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INTEGRATIONS AND APPLICATIONS OF DEFINITE INTEGRALS
1. (d) Put sec x2 = t 2x sec x2 tan x2 dx = dt
x sin x2 esec x2 cos2 x2
dx = 1 2
et dt = 1 et + c = 1 esec x2 + C
2
2
2. (c)
Put tan x = t 2
Then
cos
x =
1 - t2 1 + t2
,
sin
x
=
2t 1 + t2
dx =
2dt 1 + t2
2dt
dx
=
3 cos x - 4 sin x + 5
1 + t2
3
1 1
- +
t2 t2
-
4
?
1
2t + t2
+5
=
2dt 3(1 - t2) - 8t + 5(1 + t2)
=
dt t2 - 4t + 2
=
dt (t - 2)2
= -
1 t-2
+
C
=
-
1 tan x
-2
+ C
2
=
1 2 - tan x
+
c
2
3. (a) Put x = t2 dx = 2t dt
cos x dx = cos t. 2t dt = 2 t cos t dt
= 2[t sin t - sin t dt] = 2 [t sin t + cos t] +c
= 2 [ x sin x + cos x ] + c
4. (b)
sin x dx sin2 x + 4 cos2 x
=
sin x dx 1 + 3 cos2 x
Put 3 cos x = t - 3 sin x dx = dt
=
- 1 dt
3 1 + t2
=
1 cot ? 1 t + c = 3
1 cot ? 1 ( 3
3 cos x) + c =
1 3
sec x tan ? 1 3 + c
5. (c) Put ex = t, then ex dx = dt
ex dx
=
(1 + ex ) (2 + ex )
dt
=
(1 + t) (2 + t)
1 - 1 dt 1 + t 2 + t
= log (1 + t) ? log (2 + t) + c = log 1 + t + c = log 1 + ex + c
2+t
2 + ex
6. (d)
( ) tan x + cot x dx =
sin x + cos x dx = 2 sin x cos x
(sin x + cos x) dx 1 - (sin x - cos x)2
Put sin x ? cos x = t
(cos x + sin x)dx = dt
2
dt = 2 sin ? 1 t + c = 1 - t2
2 sin ? 1 (sin x ? cos x) + c
7. (b) Put 1 + x2 = t, then 2x dx = dt
x3
dx =
(1 + x2 )2
x2 x dx = (1 + x2 )2
(t - 1) dt t2 2
= 1 2
1 t
-
1 t2
dt
=
1 2
log
t+
1 t
+
c
=
1 2
log (1 + x2) +
1 2 (1 + x2 )
+ c
8. (a)
tan x sec x dx =
tan x sec x dx = 2 sec x + c sec x
9. (b)
ex
x+4 (x + 5)2
dx
=
ex
x +5-1 (x + 5)2
dx
=
ex
x
1 +
5
-
(x
1 + 5)2
dx
=
ex (f (x) + f | (x)) = ex f (x) = ex
x+5
10. (a)
4x2
dx + 12x
+
45
=
1 4
dx x2 + 3x + 45
4
=
1 4
dx x + 3 2 + 45 - 9
2 4 4
= 1 4
dx x + 3 + 32
2
= 1 4
?
1
tan ? 1
x + 3 2
=
1 tan ? 1 2x + 3 + c
3
3
12
6
11. (d)
e 2log tan sin x cos x dx =
elog tan2 sin x cos x dx
[Q e log t = t]
= tan2 (sin x) cos x dx = (sec2 (sin x) ? 1).cos x dx = (sec2 t ? 1) dt
= tant ? t = tan sin x ? sin x + c 12. (b)
ex (sin2 x + 2 sin x cos x) dx = ex (f (x) + f|(x)) = ex f(x) = ex sin2 x
13. (a)
sin3 x
+
cos3 x
=
sin s2x cos2 x sin2 x cos2 x
sin x +
cos2 x
cos x sin2 x
= (sec x tan x + cosec x cot x)dx = sec x ? cosec x + c
14. (d)
1 - sin2x =
cos2 x + sin2 x - 2 sin x cos x
=
(cos x - sin x)2dx = cos x ? sin x dx = sin x + cos x
15. (b)
ex
1+ 1+
sin cos
x x
dx
=
1 + 2 sin x cos x
ex
2 2 cos2 x
2
2 dx
=
ex 1 sec2 x + tan x dx
2
2
2
= ex tan x + c 2
This is in the form ex (f (x) + f | (x)) dx Answer is ex f(x) i.e. ex tan x/2
16. (d)
1
- cos2 cos4 x
x
dx
=
sin2 x cos4 x
dx =
tan2 x sec2 x dx = tan3 x 3
+ c (Put tan x = t)
17. (a)
log 5
0
ex ex - 1 ex + 3
dx
Put ex ? 1 = t2 ex dx = 2t dt when x = log 5, t = 2 x = 0, t = 0
log5 0
ex ex - 1 ex + 3
dx =
2 0
2t2 4 + t2
dt
=
2
2 0
t2 + 4 - 4 + t2
4
dt
=
2
2 0
= 2
t
-
2
tan- 1
t 2
2 0
= 2
2
-
2
4
= 2
4 - 2
= 4 -
18. (a)
tan x 1/e
t dt 1 + t2
+
cot x 1/e
dt t (1 + t2)
1 -
4
4 + t2
dt
Put t = 1 u
dt
=
-
1 u2
du
When t = cot x, u = tan x and t = 1 , u = e e
=
tan x 1/e
t dt 1 + t2
+
tan x e
-
1 u2
du
=
1 u
1
+
1 u2
tan x 1/e
t dt
1 + t2
-
tan x e
=
tan x 1/e
t dt 1 + t2
+
e tan x
t 1 + t2
dt
=
e 1/e
1
t + t2
dt
u 1 + u2
du
( ) = 1 log 1 + t2 2
e =1 1/e 2
log (1
+
e2)
-
log
e2 + e2
1
=
1 2
log
(1 + e2) (e2 + 1)
=
1 2
log e2 = loge e = 1
e2
19. (b)
Put loge x = t, then
dx = dt x
When x = e37, t = loge e37 = 37
X = 1, ,t = loge 1 = 0
e37 1
sin ( loge x) dx = x
37 0
sin
t
dt
=
-
cos t
37 0
= - [cos 37 - cos 0] = - [-1 ? 1] = 2
20. (c) Put xex = t, then (1 + x) ex dx = dt
ex (1 + x) sin2 (xex)
dx
=
dt sin2 t
=
cosec2 t dt = - cot t + c = - cot (xex) + c
21. (c) (1 + cos x + cos2 x + ......... to ) =
1
-
1 cos x
dx
(Q
it
is
a
G.
P)
=
1 2 sin2 x
dx =
1 2
cosec2
x 2
dx = - cot
x 2
+ c
2
22. (a)
sec x tan x 10 + tan2 x
dx
=
sec x tan x dx 9 + sec2 x
=
dt 32 + t2
(Put sec x = t, sec x tan x dx = dt)
= 1 tan ? 1 t + c = 1 tan ? 1 sec x + c
3
3
3
3
23. (d)
dx a2 cos2 x + b2 sin2 x
=
sec2 x dx a2 + b2 tan 2 x
= 1 b
dt a2 + t2
[Put b tan x = t, b sec2 x = dt]
= 1 ? 1 tan ? 1 t + c = 1 tan ? 1 b tan x + c
ba
a
ab
a
24. (b)
x2
1
+
1 x2
x2
x2
+
1 x2
dx
=
1
+
1 x2
dx
x
-
1 2
+
2
x
Put x -
1 x
=
t,
then
1 +
1 x2
dx = dt
=
dt t2 + 2
=
25. (b)
1 tan ? 1 2
t 2
+ c =
1
x - 1 tan ? 1 x + c =
2
2
1 2
tan ? 1
x2 - 1 2 x
+ c
1 + sin x dx 3
=
sin2 x + cos2 x + 2 sin x cos x
6
6
66
dx =
sin x + cos x 2 dx
6
6
= - 6 cos x + 6 sin x + c = 6 sin x - cos x + c
6
6
6
6
26. (a)
sec2 x cosec2 x dx =
cos2
1 x sin2
x
dx
=
sin2 x + cos2 x cos2 x sin2 x
dx
=
= (sec2 x + cosec2 x) dx = tan x ? cot x + c
27. (c) cosec4 x dx = (1 + cot2 x) cosec2 x dx
put cot x = t -cosec2 x dx = dt
=
(1 + t2) (-dt) = -
t
+
t3 3
+ c = - cot x -
cot3 x 3
+ c
28. (a)
1 cos2
x
+
1 sin2
x
dx
1-x 1+ x
dx =
1-x 1+ x
?
1 - x dx 1-x
=
1-x
dx =
1
-
x
dx = sin ? 1 x + 1 - x2 + c
1 - x2
1 - x2 1 - x2
29. (d)
sinh logx dx =
elog x - e- log x
2
dx =
x
-1 2
/
x
dx
=
?
x2
2
-
log
x
=
x2 - log 4
x+ c
30. (b)
Take 2x - 2 = t, 2x ?1 = 2x ? 2 + 1 = t2 + 1
2 dx = dt 2 2x - 2
dx = tdt
dx
(2x - 1) 2x - 2
=
tdt (t2 + 1)t
= tan ? 1 t = tan ? 1
2x - 2 + c
31. (a) xx loge ex dx = xx (loge e + loge x) dx = xx (1 + loge x) dx = xx + c
32. (a)
cos x cos x -
dx =
4
cos x dx 1 (cos x + sin x) 2
=
2 2
cos x - sin x + cos x + sin x dx = (cos x + sin x)
1 2
cos x
cos
x
- sin x + sin x
+
1
dx
= 1 [log (cos x + sin x) + x] + c 2
33. (a)
x -1 (x + 1)3
=
(x + 1) - 2 (x + 1)3
=
1 (1 + x)2
- 2 (x + 1)3
= f (x) + f | (x)
I=
ex [f (x) + f | (x)] dx = ex f (x) = ex 1 (1 + x)2
34. (a)
Nr = l (Dr) + m
d dx
(Dr)
Sin x + 3 cos x = l (3 sin x + 4 cos x) + m (3 cos x ? 4 sin x)
= (3l ? 4m) sin x+ (4l + 3m) cos x
3l ? 4m = 1
4l + 3m = 3
9l ? 12m = 3
16l + 12m = 12
25 l = 15
l= 3 5
m= 1 5
sin x + 3 cos x 3 sin x + 4 cos x
dx
=
3 5
x
+
1 5
log (3 sin x + 4 cos x)
A + B =
3+ 5
1 5
=
4 5
35. (d)
In + In ? 2 =
4
cotn x dx +
4
0
0
cotn ? 2 x dx =
4
0
(cotn x + cotn ? 2 x) dx
4
= 0
4
cotn ? 2 x (cot2 x + 1) dx = cotn ? 2 x . cosec2 x dx 0
=
-
cot n -1
x
4
n - 1 0
=
-
n
1 -
1
-
0
=
1 1-n
36. (d)
0
dx cos2 x + 3 sin2 x
=
0
sec2 x 1 + 3 tan2 x
sec2 x 1 + 3 tan2 x
is discontinuous at x =
2
37. (d)
2 2 + x dx = 2
0 2-x
0
2 + x ? 2 + x dx = 2
2-x 2+x
0
2 + x dx = 2
4 - x2
0
2 + 4 - x2
x 4 - x2
dx
=
2
sin- 1
x 2
-
4
-
x2
2 0
= [2 sin ? 1 1 ? 0] ? [0 -
4]=2? +2=+2 2
38. (b)
1 sin2 (cos ? 1 x) dx 0
= 1 1 ? [cos(cos ? 1 x)]2 dx = 1 1 ? x2 dx = x ? x3 1
0
0
3
0
................
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