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CBSE Class 12 Math 2009 Solved Paper

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Question : 25
Total: 29
Evaluate:
π
∫
0
ecosx
ecosx+e−cosx
 dx
OR
Evaluate:
π
2
∫
0
 (2 log sin x - log sin 2x) dx
Solution:  
Let I =
π
∫
0
ecosx
ecosx+e−cosx
 dx
Using
a
∫
0
 f (x) =
a
∫
0
 f (a - x) dx
I =
π
∫
0
ecos(π−x)
ecos(π−x)+e−cos(π−x)
 dx
2I =
π
∫
0
e−cosx+ecosx
ecosx+e−cosx
 dx
I =
1
2
π
∫
0
 dx =
1
2
 [π - 0] =
π
2
OR
I =
π
2
∫
0
 (2 log sin x - log sin 2x) dx
I =
π
2
∫
0
 (log‌
sin2x
2sinx.cosx
.dx)
I =
π
2
∫
0
 log (
tanx
2
) . dx ... (i)
Using property
a
∫
0
 f (x) dx =
a
∫
0
 f (a - x) dx
We get,
I =
π
2
∫
0
 log (
tan(
π
2
−x)
2
) dx
⇒ I =
π
2
∫
0
 log (
cotx
2
) dx ... (ii)
Additing (i)&(ii)
2I =
π
2
∫
0
 [log(
tanx
2
)+log(
cotx
2
)] dx
⇒ 2I =
π
2
∫
0
 log [(
tanx
2
)(
cotx
2
)] dx
⇒ I =
1
2
π
2
∫
0
 log (
1
4
) dx
⇒ I =
1
2
 log (
1
4
)×(
π
2
)
⇒ I =
1
2
 log (
1
4
)
1
2
 × (
π
2
)
⇒ I = log (
1
2
)×(
π
2
)
⇒ I =
π
2
‌log‌
1
2
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