Adding (i) and (ii), we get

21

2

2

2

Idx= ⋅ = −

−

∫

α

πα π

α

/

⇒ I=π−α

4

(B) Let I= xxdx
− +

∫

sin^2

π^1 α

π

... (i)

⇒ I= −xx dx
−∫ + −

sin (^2 )

0

0

π^1 α

π

⇒ I xdx

x

=−∫ + x

α

π α

π sin 2

1

... (ii)

Adding (i) and (ii), we get

2222

0

Ixdx xdx==∫ ∫

−

sin sin

π

π

π

⇒ Ixdx==∫∫sin sin xdx=

2 /

0

2

0

2

2

2

πππ

(C) LetI x
xx

dx

n

= nn+

−

∫

sin

sin cos

2

22

3

2

α

π α

... (i)

=

⎛⎝⎜ − ⎞⎠⎟

⎝⎛⎜ − ⎞⎠⎟+ ⎛⎝⎜ − ⎞⎠⎟

− sin

sin cos

2

22

3

2

3

2

3

2

3

2

n

nn

x

xx

dx

π

α ππ

πα

∫∫

⇒ I x
xx

dx

n

= nn+

−

∫

cos

sin cos

2

22

3

2

α

π α

... (ii)

Adding (i) and (ii), we get

213
2

3

2

Idx= ⋅ =⎜⎛⎝ −−⎞⎠⎟

−

∫

π αα

α

π α

⇒ (^) I=^3 −
4
π α
(D) Let I x
xx
= dx
− +
−
∫
tan
tan tan cot
cot
1
1
α
α
... (i)

− +
−
∫
cot
tan cot tan
cot x
xx
dx
1
1
α
α
... (ii)
Adding (i) and (ii), we get
21
1
1
Idx==^11 −
−
−
∫ −−
tan
cot
cot tan
α
α
αα
=⎝⎜⎛π− −α⎞⎠⎟− −α
2
tan^1 tan^1
or I=⎜⎛⎝π− −α⎞⎠⎟
4
tan^1

45. A  (Q), B  (P), C  (R), D  (S)

(A) We h a v e sec
(sec tan )

x

xx

dx

∫ + 2

= +

+

∫∫sec (sec tan ) =

(sec tan )

xx x

xx

dx dt

(^33) t
(Putting t = secx + tanx)
=−^1 ++−
2
(secxxCtan )^2
(B)
cos
(sin )(sin ) ( )( )
x
xx
dx dt
−− tt

∫ 12 ∫ −− 12
(Putting t = sinx)

−
−
−
⎛
⎝⎜
⎞
⎠⎟ =
−
−
∫^1 +
2
1
1
2
tt^1
dt x
x
log|sin | C
|sin |
(C) sin− sec ( tan )

• 
  

  ⎛
  ⎝⎜
  ⎞
  ⎠⎟
  ∫^122 ==∫^2
  1
  x 2
  x
  dx t tdt Puttingx t
  = 2[t tant – log|sect| + C 1 ] = 2x tan–1x – log(1 + x^2 ) + C
  (D) Putting tanx = y^2 , so that dx ydy
  y

  
  


• 
  

  2
  () 1 4
  , we have
  (tanxxdxcot ) y y y
  y
  +=⎛⎝⎜ + ⎞⎠⎟ dy

  
  


• 
  

  ∫ ∫
  (^12)
  1 4
  = +

  
  


• 
  

  = +
  ∫∫+
  2 1
  1
  2 11
  1
  2
  4
  2
  22
  y
  y
  dy y
  yy
  / dy
  /
  = +
  − +

  
  


• 
  

  (^2) ∫^11 ∫ = −
  12
  2
  2
  (^21)
  22
  /
  (/)
  y
  yy
  dy du
  u
  uy
  y
  where
  =+= 21 −−⎝⎜⎛ − ⎞⎟⎠+
  22
  2 1
  2
  tan^11 tan tan
  tan
  u C x
  x
  C

  
  

46. (5) : Let logax=t ⇒ att=x ⇒dx=alogea dt

∴⋅∫xa− axdx

a [log ]

1

=lnaa a∫∫ttt⋅⋅−[]adt=lnaa dtt

0

(^12)
0
1
=ae− = − ⇒ ae=
(^21)
2
1
2

47. (4) : Ix x=∫(π− )((sin (sin )) cos (cos ))+ xdx

π

0

22

⇒ 22 =+∫^22

0

2

Ixxdxπ

π

(sin (sin ) cos (cos ))

/

⇒ Ixxdx=+π ∫

π

(sin (sin ) cos (cos ))

/ 22

0

2

=+π ∫

π

(sin (cos ) cos (sin ))

/ 22

0

2

xxdx

⇒ 22 = ∫ ⇒ =

0 2

2 2

IdxIπ π

π/