- (a) : Put x = tanθ ⇒ dx = sec^2 θdθ
Also asxx== 00 and == 1
4,,θθπTherefore, tan sec/
−
∫∫=1
01
2
04
xdx θθθdπ=ππ− = −
42
41
2log log 2- (c) : I dx
x
=
∫ 0 2 +2
cosπ/=
++−∫
dx
2 xxxx
22
2220 22222sin cos cos sinπ/=
+=
+∫∫
dx
xxxsin cos xdxsectan//0 222 20 222 3 22(^32)
ππ
Putt=tanx ⇒dt= sec xdx,then
2
1
22
2
I dt
t
- = ⎛
⎝⎜
⎞
∫ ⎠⎟
2 −
3
2
3
1
0 2 3
1
tan^1
- (a) : Put x = atanθ ⇒ dx = asec^2 θdθ
∴ (^) I aa
a
= ∫ ⋅ d
44 2
88
0
(^4) tan sec
sec
/ θθ
θ
θ
π
==−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
∫∫⎥
11
3
42
3
0
4
46
0
4
a
d
a
sin cos (sin sin )d
//
θθθ θ θθ
ππ
= − − −
⎡
⎣
⎢
⎢
⎤
⎦
⎥
∫ ⎥
112
4
12
(^38)
23
0
4
a
(cos) (cos)d
/ θθ
θ
π
=++^1 ∫ −
8
3 121 222^2
0
4
a
( cos )( cos cos )d
/
θθθθ
π
=^1 ∫ −−+
8
3 12232 2
0
4
a
( cos cos cos )d
/
θθθθ
π
=^1 ∫ −− +
32
3 2224 6
0
4
a
( cos cos cos )d
/
θθθθ
π
= ⎡ −−+
⎣⎢
⎤
⎦⎥
1
32
2 2
2
4
2
6
(^360)
4
a
θ θθθ
sin sin sin π/
=^1 ⎛⎝⎜ − ⎞⎠⎟
16 4
1
a^33
π
- (a) : Let I xx
x
= + dx
∫ +sin cos
sin/0 916 2π 4Put sinx – cosx = t ⇒ (sinx + cosx)dx = dtI dt
tdt
t=
+ −=
−−∫∫ 1 9 16 1^2 25 16−01 20
()I dt
tx
x=
⎛
⎝⎞
⎠ −= +
−⎡
⎣⎢⎤
−∫ ⎦⎥−1(^165)
4
1
16
2
5
54
1 2 2 54
0
1
0
log
=^1 − +=
40
1191
20
[log log log ] log 3
- (b) : Let I xxdx
xx
=
∫ ++sin cos
cos cos/0 22
32πPut cosx = t ⇒ –sinxdx = dt, thenI tdt
tt tt= dt
++=^2
+−
+⎡
⎣⎢⎤
∫ 2 0 ∫ ⎦⎥101
32 21
1
=+[ log(tt) log(− +=)]^10 [ log −−log log ]22 123222=[log 98 −log ] log= ⎛⎝⎜^9 ⎟⎠⎞
8- (a) : Let Imxnxdx= (^2) ∫
0
sin sin
π
=∫[cos(mnx−−) cos(mnxdx+ ) ]
0
π
= −
−
− +
- ⎡
⎣⎢
⎤
⎦⎥
sin( )
()
sin( )
()
mnx
mn
mnx
mn 0
π
= sin( )
()
sin( )
()
mn
mn
mn
mn
−
−
− +
⎡
⎣⎢
⎤
⎦⎥
ππ= 0
Since, sin(m – n)π = 0 = sin(m + n)π for m ≠ n
- (c) :
dx
() 1 a cos xxsin a cos xxsin
22
2
220 +^2 ⎛^22 +^22
⎝⎞
⎠− −⎛
⎝⎞
⎠∫
π=
− ++∫
dx(^0) ( 1 a) cos^222 x ( 1 a) sin^222 x
π
− ++
⎧
⎨
⎩
⎫
⎬
⎭
∞
∫
2
(){()/()} 1112220 2
;tan
a
dt
aat
where t = x
−
−
⎡ ⎛⎝⎜ ⋅ ⎞⎠⎟
⎣⎢
⎤
⎦⎥
(^2) − ∞
1
1
1
1
(^21)
1
() 0
()
()
tan
a
a
a
a
a
t
−
∞− =
−
(^2) −−
1
0
(^21)
11
()aa[tan tan ]^2
π
- (b)