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Question : 79 of 160
Marks:
+1,
-0
Solution:
Let
I=xf(sin‌x)dx ⇒‌‌I=(π−x)f[sin(π−x)]dx ⇒‌‌I=πf(sin‌x)dx−xf(sin‌x)dx ⇒‌‌2I=π‌f(sin‌x)dx⇒2I=π‌(‌)×2f(sin‌x)dx ⇒‌‌2I=2π‌f(sin‌x)dx ‌‌[∵ if
f(2a−x)=f(x), then
f(x)dx=2‌f(x)dx] ⇒‌‌I=π‌f(sin‌x)dx
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