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CBSE Class 12 Math 2011 Solved Paper

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Question : 26
Total: 29
Evaluate:
π
2
∫
0
 2 sin x cos x tan−1 (sin x) dx
OR
Evaluate:
π
2
∫
0
xsinxcosx
sin4x+cos4x
 dx
Solution:  
Consider the given integral
I =
π
2
∫
0
 2 sin x cos x tan−1 (sin x) dx
Let t = sinx
⇒ dt = cos x dx
When x =
π
2
 , t = 1
When x = 0 , t = 0
Now, ∫ 2 sin x cos x tan−1 (sin x) dx
= ∫ 2t tan−1 t dt
= [tan−1t] ∫ 2t dt - ∫ [
d
dt
(tan−1t)‌∫2tdt] dt
[tan−1t] [2.
t4
2
] - ∫ (
1
1+t2
×2.
t2
2
) dt
= t2tan−1 t - ∫
t2
1+t2
 dt
= t2tan−1 t - ∫ [1−
1
1+t2
] dt
= t2tan−1 t - t + $$tan^{-1} t
∴ I =
π
2
∫
0
 2 sin x cos x tan−1 (sin x) dx
= [t2tan−1t−t+tan−1t]01
= [12tan−11−1+tan−11] - [02tan−10−0+tan−10]
= [1×
π
4
−1+
π
4
] - 0
=
π
4
−1+
π
4
=
π
2
 - 1
OR
I =
π
2
∫
0
xsinxcosx
sin4x+cos4x
 dx ... (1)
Using the property
a
∫
0
 f (x) dx =
a
∫
0
 f (a - x) dx
I =
π
2
∫
0
(
π
2
−x)sin(
π
2
−x)cos(
π
2
−x)
sin4(
π
2
−x)+cos4(
π
2
−x)
dx
⇒ I =
π
2
∫
0
(
π
2
−x)cosxsinx
cos4x+sin4x
 dx ... (2)
Adding (1) and (2)
2I =
π
2
∫
0
(
π
2
.sinxcosx)
sin4x+cos4x
 dx
⇒ I =
π
4
π
2
∫
0
|
sinxcosx
sin4x+cos4x
| dx
=
π
4
π
2
∫
0
|
sinxcosx
cos4x
sin4x
cos4x
+1
| dx
π
4
‌
π
2
∫
0
tanxsec2x
tan4x+1
 dx
Put tan2 x = z
∴ 2 tan x sec2 x dx = dz
⇒ tan x sec2 x dx =
dz
2
When x = 0 , z = 0 and when x =
π
2
 , z = ∞
∴ I =
π
4
‌
∞
∫
0
dz
2
z2+1
⇒ I =
π
8
‌
∞
∫
0
dz
1+z2
=
π
8
|tan−1(z)|0∞
=
π
8
tan−1∞−tan−10
=
π
8
(
π
2
−0)
=
π2
16
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