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Question Number 17210 by Arnab Maiti last updated on 02/Jul/17

prove that \int_{0}^{Π} f (\sin x) dx = 2 \times \int_{0}^{\frac{Π}{2}} f (\sin x) dx

Answered by mrW1 last updated on 02/Jul/17

let I = \int_{\frac{π}{2}}^{π} f (\sin x) dx

t = π - x

x = π - t

\sin x = \sin (π - t) = \sin t

dx = - dt

I = \int_{\frac{π}{2}}^{π} f (\sin x) dx = \int_{\frac{π}{2}}^{0} f (\sin t) (- dt)

= - \int_{\frac{π}{2}}^{0} f (\sin t) dt

= \int_{0}^{\frac{π}{2}} f (\sin t) dt

= \int_{0}^{\frac{π}{2}} f (\sin x) dx

\Rightarrow \int_{0}^{π} f (\sin x) dx = \int_{0}^{\frac{π}{2}} f (\sin x) dx + \int_{\frac{π}{2}}^{π} f (\sin x) dx

= \int_{0}^{\frac{π}{2}} f (\sin x) dx + \int_{0}^{\frac{π}{2}} f (\sin x) dx

= 2 \int_{0}^{\frac{π}{2}} f (\sin x) dx

Commented by Arnab Maiti last updated on 02/Jul/17

Thank u sir . I really appriciate .

Commented by Arnab Maiti last updated on 02/Jul/17

Sir can you give me some more questions

like that ? Please do .

Commented by mrW1 last updated on 02/Jul/17

But I {don}^{'} t have any such questions .

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