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$\int_{0}^{\pi}x\; f(sin\: x)\: dx\;$ is equal to

Option 1)
$\pi \int_{0}^{\pi }f(cos\, x)dx\;$

Option 2)
$\; \; \pi \int_{0}^{\pi }f(sin\, x)dx\;$

Option 3)
$\; \frac{\pi }{2}\int_{0}^{\pi /2}\! \! f(sin\, x)dx\; \;$

Option 4)
$\; \pi \int_{0}^{\pi /2}\! \! f(cos\, x)dx$

Answers (1)

best_answer

As learnt in concept

Properties of Definite Integration -

$\int_{0}^{2a}f(x)dx= \int_{0}^{a}\left [ f(x)+f(-x) \right ]dx$

$= \left\{\begin{matrix} 2\int_{0}^{a}f(x)dx & if f(2a-x)=f(x)\\ 0 &if f(2a-x)=-f(x) \end{matrix}\right.$

-

Properties of Definite integration -

$\int_{a}^{b}f\left ( x \right )dx= \int_{a}^{b}f\left ( a+b-x \right )dx$

When $\int_{0}^{b}f\left ( x \right )dx= \int_{0}^{b}f\left ( b-x \right )dx$

- wherein

Put the $\left ( a+b-x \right )$ at the place of x in $f\left ( x \right )$

$I=\int_{0}^{\pi }xf(sinx)dx$

$I=\int_{0}^{\pi }(\pi -x)f(sinx)dx$

$\left [ \int_{0}^{a} \right f(x)dx)=\int_{0}^{a}f(a-x)dx$

$2I=\int_{0}^{\pi }\pi f(sinx)dx$

$2I=\pi \int_{0}^{\frac{\pi }{2}}\left ( f(sinx) +f(sin(\pi -x))\right )dx$

$I=\pi \int_{0}^{\frac{\pi }{2}}f(sinx)dx$

Also $I=\pi \int_{0}^{\frac{\pi }{2}}f(cosx)dx$

Option 1)

$\pi \int_{0}^{\pi }f(cos\, x)dx\;$

This option is incorrect

Option 2)

$\; \; \pi \int_{0}^{\pi }f(sin\, x)dx\;$

This option is incorrect

Option 3)

$\; \frac{\pi }{2}\int_{0}^{\pi /2}\! \! f(sin\, x)dx\; \;$

This option is incorrect

Option 4)

$\; \pi \int_{0}^{\pi /2}\! \! f(cos\, x)dx$

This option is correct

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