏 = 6 = major axis 𝑎 = 4 = 𝑚𝑖𝑛𝑜𝑟 𝑎𝑥𝑖𝑠

0

y 𝑥

b a

A=𝜋𝑎𝑏

x


84

EXERCISE 13.2

WORK

1.

𝑏

𝑤=

𝑓 𝑥 𝑑𝑥 𝑎

𝑓 𝑥 = 𝑘𝑥 40 𝑙𝑏 = 𝑘 𝑤=

;

𝑤𝑕𝑒𝑟𝑒 𝑥 =

1 𝑓𝑡, 2

𝑓 𝑥 = 40 𝑙𝑏 ; 𝑎 = 0,

𝑏 = 14 − 10 = 4

1 𝑓𝑡 , 𝑘 = 80 2

4 80𝑥𝑑𝑥 0

𝒘 = 𝟔𝟒𝟎 𝒍𝒃 − 𝒇𝒕

3.

𝑏

𝑤=

𝑓 𝑥 𝑑𝑥 𝑎

𝑓 𝑥 = 𝑘𝑥 ; 𝑤𝑕𝑒𝑟𝑒 𝑥 = 𝑤=

1 𝐿 𝑓𝑡, 10

𝑓 𝑥 = 5 𝑙𝑏

𝑎 = 0, 𝑏 = 𝐿

𝐿 50 𝑥𝑑𝑥 0 𝐿

𝒘 = 𝟐𝟓𝑳 𝒇𝒕 − 𝒍𝒃


85

EXERCISE 13.2

WORK

5. 𝑊 = 𝐹𝑆 𝑑𝑤 = 𝑤 𝑑𝑣 60 − 𝑥 𝑑𝑤 = 𝜋𝑟 2 𝑤 60 − 𝑥 𝑑𝑥 𝑑𝑤 = 9𝜋𝑤(60 − 𝑥)𝑑𝑥 𝑤

10

𝑑𝑤 = 9𝜋𝑤 0

60 − 𝑥 𝑑𝑥 0

𝑤 = 9𝜋𝑤 60𝑥 − 𝑥 2 𝑤 = 9𝜋𝑤 60𝑥 −

10 0

𝑥 2 10 2 0

𝑤 = 9𝜋𝑤 600 − 50 𝒘 = 𝟒𝟗𝟓𝟎𝒘𝝅 𝒇𝒕. 𝒍𝒃


86

EXERCISE 13.2

WORK

9.

𝑏

𝑤=𝑤

𝑕𝑑𝑉 𝑎

𝑉𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 = 𝑙 𝑥 𝑤 𝑥 𝑕; 𝑤𝑕𝑒𝑟𝑒 𝑙 = 10 𝑓𝑡, 𝑤 = 2𝑥, 𝑕 = 𝑑𝑦 𝑥2 + 𝑦2 = 𝑟2 ; 𝑥 =

𝑟 2 − 𝑦 2 ; 𝑤𝑕𝑒𝑟𝑒 𝑟 = 2

2

𝑤=𝜋

6−𝑦

10 𝑓𝑡 2𝑥 𝑑𝑦

−2

2

𝑤 = 20𝜋

6−𝑦

22 − 𝑦 2 𝑑𝑦

−2

𝒘 = 𝟐𝟒𝟎𝝅𝒘 𝒇𝒕 − 𝒍𝒃


87

EXERCISE 13.3

FIRST MOMENT OF A PLANE AREA

1. 𝑦 2 = 4𝑥, 𝑡𝑕𝑒 𝑥 − 𝑎𝑥𝑖𝑠 𝑎𝑛𝑑 𝑥 = 4 𝑀𝑥 =

5. 𝑦 2 = 4𝑥 𝑎𝑛𝑑 𝑥 2 = 4𝑦

1 4 4𝑥𝑑𝑥 2 0

𝑴𝒙 = 𝟏𝟔 𝑀𝑦 =

4 𝑥 0

4𝑥 𝑑𝑥

𝑴𝒚 = 𝟐𝟓. 𝟔

𝑦2 = 4𝑥

4𝑥 =

𝑥4 16

64𝑥 − 𝑥 4 = 0

𝑥=4

𝑥 64 − 𝑥 3 = 0 𝑥1 = 0, 𝑥2 = 4 1

𝑀𝒙 = 2 3. 𝑥 = 4

𝑴𝒙 =

4 𝑜

4𝑥 −

𝑥4 16

𝑑𝑦

𝟒𝟖 𝟓

𝑏

𝑀𝛌 = 𝑀𝛌 = 𝑀𝛌 = 𝑀𝛌 = 𝑴𝛌 =

𝑙𝑑𝐴 𝑎 4

𝑥 4 − 𝑥 𝑑𝑦 𝑜 4 𝑜 4

4𝑥 − 𝑥 2 𝑑𝑦 𝑦2 −

𝑜

𝑦4 𝑑𝑦 16

𝟐𝟓𝟔 𝟏𝟓 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy

88

EXERCISE 13.3

FIRST MOMENT OF A PLANE AREA

7. 𝑀 =

3 0

3−𝑦 [

=

3 0

3−𝑦

27𝜋 −9−9 4 27𝜋 𝑀= − 18 4 27𝜋−72 𝑀= 4

9 − 𝑦 2 − 3 − 𝑦 𝑑𝑦

𝑀=

3 + 𝑦 3 − 𝑦 − (3 − 𝑦)2 𝑑𝑦

3 1 3 (3 − 𝑦)2 (3 + 𝑦)2 − (3 − 𝑦)2 𝑑𝑦 0 3 3 3 = 0 3 9 − 𝑦 2 𝑑𝑦 − 0 𝑦 9 − 𝑦 2 𝑑𝑦 − 0 (3 − 𝑦)2 𝑑𝑦

=

3 9 0 9−𝑦 2 3

*𝐴 =3 𝑐𝑜𝑠𝜃 =

− 𝑦2

𝑑𝑦

𝟗 [𝟑𝝅 − 𝟖] 𝟒

𝑦

𝑠𝑖𝑛𝜃 = 3

9 − 𝑦2

3𝑐𝑜𝑠𝜃 =

𝑴=

𝑦

3𝑠𝑖𝑛𝜃 = 𝑦; 𝜃 = 𝑎𝑟𝑐𝑠𝑖𝑛 3

3𝑐𝑜𝑠𝜃𝑑𝜃 = 𝑑𝑦 𝜋 𝑦 = 3; 𝜃 = 2

𝟗. 𝑥 = 4𝑦 − 𝑦 2 , 𝑦 = 𝑥

𝑦 = 0; 𝜃 = 0 =3 = 27

𝜋 2

0

𝜋 2

3𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝜃𝑑𝜃

0

= 27

𝜋 2

𝑐𝑜𝑠 2 𝜃𝑑𝜃 = 27

0

𝜋

1 + 𝑐𝑜𝑠2𝜃 𝑑𝜃 2

𝜃 𝑠𝑖𝑛2𝜃 2 𝜋 27𝜋 + = 27 = 2 4 4 4 0 3

*𝐵 = −

𝑦 9 − 𝑦 2 𝑑𝑦

0

𝑢 = 9 − 𝑦2 𝑑𝑢 = −2𝑦𝑑𝑦 − = = =

1 2

𝑑𝑢 2

9−𝑦 2 3 2

= 𝑦𝑑𝑦 | 30 3 2

1 2 ( )[(9 − 9) − 2 3 1 −27 = −9 3

*𝐶 =−

@ 𝑦 = 3; 𝑢 = 0 𝑦 = 0; 𝑢 = 9

3 (3 − 0

3 2

𝑑

𝑀𝑦 =

1 2

𝑀𝑦 =

1 3 2 0

𝑴𝒚 =

𝟓𝟒 𝟓

(9 − 0) ]

𝑐

𝑥𝑟2 − 𝑥𝑙2 𝑑𝑦 4𝑦 − 𝑦 2

2

− 𝑦 2 𝑑𝑦

𝑦)2 𝑑𝑦

𝑢 =3−𝑦 𝑑𝑢 = −𝑑𝑦 = =

(3−𝑦)3 3 |0 3 3 0 3 − 3 3

= −9


89

EXERCISE 13.4

CENTROID OF A PLANE AREA

1. 𝑥 + 2𝑦 = 6, 𝑥 = 0, 𝑦 = 0 Solving for A 𝑑𝐴 = 𝑦𝑑𝑥 6 𝑑𝐴 0

=

6 0

𝐴 = [3𝑥 −

𝑥

3 − 2 𝑑𝑥

𝑥2 6 ] 4 0

𝐴= 3 6 −

36 4

𝑨 = 𝟗 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔

Solving for 𝑥

Solving for 𝑦

𝐴𝑥 =

6 𝑋𝑐 0

𝑑𝐴

𝐴𝑥 =

6 𝑥 0

3−

𝑥2 2

𝑑𝑥

𝐴𝑦 = 2

𝐴𝑥 =

6 0

3𝑥 −

𝑥2 2

𝑑𝑥

𝐴𝑦 =

3𝑥 2 2

𝐴𝑥 = [

−

𝐴𝑦 =

6 𝑌𝑐 0 1

𝑥3 6 ] 3 0

𝑑𝐴

6 𝑥 𝑥 (3 − 2 ) (3 − 2 )𝑑𝑥 0

1 6 (9 − 2 0 1

3

3𝑥 +

𝑥2 )𝑑𝑥 4 𝑥3

𝐴𝑦 = 2 [9𝑥 − 2 𝑥 2 + 12 ] 60 1

9𝑥 = 18

𝑦 = 3 (3)

𝒙 = 𝟐 𝒖𝒏𝒊𝒕𝒔

𝒚 = 𝟏 𝒖𝒏𝒊𝒕

Centroid: (2,1)


90

EXERCISE 13.4


3. 𝑦 = 𝑠𝑖𝑛𝑥, 𝑦 = 0 𝑓𝑟𝑜𝑚 𝑥 = 0 − 𝜋

A= 𝑦𝑑𝑥 =

𝜋 0

𝑠𝑖𝑛𝑥𝑑𝑥

= −𝑐𝑜𝑠𝑥 A=2 𝑦

𝑀𝑥 = =

1

𝜋 0

1

𝜋 0

=2

𝑥𝑐𝑑𝑎; 𝑥𝑐 = 𝑥

=

𝜋 0

𝑦 2 𝑑𝑥

=

𝜋 0

𝑠𝑖𝑛2 𝑥 𝑑𝑥

𝑢 = 𝑥 ; 𝑑𝑣 = 𝑠𝑖𝑛𝑥

1 𝜋 1−𝑐𝑜𝑠 2𝑥 ( ) 𝑑𝑥 2 0 2 1 𝑥

= 2 (2 − 2 1 𝑥

= 2 (2 −

𝑠𝑖𝑛 2𝑥 2

𝑥𝑦𝑑𝐴 𝑥𝑠𝑖𝑛𝑥𝑑𝑥

𝑑𝑢 = 𝑑𝑥 ; 𝑣 = −𝑐𝑜𝑠𝑥 𝑑𝑥

= −𝑐𝑜𝑠𝑥 − −𝑐𝑜𝑠𝑥𝑑𝑥

𝑠𝑖𝑛 2𝑥 ) 4

= [−𝑥𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥]

𝜋 4

𝑀𝑥 = (2) 𝑥 =

𝜋 0

𝑀𝑦 =

𝜋 𝑦 ( )𝑦𝑑𝑥 0 2

=2

=

𝑦𝑐𝑑𝑎; 𝑦𝑐 = 2

= −𝜋𝑐𝑜𝑠𝜋 + 𝑠𝑖𝑛𝜋 + 0 − 𝑠𝑖𝑛0

𝜋 2

=𝜋 𝜋

𝑦 = ( 4 )(2) =

Centroid:

𝜋 8

𝝅 𝝅 , 𝟐 𝟖


91

EXERCISE 13.4


7. 𝑦 2 = 𝑥 3 , 𝑦 = 2𝑥 4

𝐴=

3

(2𝑥 − 𝑥 2 )𝑑𝑥 0

2 5 𝐴 = [𝑥 2 − 𝑥 2 ] 5 𝐴 = [16 − 𝐴=

64 ] 5

16 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠 5

4

𝐴𝑥 =

𝑦𝑥𝑑𝑥

𝐴𝑦 =

0 4

𝐴𝑥 =

3

(2𝑥 − 𝑥 2 )𝑥𝑑𝑥

𝐴𝑦 =

0 4

𝐴𝑥 =

5

(2𝑥 2 − 𝑥 2 )𝑑𝑥

𝐴𝑦 =

0

2 2 7 𝐴𝑥 = [ 𝑥 3 − 𝑥 2 ] 3 7 𝑥 =

𝐴𝑦 =

7 5 2 2 [ (4)3 − (4)2 16 3 7

𝑥=

5 128 257 [ − ] 16 3 7

𝑥=

40 𝑢𝑛𝑖𝑡𝑠 21

𝑦=

𝑪𝒆𝒏𝒕𝒓𝒐𝒊𝒅:

1 2 1 2 1 2

4

𝑦 2 𝑑𝑥

0 4

5

[(2𝑥)2 − 𝑥 2 ]𝑑𝑥

0 4

(4𝑥 2 − 𝑥 3 ) 𝑑𝑥

0

1 4 3 𝑥4 4 𝑥 − 2 3 4 0

10 𝑢𝑛𝑖𝑡𝑠 3

𝟒𝟎 𝟏𝟎 , 𝟐𝟏 𝟑


92

EXERCISE 13.4


9. 𝑥 2 + 𝑦 2 = 25,

𝑥+𝑦 =5

25𝜋 − 50 4 25 𝐴= (𝜋 − 2) 4 𝐴=

5

𝑀𝑦 = 5

=

25 − 𝑥 2 − 5𝑥 𝑑𝑥

𝑥 0

𝑥 25 −

𝑥 2 𝑑𝑥

5

−

0

5 x 25 − 𝑥 2 − 5𝑥 𝑑𝑥

𝐴= 𝐴=

0 5 0

25 − 𝑥 2 𝑑𝑥 − 5

𝐴∶

5 0

A 25 − 𝑥 2 𝑑𝑥

5 𝑑𝑥 0

+

B

25 − 5 𝑥𝑑𝑥 0

C

25 − 𝑥 2

5 cos 𝜃 =

𝑥 5 sin 𝜃 = 𝑥 ; 𝜃 = arcsin 5 𝜋 5 cos 𝜃 = 𝑑𝑥 @𝑥 = 5 ; 𝜃 = 2 𝑥 =0; 𝜃 =0 =

𝜋 2

5 cos 𝜃 ∙ 5 cos 𝜃

0

= 25

𝜋 2

0

1 = 25 2

1 + 𝑐𝑜𝑠2𝜃 cos 𝜃𝑑𝜃 → cos 𝜃 = 2 2

𝜋 2

0

1 𝑑𝜃 + 2

2

𝜋 2

0

∏/2 𝜋 +0 4 0 25𝜋 = 4 5 𝐵 ∶ − 5 0 𝑑𝑥 5 = −5𝑥 0 = −25

= 25

𝑥2

𝐶∶ 2 25 = 2

5 0

25𝜋 25 − 25 + 4 2 25 𝐴 = 25𝜋 − 2 ∴𝐴=

cos 2𝜃𝑑𝜃

𝑥2

𝑢 = 25 − 𝑥 2 𝑑𝑢 = −2𝑥𝑑𝑥 =

1 25−𝑥 2 2 −2 3

−

2

3

=−

25−𝑥 2 2

5𝑥𝑑𝑥 + 0

3

5

5

5𝑥 2 2

+

𝑥3 3

5

5𝑥 2 2

𝑥 2 𝑑𝑥

0

5 0

𝑥3 3 0

− + 0 125 125 125 = − − + − − −0+0 3 2 3 3 125 250 375 + 500 =− + = − 2 3 6 125 𝑀𝑦 = 6 2 1 5 𝑀𝑥 = 25 − 𝑥 2 − 5𝑥 2 𝑑𝑥 2 0 5 1 1 5 = 25 − 𝑥 2 𝑑𝑥 − 5 − 𝑥 2 𝑑𝑥 2 0 2 0 1 𝑥3 𝑥3 = 25𝑥 − − 1/2 25𝑥 + − 5𝑥 2 3 3 =

3

1 125 1 125 125 − − 125 + − 125 2 3 2 3

− 0

125 6 ∴ 𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑: 𝑀𝑥 =

𝑀𝑥 𝑥= = 𝐴 𝑥=

125 6 25 𝜋−2 4

=

125 4 ∙ 6 25 𝜋 − 2

10 3 𝜋−2

𝑀𝑥 𝑦= = 𝐴 𝑦=3

10 𝜋−2

125 6 25 𝜋−2 4

𝑪𝒆𝒏𝒕𝒓𝒐𝒊𝒅 𝒊𝒔 𝒂𝒕

𝟏𝟎 𝟏𝟎 , 𝟑 𝝅−𝟐 𝟑 𝝅−𝟐


2

93

EXERCISE 13.5

CENTROID OF A SOLID OF REVOLUTION

1. 𝑦 2 = 𝑥 ; 𝑦 = 3 ; 𝑥 = 0 ; 𝑎𝑏𝑜𝑢𝑡 𝑡𝑕𝑒 𝑦 − 𝑎𝑥𝑖𝑠

𝟑. 𝑥 2 𝑦 = 4, 𝑥 = 1, 𝑥 = 4, 𝑦 = 0 𝑎𝑏𝑜𝑢𝑡 𝑥 − 𝑎𝑥𝑖𝑠 𝑏

3−𝑦

𝑀𝑥𝑧 =

𝑌𝑐𝑑𝑣 𝑎

𝑦

𝑦2 = 𝑥 𝑑𝑥

4

𝑦 2𝜋 𝑥𝑦𝑑𝑥 = 2 2

= 2𝜋 1 4

=𝜋 1

= 16𝜋 𝑀𝑥𝑧 =

𝑌𝑐 𝑑𝑉

;

1

=

𝑥 𝑦 0 0 9 3

4

𝑦 𝑥𝑑𝑥 = 𝜋 1

1

𝑑𝑥 𝑥 −2 = 16𝜋 𝑥3 −2

4

1

2

4 𝑥2

2

16 𝑥 4 𝑥𝑑𝑥 = 16𝜋 𝑥 4

𝑀𝑥𝑧 𝑦= 𝑉

4

𝑥𝑑𝑥

𝑥 𝑥𝑑𝑥 𝑥4

4

−8𝜋 𝑥2

= 1

4 1

15𝜋 2 𝑏

𝑙 2 𝑑𝑥

𝑉=𝜋 9

𝑉 = 2𝜋

𝑥𝑦𝑑𝑥 = 2𝜋 0

𝑌𝑐 𝑑𝑉 = 2𝜋

𝑀𝑥𝑧 = 381.70

4

𝑥 3 − 𝑦 𝑑𝑥 0

9

𝑀𝑥𝑧 =

𝑎

9

0

3+𝑦 2

4 𝑥2

=𝜋 1

𝑥

𝑥 𝑑𝑥

𝑥 1

𝑀𝑥𝑧 381.70 𝑦= = 𝑉 152.68 𝑦 = 2.5 𝟎, 𝟐. 𝟓, 𝟎

𝑑𝑥 = 𝜋 1

4

= 16𝜋

𝑉=

21𝜋 4

𝑦=

𝑀𝑥𝑧 𝑉

= 0,

10 , 7

=

4

2

−4

16 𝑑𝑥 = 16𝜋 𝑥4

𝑥 −3 𝑑𝑥 = 16𝜋 −3

4

= − 1

4

1

𝑑𝑥 𝑥2

16𝜋 3𝑥 3

4 1

15𝜋 2 21𝜋 4

0


94

EXERCISE 13.5

𝟕. 𝑥 2 = 4𝑦 x 0 1 2 4

,

4

4𝑥

2

0 4

=𝜋 0

=

𝑥2 − 4

y 0 1 2 4

2

𝑑𝑥

𝑥4 4𝑥 − 𝑑𝑥 16

96𝜋 5 4

𝑀𝑥𝑧 = 2𝜋 0

=

𝑦 2 = 4𝑥 𝑎𝑏𝑜𝑢𝑡 𝑥 − 𝑎𝑥𝑖𝑠 x 0 1/4 1 4

y 0 ¼ 1 4

=𝜋


4𝑥 + 2

𝑥2 4

𝑥

4𝑥 −

𝑥2 𝑑𝑥 4

128𝜋 3

𝑀𝑥𝑧 𝑉

=𝑦=

𝒚 = 𝟎,

128 𝜋 3 96𝜋 5

𝟐𝟎 ,𝟎 𝟗


95

EXERCISE 13.5


𝟏𝟏. 𝑦 2 = 4𝑥, 𝑦 = 𝑥 𝑎𝑏𝑜𝑢𝑡 𝑥 = 0

X 0 1/4 1 4

Y 0 1 2 4

X 1 2 3 4

Y 1 2 3 4

4

𝑉 = 2𝜋

𝑋( 4𝑋 − 𝑋)𝑑𝑥 0

𝑉 = 26.80829731 𝑐𝑢. 𝑢𝑛𝑖𝑡𝑠 𝑀𝑥𝑧 = 2𝜋

𝑦𝑐 𝑥𝑑𝑥 4

𝑀𝑥𝑧 = 2𝜋 ( 0

4𝑥 + 𝑥 )𝑥 2

4𝑥 − 𝑥 𝑑𝑥

𝑀𝑥𝑧 = 64𝜋/3 𝑦=

𝑀𝑥𝑧 𝑉

= 2.5

y=(0, 2.5, 0)


96

EXERCISE 13.6

MOMENT OF INERTIA OF A PLANE AREA

1. 2𝑥 + 𝑦 = 6 , 𝑥 = 0 , 𝑦 = 0 ; 𝑎𝑏𝑜𝑢𝑡 𝑥 − 𝑎𝑥𝑖𝑠

5. 𝑥 = 2 𝑦 , 𝑥 = 0, 𝑦 = 4

dy

x

y

4-y

(4,4)

dx 𝑥 0 3

𝑦 6 0

𝐼𝑥 =

6 2 𝑦 𝑥𝑑𝑦 0

= 6

6 2 6−𝑦 𝑦 0 2

𝑥 0 4

𝑑𝑦

1 6𝑦 2 − 𝑦 3 𝑑𝑦 2 0 1 𝑦4 = 2𝑦 3 − 2 4

𝑦 0 4

=

1 = 2 6 2

6 3 − 4

4

𝐼𝑦 =

0 4

4

𝑟 2 𝑑𝐴

=

𝑥 2 4 − 𝑦 𝑑𝑥

0

= 𝟓𝟒

=

3

3. 𝑦 = 𝑥 , 𝑥 = 8 , 𝑦 = 0 ; 𝑤𝑖𝑡𝑕 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑦 = 0

4 2 𝑥 0

𝑰𝒚 =

𝑥2

4−2

𝑑𝑥

𝟓𝟏𝟐 𝟏𝟓

dy x

𝐼𝑥 = =

2 2 𝑦 𝑥𝑑𝑦 0 2 5

𝑦 𝑑𝑦

0

𝑰𝒙 =

= =

2 2 3 𝑦 (𝑦 )𝑑𝑦 0 6 6

𝑦 6

=

2 6

𝟑𝟐 𝟑 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy

97

EXERCISE 13.6


7. 𝑦 2 = 8𝑥 , 𝑦 = 2𝑥

9. 𝑦 = 4𝑥 2 , 𝑦 = 4𝑥 ; 𝑤𝑖𝑡𝑕 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑦 − 𝑎𝑥𝑖𝑠 𝑦 = 4𝑥 2 𝑦 = 4𝑥

X1 X2

(1,4) dy

dx

y (0,0)

𝑥 0 1

𝑥 𝑦 𝑥 𝑦 0 0 0 0 1 2 2 1 2 2 4 2 4

𝑦 0 4

𝑏

𝐼𝑦 = 4

𝐼𝑥 =

0

𝑦 2 (𝑥𝑟 − 𝑥𝑙 ) 𝑑𝑦

4

𝐼𝑥 = 𝐼𝑥 = 𝑰𝒙 =

0

2

𝑦 𝑦 𝑦 2 ( − ) 𝑑𝑦 2 8

4 𝑦3 ( 0 2

−

𝑥 𝑦 0 0 1 4

𝐼𝑦 = Iy =

𝑎

𝑥 2 (𝑦𝑢 − 𝑦𝑙 ) 𝑑𝑥

1 2 𝑥 (4𝑥 0

− 4𝑥 2 ) 𝑑𝑥

1 5

𝑦4 ) 𝑑𝑦 8

𝟑𝟐 𝟓


98

EXERCISE 13.6


11. 𝑦 2 = 8𝑥 , 𝑥 = 0 , 𝑦 = 4 , with respect to 𝑦 = 4

13. 𝑦 = 𝑥 , 𝑦 = 2𝑥 , 𝑥 + 𝑦 = 6, 𝑤𝑖𝑡𝑕 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑥 = 0 𝑥+𝑦 = 6

(6 − 𝑥 − 2𝑥) 6 − 3𝑥 =

4

𝐼𝑥 = =

1 8

4−𝑦

2

0 4

𝑦2 𝑑𝑦 8

16𝑦 2 − 8𝑦 3 + 𝑦 4 𝑑𝑦

0

=

1 16𝑦3 8 3

=

𝟔𝟒 𝟏𝟓

− 2𝑦 4 +

4 𝑦5 5 0

𝑥 𝑦 0 0 1 1 2 2

𝑦 = 2𝑥

𝑥 𝑦 0 0 1 2 2 4

𝑥 0 1 2

𝑦 0 5 4

𝒃

𝑰𝒚 = 𝑰𝒚 = 𝑰𝒚 =

𝒂

𝒙𝟐 𝒀𝒖 − 𝒀𝒍 𝒅𝒙

𝒃 𝟐 𝒙 𝒂

𝟔−𝒙 −

𝒙 𝟐

𝒅𝒙

𝟏𝟗 𝟐


99

EXERCISE 13.7

MOMENT OF INERTIA OF A SOLID OF REVOLUTION

1. 𝑦 = 2 𝑥 , 𝑦 = 0 , 𝑥 = 4 ;about 𝑥 = 0

3. 𝑏𝑥 + 𝑎𝑦 = 𝑎𝑏 , 𝑥 = 0 , 𝑦 = 0 ;about the y-

axis

4

𝐼𝑦 = 2𝜋 4

= 4𝜋

𝑥 3 2 𝑥 − 0 𝑑𝑥

0

𝑥

7

2 𝑑𝑥

=

0 9

2𝑥 = 4𝜋 9

= 4𝜋

= 4𝜋 =

2 4 9 1024 9

𝟒𝟎𝟗𝟔𝝅 𝟗

2

𝑎

𝐼𝑦 = 2𝜋

4

= 0

9

2

−

2 0 9

9

2

4

2𝑏𝜋 𝑎

0 𝑎

𝑥 3 𝑎 − 𝑥 4 𝑑𝑥

0

2𝑏𝜋 𝑎 𝑎

𝑎

𝑎

𝑥 3 𝑑𝑥 −

0

2𝑏𝜋 𝑥4 = 𝑎 𝑎 4

0

𝑎𝑏 − 𝑏𝑥 − 0 𝑑𝑥 𝑎

𝑥3

𝑥 4 𝑑𝑥

0 𝑎 0

=

2𝑏𝜋 𝑎5 𝑎5 − 𝑎 4 5

=

2𝑏𝜋 𝑎

=

𝒂𝟒 𝒃𝝅 𝟏𝟎

𝑥5 − 5

𝑎 09

𝑎5 20


100

EXERCISE 13.7


5. 2𝑥 + 3𝑦 = 6 , 𝑥 = 0 , 𝑦 = 0 ; about the x-

9. 𝑥𝑦 = 4 , 𝑦 = 𝑥 , 𝑦 = 1 ; about𝑦 = 0

axis

𝐼𝑥 =

3

𝜋 2

4

6 − 2𝑥 3

0

− 0 𝑑𝑥 2

3 16𝑥 4 −192𝑥 3 +864𝑥 2 −1728𝑥+1296

=

𝜋 2 0

=

𝟐𝟒𝝅 𝟓

81

𝑑𝑥

𝐼𝑥 = 2𝜋 2

= 2𝜋

𝑦3

1

4 − 𝑦 𝑑𝑦 𝑦

4𝑦 2 − 𝑦 4 𝑑𝑦

1

7. 𝑦 2 = 3𝑥 , 𝑦 = 𝑥 ; about𝑥 = 0

𝑦3 = 2𝜋 4 3 = 2𝜋 =

3

𝐼𝑦 = 2𝜋

28 3

−

2

𝑦5 − 5 1

2 1

31 5

𝟗𝟒𝝅 𝟏𝟓

𝑥 3 𝑥 3 − 𝑥 𝑑𝑥

0

3

= 2𝜋

𝑥

7

3 − 𝑥 4 𝑑𝑥

2

0 3

= 2𝜋

3

𝑥

7

0

= 2𝜋 54 − =

3 2 𝑑𝑥

−

𝑥 4 𝑑𝑥

0

243 5

𝟓𝟒𝝅 𝟓


101

EXERCISE 13.7


11. 𝑦 = 𝑥 2 , 𝑦 = 2𝑥 ;about the y-axis

2

𝐼𝑦 = 2𝜋 2

= 2𝜋

𝑥 3 2𝑥 − 𝑥 2 𝑑𝑥

0 1

2𝑥 4 − 𝑥 5 𝑑𝑥

𝐼𝑦 = 2𝜋

0 2

= 2𝜋 2

𝑥 4 𝑑𝑥 −

0

=

64 5

−

2 0

𝑥5 = 2𝜋 2 5 = 2𝜋

13. 𝑦 = 𝑥 3 , 𝑥 = 1 , 𝑦 = 0 ; about 𝑥 = −1

2

𝑥6 − 6 0

32 3

2

𝑥 5 𝑑𝑥

1

= 2𝜋

𝑥+1

3

𝑥 3 − 0 𝑑𝑥

0

𝑥 6 + 3𝑥 5 + 3𝑥 4 + 𝑥 3 𝑑𝑥

0

= 2𝜋

𝑥7 7

0

=

+

𝑥6 2

+

3𝑥 5 5

+

1 𝑥4 4 0

𝟐𝟎𝟗𝝅 𝟕𝟎

𝟔𝟒𝝅 𝟏𝟓


102

EXERCISE 13.7


15. 𝑦 = 2𝑥 , 𝑥 = 1 , 𝑦 = 0 ; about 𝑥 = 2

1

𝐼𝑦 = 2𝜋 1

= 4𝜋

2−𝑥

3

2𝑥 𝑑𝑥

0

8𝑥 − 12𝑥 2 + 6𝑥 3 − 𝑥 4

0

= 4𝜋 4𝑥 2 − 4𝑥 3 + =

3𝑥 4 2

−

1 𝑥5 5 0

𝟐𝟔𝝅 𝟓


103

262659041 Math 29 Problem Set Compilation FIXED

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