-0- xfsin xdx is equal to

Let I=∫_0^π xf(sin x)dx————(1) ⇒ I=∫_0^π(π x)f(sin(π x))dx ⇒ I=∫_0^π(π x)f(sin x)dx———— (2) Adding (1) and (2) 2I=∫_0^π π f(sin x)dx =2π∫_0^π/2f(sin x)dx ⇒ I= ...

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Byju's Answer

Standard XII

Mathematics

Inequalities of Integrals

∫0π xfsin xdx...

Question

$\int_{0}^{π} x f (s i n x) d x$ is equal to

π \int_{0}^{π} f (s i n x) d x

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\frac{π}{2} \int_{0}^{\frac{π}{2}} f (s i n x) d x

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π \int_{0}^{\frac{π}{2}} f (c o s x) d x

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π \int_{0}^{x} f (c o s x) d x

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Solution

The correct option is C $π \int_{0}^{\frac{π}{2}} f (c o s x) d x$
$L e t I = \int_{0}^{π} x f (s i n x) d x - - - - - - - - - - - - (1)$
$\Rightarrow I = \int_{0}^{π} (π - x) f (s i n (π - x)) d x$
$\Rightarrow I = \int_{0}^{π} (π - x) f (s i n x) d x - - - - - - - - - - - - - (2)$
$A d d i n g (1) a n d (2)$
$2 I = \int_{0}^{π} π f (s i n x) d x$
$= 2 π \int_{0}^{\frac{π}{2}} f (s i n x) d x$
$\Rightarrow I = π \int_{0}^{\frac{π}{2}} f [s i n (\frac{π}{2} - x)] d x$
$= π \int_{0}^{\frac{π}{2}} f (c o s x) d x$

Suggest Corrections

∫π0 xf(sin x)dx is equal to

$\int_{0}^{π} x f (s i n x) d x$ is equal to