The correct option is C π∫_0^π/2f(cos x)dxI=∫_0^π xf(sin x)dx=∫_0^π (π x)f [sin (π x)]dx=∫_0^π(π x)f(sin x)dx=π∫_0^π f(sin x)dx I ⇒ 2I = π∫_0^π f(sin x)dx=2 ...

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Inequalities of Integrals

∫0π× fsin x d...

Question

$\int_{0}^{π} x f (s i n x) d x =$

\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}^{π} f (c o s x) d x

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π \int_{0}^{π} f (s i n x) d x

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Solution

The correct option is B $π \int_{0}^{\frac{π}{2}} f (c o s x) d x$
$I = \int_{0}^{π} x f (s i n x) d x = \int_{0}^{π} (π - x) f [s i n (π - x)] d x = \int_{0}^{π} (π - x) f (s i n x) d x = π \int_{0}^{π} f (s i n x) d x - I$
$\Rightarrow 2 I = π \int_{0}^{π} f (s i n x) d x = 2 π \int_{0}^{\frac{π}{2}} f (s i n x) d x$
$\Rightarrow I = π \int_{0}^{\frac{π}{2}} f (s i n x) d x = π \int_{0}^{\frac{π}{2}} f [s i n (\frac{π}{2} - x)] d x = π \int_{0}^{\frac{π}{2}} f (c o s x) d x$

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∫π0 xf(sin x)dx=

The correct option is B π∫π20f(cos x)dxI=∫π0 xf(sin x)dx=∫π0(π−x)f[sin(π−x)]dx=∫π0(π−x)f(sin x)dx=π∫π0f(sin x)dx−I ⇒2I=π∫π0 f(sin x)dx=2π∫π20 f(sin x)dx ⇒I=π∫π20f(sin x)dx=π∫π20f[sin(π2−x)]dx=π∫π20f(cos x)dx

$\int_{0}^{π} x f (s i n x) d x =$