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Question : 157 of 180
Marks:
+1,
-0
Solution:
We have,
xf(sin‌x)‌dx=A‌f(sin‌x)‌dxLet
‌‌I=x⋅f(sin‌x)‌dx . . . (i)
⇒I=(π−x)f(sin‌(π−x))‌dx⇒‌‌I=(π−x)f(sin‌x)‌dx . . . (ii)
Adding Eqs. (i) and (ii), we get
2I=π‌f(sin‌x)‌dx⇒I=‌‌f(sin‌x)‌dx⇒‌‌I=‌×2‌f(sin‌x)‌dx⇒‌‌I=π‌f(sin‌x)‌dx∴ On comparing, we get
A=Ï€
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