MATHS
Definite integrals Newton-Leibnitz formula. d Let dx (F(x)) = f(x) x (a, b). Then
b
f (x) dx =
lim F(x) – lim F(x). x a
x b
a
b
Note : 1.
If a > b, then
f (x) dx =
lim F(x) – lim F(x). x a
x b
a
b
2.
If F(x) is continuous at a and b, then
f (x) dx = F(b) – F(a) a
2
Example # 1
Evaluate
1
Solution
1 1 1 = – ( x 1)( x 2) x 1 x2
2
1
dx ( x 1)( x 2)
dx ( x 1)( x 2) =
(by partial fractions)
n( x 1) n( x 2)12
9 = n3 – n4 – n2 + n3 = n 8
Self practice problems : Evaluate the following 2
(1)
1
5x 2 x 2 4x 3
4
(2)
2 sec
2
dx
x x 3 2 dx
0
3
(3)
x
1 sec x
dx
0
Answers :
(1)
5–
5 9 n 5 n 3 4 2 2
(2)
4 + +2 1024 2
"manishkumarphysics.in"
1
MATHS 2 – + 2 n 3 3 18
(3)
b
Property (1)
b
f (t) dt
f ( x ) dx =
a
a
i.e.
Property (2)
definite integral is independent of variable of integration.
b
a
a
b
f (x ) dx = – f (x ) dx b
Property (3)
c
f ( x ) dx =
a
Example # 2
b
f ( x ) dx +
a
f (x ) dx, where c may lie inside or outside the interval [a, b]. c
x3 : x 3 If f(x) = 2 , then find 3 x 1 : x 3 5
Solution
2 3
3
f ( x ) dx =
2
5
f (x ) dx. 2
5
f ( x ) dx +
2
f (x ) dx 3
3
x2 = ( x 3 ) dx + (3 x 1) dx = 2 3 x + x 3 x 2 2 3 3
5
=
2
94 + 3 (3 – 2) + 53 – 33 + 5 – 3 2
=
5 3
211 2
8
Example # 3
Evaluate | x 5 | dx. 2
5
8
Solution
| x 5 | dx =
8
( x 5) dx +
2
2
( x – 5) dx
=9
5
2
5
2
0
0
5
(2x 1) dx = (2x 1) + (2x 1)
Example # 4
Show that
Solution
L.H.S. = x2 + x ]20 = 4 + 2 = 6 R.H.S. = 25 + 5 – 0 + (4 + 2) – (25 + 5) = 6 L.H.S. = R.H.S
Self practice problems : Evaluate the following 2
(4)
| x
2
2x 3 | dx.
0
"manishkumarphysics.in"
2
MATHS 3
(5)
[x] dx , where [x] is integral part of x. 0
t dt. 9
(6)
0
Answers :
(4)
4
Evaluate
e x ex
1
e x ex 1 ex
1 1
=
0
1
dx =
0
13
if f ( x ) f ( x ) i.e. f ( x ) is even if f ( x ) f ( x ) i.e. f ( x ) is odd
dx
1 ex
1
Solution
(6)
1
Example # 5
3
a 2 f x dx , f ( x ) dx = ( f ( x ) f ( x )) dx = 0 0, 0 a a
a
Property (4)
(5)
e x e x ex e x 1 ex 1 ex
e x e x e x (e x e x ) x 1 ex dx = e 1
dx
1
(e x e x ) dx = e – 1 +
0
e2 1 (e 1 1) = e 1
2
Example # 6
Evaluate
cos x dx.
Solution
2
2
2
2
0
cos x dx = 2 cos x 1
Example # 7
Evaluate
log
1
Solution i.e.
( cos x is even function)
2x dx. 2 x
e
2x Let f(x) = loge 2x 2x 2x = – log = – f(x) f(–x) = loge e 2x 2x f(x) is odd function 1
dx = 2
log
1
2x dx = 0 2 x
e
Self practice problems : Evaluate the following
"manishkumarphysics.in"
3
MATHS 1
(7)
| x | dx.
1
2
(8)
sin
2 2
(9)
2
7
x dx.
cos x
Answers :
Property (5)
Further
dx.
1 ex
(7)
1
b
b
a
a
a
a
0
0
Prove that 2
Let =
0
g (sin x ) dx = g (sin x ) g (cos x )
1
=
0
2
0
g (cos x ) dx = . 4 g (sin x ) g (cos x )
g (sin x ) dx g (sin x ) g (cos x )
g sin x 2 = g sin x g cos x 2 2
2
(9)
f (x ) dx = f (a x) dx
0
Solution
0
f (x ) dx = f (a b x ) dx
2
Example # 8
(8)
2
g (cosx)
g (cosx) g (sinx)
dx
0
on adding, we obtain 2
2 =
g (sin x )
g (cos x )
g (sin x ) g (cos x ) g (cos x) g (sin x) dx 0
2
=
dx
0
=
4
Self practice problems: Evaluate the following
(10)
x
1 sin x
dx.
0
"manishkumarphysics.in"
4
MATHS 2
(11)
x
sin x cos x
dx.
0
2
(12)
x sin x cos x
sin
4
0
3
(13)
1
dx tan x
6
Answers :
(10)
(11)
loge 1 2
2 2
(12)
2 16
(13)
12
a 2 f x dx , if f (2a x) f (x) = ( f ( x ) f (2a x )) dx = 0 0 0, if f (2a x) f (x)
2a
Property (6)
dx.
x cos 4 x
a
f (x) dx
0
Example # 9
Evaluate
sin
3
x cos 3 x dx.
0
Solution
Let f(x) = sin3x cos3x
f( – x) = – f(x)
sin
3
x cos 3 x dx = 0
0
Example # 10 Evaluate
dx
1 2 sin
2
0
Solution
1 2 sin2 x f( – x) = f(x)
1 2 sin
sec 2 x dx 2
x
2 =
3
2
x
=2
0
sec 2 x dx 1 tan2 x 2 tan2 x
tan 3 tan x 1
2 0
is undefined, we take limit 2 1 1 Lt tan 3 tan x tan 3 tan 0 3 x 2
2 =
x
2
dx
1 2 sin
=2
1 3 tan 0
tan
2
0
2
=2
2
dx
0
dx.
1
Let f(x) =
x
"manishkumarphysics.in"
5
MATHS = 3 2
2 =
3
Alternatively :
dx
1 2 sin
2
0
x
=
cos ec 2 x
cos ec 2 x 2
0
dx =
0
cos ec 2 x dx cot 2 x 3
Observe that we are not converting in terms of tan x as it is not continuous in (0, )
1 cot x 1 = – tan 3 3 3 0
1 =–
cot x 1 cot x Lt tan 1 Lt tan x 3 x 0 3
2 2 = 3 3
1 =–
2
n sin x
Example # 11 Prove that
dx =
0
2
2
0
0
n cos x dx = n (sin 2x) dx = –
n 2 . 2
2
Solution
Let =
n sin x dx
..........(i)
0
=
2
dx
n sin 2 x
(by property P – 5)
0
2
=
n (cos x ) dx
..........(ii)
0
Adding (i) and (ii) 2
2
0
0
sin 2x dx 2
n (sin x . cos x ) dx = n
2=
2
2
0
0
n (sin 2x ) dx – n 2 dx
2=
2 = 1 –
n 2 2
..........(iii)
2
where
=
n (sin 2x ) dx 0
put 2x = t
dx =
L.L:x=0 U.L:x= 2
t=0
t=
1 dt 2
1 =
n (sin t ) · 0
=
1 ×2 2
1 dt 2
2
n (sin t ) dt
(by using property P – 6)
0
"manishkumarphysics.in"
6
MATHS
1 =
(iii) gives = –
n 2 2
Self practice problems : Evaluate the following
1 n x x 1 x 2 dx.
(14)
0
1
(15)
sin 1 x dx. x
0
(16)
xn sin x
dx.
0
Answers :
Property (7)
n 2 (16) 2
n 2 (15)
(14)
–
2 n 2 2
If f(x) is a periodic function with period T, then T
nT
(i)
f (x ) dx, n z
f ( x ) dx = n
0
0
a nT
(ii)
T
f ( x ) dx = n
a
f (x ) dx, n z, a R 0
T
nT
(iii)
f ( x ) dx = (n – m)
f (x ) dx, m, n z 0
mT
a nT
a
nT
0
b nT
b
a nT
a
f ( x ) dx = f (x ) dx, n z, a R
(iv)
(v)
f ( x ) dx = f (x ) dx, n z, a, b R 2
Example # 12 Evaluate
e
{x}
dx.
1
1 3
2
Solution
e { x } dx =
1
1
n v
Example # 13 Evaluate
| cos x | dx , 0
1
e { x } dx = 3 e { x } dx = 3 0
1
e
{x}
dx = 3(e – 1)
0
< v < and n z. 2
"manishkumarphysics.in"
7
MATHS n v
Solution
0
| cos x | dx +
0
2
=
n v
v
| cos x | dx =
v
v
cos x
cos x
–
| cos x | dx
dx + n
/2
0
| cos x | dx
0
2
= (1 – 0) – (sin v – 1) + 2n
cos x
dx
0
= 2 – sin v + 2n (1 – 0) = 2n + 2 – sin v
Self practice problems : Evaluate the following 2
(17)
e
{3 x }
dx.
1
2000
(18)
0
dx 1 e sin x
5 4
(19)
Answers :
Property (8)
dx.
sin 2x sin x cos 4 x 4
(17)
3 (e – 1)
If (x) f(x) (x)
(18)
for
b
b
dx.
( x ) dx
(19)
4
a x b, then
b
f ( x ) dx
a
a
1000
(x) dx a
b
Property (9)
If m f(x) M for a x b, then m (b – a)
f (x ) dx M (b – a) a
b
Further if f(x) is monotonically decreasing in (a, b), then f(b) (b – a) <
f (x ) dx < f(a) (b – a) and if f(x) a
b
is monotonically increasing in (a, b), then f(a) (b – a) <
f (x ) dx < f(b) (b – a) a
b
Property (10)
a
b
f ( x ) dx
f ( x ) dx
a
"manishkumarphysics.in"
8
MATHS b
Property (11) If f(x) 0 on [a, b], then
f (x) dx 0 a
Example # 14 For x (0, 1) arrange f1(x) = 1
order and hence prove that < 6
Solution
4 x2
, f2(x) =
4x x 2
0
1 4 2x 2
1 and f3(x) =
4 x2 x3
in ascending
dx
0 < x3 < x2 2 4 – 2x < 4 –x2 – x3 < 4 – x2
3
<
4 2
.
x2 < x2 + x3 < 2x2
–2x2 < – x2 – x3 < –x2
4 2x 2 < 4 x 2 x 3 < 4 x 2 f1(x) < f3(x) < f2(x) for x (0, 1) 1
1
1
0
sin–1 < 6
1
f1( x ) dx <
f3 ( x ) dx <
0
x 2
0
1
dx
< 0
4x x 2
3
Example # 15 Estimate the value of
<
0
f(x) =
<
2
1
sin–1
x 2 0
dx
Let f(x) =
1
4 x2 x3
0
2
Solution
2
0
1
1
f (x) dx
4 2
sin x dx. x
sin x x
x cos x sin x
=
(cos x )( x tan x )
<0 x x2 f(x) is monotonically decreasing function. f(0) is not defined, so we evaluate 2
Lt sin x = 1. Take f(0) = Lt f(x) = 1 f(x) = x 0 x 0 x
Lt
x 0
2 f = 2 2 . 0 < 2 2
1<
0
2
0
sin x dx < 1 . 0 2 x
sin x dx < 2 x
1
Example # 16 Estimate the value of
e
x2
dx using (i) rectangle, (ii) triangle.
0
Solution
(i)
By using rectangle
"manishkumarphysics.in"
9
MATHS 1
Area OAED <
e
x2
dx < Area OABC
0
1
1<
e
x2
dx < 1 . e
0
1
1<
2
e x dx < e
0
(ii) By using triangle 1
Area OAED <
e
x2
dx < Area OAED + Area of triangle DEB
0
1
1
x2
1 1 < e dx < 1 + . 1. (e – 1) 2 0
1<
0
1
Example # 17 Estimate the value of
e
e
x2
dx < e 1 2
1
x2
dx by using
0
e dx . x
0
For x (0, 1), e x < ex 2
Solution
1
1×1<
1
2
e x dx <
0
e dx x
0
1
1<
e
x2
dx < e – 1
0
Self practice problems : 1
(20)
Prove the following :
1
e x cos 2 x dx <
0
Prove the following : 0 <
sin
n 1
2
x dx <
0
(22)
Prove the following : e
2
0
2
(21)
e x cos 2 x dx
1 4
sin
2
x dx , n > 1
0
1
<
e
x2 x
dx < 1
0
"manishkumarphysics.in"
10
MATHS 1
(23)
1 Prove the following : – 2
x 3 cos x 2 x2
dx <
sin x dx <
2
0
2
(24)
Prove the following : 1 <
0
2
(25)
Prove the following : 0 <
x dx
16 x
3
0
<
1 2
1 6
h( x )
Leibnitz Theorem :
If F(x) =
f (t) dt , then
g( x )
dF( x ) = h(x) f(h(x)) – g(x) f(g(x)) dx
Proof :
Let P(t) =
f (t) dt h( x )
F(x) =
f (t) dt = P(h(x)) – P(g(x))
g( x )
dF( x ) = P(h(x)) h(x) – P(g(x)) g(x) dx = f(h(x)) h(x) – f(g(x)) g(x)
x2
Example # 18 If F(x) =
sin t dt , then find F(x).
x
Solution
F(x) = 2x .
e3 x
Example # 19 If F(x) =
e2 x
sin x 2 – 1 .
sin x
t dt , then find first and second derivative of F(x) with respect to n x log e t
at x = n 2. Solution
2x e3x dF( x ) dx dF( x ) 3x 2x e 3 . e . 2 . e x = e6x – e4x. d n x = dx d ( n x ) = n e 3 x n e 2x
d 2F( x ) d ( n x ) 2
d d 1 = d (n x ) (e6x – e4x) = (e6x – e4x) × = (6 e6x – 4 e4x) x dx dnx / dx
First derivative of F(x) at x = n 2 Second derivative of F(x) at x = n 2
(i.e. ex = 2) is 26 – 24 = 48 (i.e. ex = 2) is (6 . 26 – 4 . 24) . n 2 = 5 . 26 . n 2.
"manishkumarphysics.in"
11
MATHS x 2 e t dt 0
2
Example # 20 Evaluate xLt
x
e
2t 2
.
dt
0
x 2 e t dt Lt 0
2
Solution
x
form
x
e
2t 2
dt
0
Applying L Hospital rule x
2
2 . e t dt . e x
Lt
=
2
0
x
1 . e 2x
2
x
2
2 . e t dt
Lt
=
0
x
x
Example # 21 If f(x) =
log ex
ex
f(x) =
n x
=
x
2 . ex
2
2x . e x
2
=0
dt x t , then find f(x). x
Solution
=
2
Lt
1
x t 2
x
1 1 1 1 = + – x n x ( x t ) n x 2x x x n x
dt + 1 . 1 – 1 2x x
1 1 1 1 – x n x + – x x n x 2x 2x
=
x 1 n x 1 1 – x x n x = x x n x x
x
dt n ( x t ) Alternatively : f(x) = (treating ‘t’ as constant) xt n x n x x
f(x) = n 2x – n (x + n x) f(x) =
1 n x 1 1 1 – x n x 1 = x x n x x x 1
Example # 22 Evaluate
0
xb 1 , ‘b’ being parameter.. nx 1
Solution
Let (b) =
0
d (b ) = db
1
0
xb 1 nx dx
x b nx dx + 0 – 0 nx
"manishkumarphysics.in"
12
MATHS (using modified Leibnitz Theorem) 1
=
0
x b 1 x dx = b 1
1
b
= 0
1 b 1
(b) = n (b + 1) + c b=0 (0) = 0 c=0 (b) = n (b+1) 1
tan 1(ax )
Example # 23 Evaluate
1
Let (a) =
Solution
0
1
tan 1(ax ) x 1 x2
dx 1
1
x
d (a) = da
dx , ‘a’ being parameter..
x 1 x2
0
(1 a 2 x 2 ) x 1 x 2
0
dx =
0
Put x = sin t L.L. : x = 0
dx = cos t dt t=0
U.L. : x = 1
t=
2
d (a) = da
1 a
=
1 cos t dt = cos t sin t 2
2
0
1
=
1 a2
(a) =
But (0) = 0
(a) =
.
(1 a x ) 1 x 2
2
1 a 0
) tan 2 t
=
1 a2
dt 2
sin 2 t
1
sec 2 t dt
1 (1 a
dx 2 2
2
1 2
0
2
–1
tan
1 a 2 tan t 2 0
2
2 n a 1 a + c 2 c=0 2 n a 1 a 2
Self Practice Problems : x3
(26)
If f(x) =
cos t dt, find f(x).
0
x
(27)
g(x)
If f(x) = e
and g(x) =
t
1 t
4
dt, then find the value of f(2).
2
y
(28)
If x =
0
dt 1 4t 2
and
d2 y dx 2
= Ry, then find R
"manishkumarphysics.in"
13
MATHS x2
(29)
If f(x) =
x
2
sin t dt , then find f(x).
x
x
(30)
If (x) = cos x –
( x t ) (t) dt, then find the value of (x) + (x). 0
(n t ) x
(31)
Find the value of the function f(x) = 1 + x +
2
2n t dt, where f(x) vanishes.
1
x2
(32)
Evaluate xLt 0
cos t
2
dt
0
x sin x
.
(33)
Evaluate
n (1 b cos x) dx, ‘b’ being parameter.. 0
Answers :
2 17
(26)
3x2
(29)
x2 (2x sin x2 – sin x) + (cos x – cos x2) 2x
(31)
1+
cos x 3
2 e
(27)
(32)
1
(28)
(33)
4
(30)
– cos x
1 1 b2 n 2
Definite Integral as a limit of sum Let f(x) be a continuous real valued function defined on the closed interval [a, b] which is divided into n parts as shown in figure.
The point of division on x-axis are a, a + h, a + 2h ..........a + (n – 1)h, a + nh, where
ba = h. n
Let Sn denotes the area of these n rectangles. Then, Sn = hf(a) + hf(a + h) + hf(a + 2h) + ........+hf(a + (n – 1)h) Clearly, Sn is area very close to the area of the region bounded by curve y = f(x), x–axis and the ordinates
"manishkumarphysics.in"
14
MATHS x = a, x = b. b
Hence
f (x) dx =
Lt
Sn
n
a
n 1
n 1
b
h f (a rh) =
f ( x ) dx = Lt
n
r 0
a
Lt
n
ba (b a ) r a n f n
r0
Note : 1. We can also write n
b
Sn = hf(a + h) + hf (a + 2h) + .........+ hf(a + nh) and
f ( x ) dx = Lt
a
n 1
1
2.
If a = 0, b = 1,
f ( x ) dx = n Lt
0
ba ba a r n f n
n r 1
1 r
n f n
r 0
Steps to express the limit of sum as definte integral : Step 1. Replace
r 1 by by x, by dx and n Lt n n
r Step 2. Evaluate n Lt n by putting least and greatest values of r as lower and upper limits respectively.. pn
For example n Lt
r 1
1 r f n n =
p
f (x) dx 0
r r ( n Lt = 0, n Lt = p) n n r 1 r np
Example # 24 : Evaluate
Lt
1 1 1 1 1 n 2 n 3 n ......... 2n
Lt
1 1 1 1 1 n 2 n 3 n ......... 2n
n
Solution
n n
= n Lt
r n r 1 n
= n Lt
1
r 1
1 1 = n r 1 n
1
0
dx 1 x 1 = n ( x 1)0 = n 2
n2 n3 3 n 1 2 2 ......... Example # 25 : Evaluate n Lt . 2 2 2 2 5 n n 2 n 3 n 1
"manishkumarphysics.in"
15
MATHS 2n
n
Lt
Solution
n
r 1
2n
nr 2
Lt r2 = n
r 1
r n 2 r 1 n 1
1 n
Lt
r = 0, when r = 1, lower limit = 0 n
Lt
r = n
n
and
n 2
2
1 x
1 x
Lt
n
2
dx =
0
2n = 2, when r = 2n, upper limit = 2 n
1
1 x
2
dx +
0
1 2
2
2x
1 x
2
dx
0
2
–1
1 2 + 2 log e (1 x ) 0
2
= [ tan x ]
= tan–1 2 +
0
1 n 5 2
Example # 26 : Evaluate 1
Lt
n
n! n n . n 1
Solution
n! n Let y = n Lt n n
n! 1 n y = n Lt n n n n 1 . 2 . 3 ........n 1 = n Lt n nn n
1 n 1 n 2 n 3 ..... n n = n Lt n n n n n
=
Lt
n
1
=
0
1 n
n
r
n n r 1
1
nx dx x n x x 0
= (0 – 1) – Lt x n x + 0 x 0
= – 1 – 0 = –1
y=
1 e
Self Practice Problems : Evaluate the following limits
"manishkumarphysics.in"
16
MATHS Lt
1 1 1 1 .... 2 2 2 2 n n n n 2n n n2
Lt
1 1 1 1 1 n 2 n 3 n ....... 5n
Lt
2 3 n 1 3 sin 2 sin 3 3 sin 3 ........ n sin3 2 4n 4n 4n 4n n
(34)
n
(35)
n
(36)
n
n 1
Lt
(37)
n
1 n r2 2
r 0
n n n n ...... 1 n3 n6 n9 n 3(n 1)
3 n
Lt
(38)
n
Answers :
(34)
2
(37)
2
2 1
(35)
(38)
n 5
(36)
2 9 2
(52 – 15)
2
Reduction formulae in definite Integrals: 2
1.
If n =
sin 0
n
n 1 x dx , then show that n = n n–2
2
Proof : n =
sin
n
x dx
0
n = sinn1 x cos x
2 0
2
+
(n 1) sin
n 2
x . cos 2 x dx
0
2
= (n – 1)
sin
n2
x . (1 sin 2 x ) dx
0
2
= (n – 1)
sin
n2
2
x dx (n 1)
0
sin
n
x dx
0
n + (n – 1) n = (n – 1) n–2 n 1 n = n n–2
"manishkumarphysics.in"
17
MATHS 2
sin
Note : 1.
2
n
x dx =
0
cos
n
x dx
0
n 1 n 3 n 5 ..... or n = 0 1 n n2 n4
2.
, =1 2 1
according as n is even or odd. 0 =
n 1 n 3 n 5 1 ........ . , if n is even 2 2 n n 2 n 4 Hence n = n 1 n 3 n 5 2 ........ . 1 , if n is odd n n 2 n 4 3
4
2.
If n =
tan
n
x dx , then show that n + n–2 =
0
4
n =
Solution
(tan x)
n2
1 n 1
. tan2x dx
0
4
=
(tan x)
n2
(sec2x – 1) dx
0
4
=
(tan x)
n2
4
sec2x dx –
0
(tan x)
n2
dx
0
(tan x )n 1 4 – n–2 = n 1 0
n =
1 – n–2 n 1
n + n–2 =
1 n 1
2
3.
If m,n =
sin 0
m
x . cosn x dx , then show that m,n =
m 1 m n m–2, n
"manishkumarphysics.in"
18
MATHS 2
Solution
m,n =
sin
m 1
x (sin x cosn x ) dx
0
sinm 1 x . cosn 1 x 2 + = n 1 0
m 1 = n 1
m 1 = n 1
2
sin
m2
2
0
cos n 1 x (m – 1) sinm–2 x cos x dx n 1
x . cosn x . cos 2 x dx
0
2
sin
m2
x . cosn x sinm x . cos n x dx
0
m 1 m 1 = – m–2,n n 1 n 1 m,n m 1 m 1 = 1 n 1 m,n n 1 m–2,n m 1 m,n = m n m–2,n
Note : 1.
m 1 m 3 m 5 ........ or according as m is even or odd. m,n = 0,n 1,n mn mn2 mn 4 2
0,n =
cos
2
n
x dx and 1,n =
0
2.
sin x . cos
n
x dx =
0
1 n 1
Walli’s Formula
m,n
(m 1) (m 3) (m 5 ) .........( n 1) (n 3) (n 5)....... when both m, n are even (m n) (m n 2) (m n 4)........ 2 = (m 1) (m 3 ) (m 5) .........( n 1) (n 3) (n 5)....... otherwise (m n) (m n 2) (m n 4)........
2
Example # 27 : Evaluate
sin
2
x cos 2 x(sin x cos x ) dx .
2
2
2
Solution
Given integral =
sin
2
3
x cos 2 x dx +
sin
2
x cos 3 x dx
2
"manishkumarphysics.in"
19
MATHS 2
=0+2
sin
2
x cos 3 x dx
( sin3x cos2x is odd and sin2x cos3x is even)
0
1. 2 4 = 2. 5 . 3 . 1 = 15
Example # 28 : Evaluate
x sin
5
x cos 6 x dx .
0
Let =
Solution
x sin
5
x cos 6 x dx
0
=
( x) sin ( x) cos 5
6
( x ) dx
0
=
sin
5
x sin
6
. cos x dx –
0
5
x . cos 6 x dx
0
2
sin
2 = . 2
5
x . cos 6 x dx
0
=
=
4 . 2 . 5 . 3 .1 11 . 9 . 7 . 5 . 3 . 1
8 693 1
Example # 29 : Evaluate
x
3
(1 x )5 dx .
0
Solution L.L U.L.
Put x = sin2 :x=0
=0
:x=1
=
x
2
2
1
dx = 2 sin cos d
3
(1 x )5 dx =
0
sin
6
(cos 2 )5 2 . sin . cos d
0
2
=2.
sin
7
cos11 d
0
6 . 4 . 2 . 10 . 8 . 6 . 4 . 2 = 2 . 18 . 16 . 14 . 12 . 10 . 8 . 6 . 4 . 2
=
1 504
"manishkumarphysics.in"
20
MATHS Self Practice Problems: Evaluate the following 2
(39)
sin
5
x dx .
5
x cos 4 x dx .
0
2
(40)
sin 0
1
(41)
x
6
sin 1 x dx .
0
a
(42)
x a2 x 2
7 2
dx .
0
2
(43)
x
3/2
2 x dx.
0
Answers :
(39)
8 15
(43)
2
(40)
8 315
(41)
16 – 245 14
"manishkumarphysics.in"
(42)
a9 9
21