∫_0^πx f (sin x)dx = π/2∫_0^π f(sin x)dx Since ∫_0^π xf (x)dx =1/2a ∫_0^πf(x)dx, if f(a x)=f(x) ...

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Integration of Piecewise Continuous Functions

∫0πx f sin xd...

Question

$\int_{0}^{π} x f (s i n x) d x =$

π \int_{0}^{π} f (s i n x) d x

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\frac{π}{2} \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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Solution

The correct option is B $\frac{π}{2} \int_{0}^{π} f (s i n x) d x$
$\int_{0}^{π} x f (s i n x) d x = \frac{π}{2} \int_{0}^{π} f (s i n x) d x$
$S i n c e \int_{0}^{π} x f (x) d x = \frac{1}{2} a \int_{0}^{π} f (x) d x, i f f (a - x) = f (x)$

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Similar questions

\int_{0}^{π} x f (s i n x) d x =

\int_{0}^{π} \frac{x}{a^{2} {cos}^{2} x + b^{2} {sin}^{2} x} d x

f : (0, 1] \to R

\int_{0}^{π} f (sin x) dx = 2018, then \int_{0}^{π} x f (sin x) dx

\int_{0}^{2 π} {sin}^{3} x d x = 0

f (x) = {sin}^{3} x

f (- x) = - f (x)

\int {sin}^{n} d x = - \frac{cos x {sin}^{n - 1} x}{n} + f (n) \int s i n^{n - 2} x d x

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