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Question : 76 of 160
Marks:
+1,
-0
Solution:
Let
I=‌‌dx.
‌+‌ As
‌ is an odd function.
So,
‌‌dx=0As
‌ is an even function.
So,
‌‌dx=2‌‌‌dx I=4‌‌‌dx......(i)
=4‌‌| (π−x)sin‌(π−x) |
| 1+cos2(π−x) |
⇒‌‌I=4‌‌.....(ii)
On adding Eqs. (i) and (ii),
2I=4π‌‌‌dx Let
cos‌x=t⇒−sin‌x‌dx=dtWhen
x→0,t→1 and
x→π,t→−1‌I=−2π‌‌‌dt‌I=2π‌‌‌dt=2π[tan−1t]−11‌=π2
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