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Question Number 17210 by Arnab Maiti last updated on 02/Jul/17
prove that ∫_0 ^( Π) f(sin x)dx=2×∫_0 ^( (Π/2)) f(sin x)dx
$$\mathrm{prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\:\Pi} {f}\left(\mathrm{sin}\:\mathrm{x}\right)\mathrm{dx}=\mathrm{2}×\int_{\mathrm{0}} ^{\:\frac{\Pi}{\mathrm{2}}} {f}\left(\mathrm{sin}\:\mathrm{x}\right)\mathrm{dx} \\ $$
Answered by mrW1 last updated on 02/Jul/17
let I=∫_(π/2) ^π f(sin x)dx t=π−x x=π−t sin x=sin (π−t)=sin t dx=−dt I=∫_(π/2) ^π f(sin x)dx=∫_(π/2) ^0 f(sin t) (−dt) =−∫_(π/2) ^0 f(sin t) dt =∫_0 ^(π/2) f(sin t) dt =∫_0 ^(π/2) f(sin x) dx ⇒ ∫_0 ^π f(sin x)dx=∫_0 ^(π/2) f(sin x)dx+∫_(π/2) ^π f(sin x)dx =∫_0 ^(π/2) f(sin x)dx+∫_0 ^(π/2) f(sin x)dx =2∫_0 ^(π/2) f(sin x)dx
$$\mathrm{let}\:\mathrm{I}=\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \mathrm{f}\left(\mathrm{sin}\:\mathrm{x}\right)\mathrm{dx} \\ $$$$\mathrm{t}=\pi−\mathrm{x} \\ $$$$\mathrm{x}=\pi−\mathrm{t} \\ $$$$\mathrm{sin}\:\mathrm{x}=\mathrm{sin}\:\left(\pi−\mathrm{t}\right)=\mathrm{sin}\:\mathrm{t} \\ $$$$\mathrm{dx}=−\mathrm{dt} \\ $$$$\mathrm{I}=\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \mathrm{f}\left(\mathrm{sin}\:\mathrm{x}\right)\mathrm{dx}=\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \mathrm{f}\left(\mathrm{sin}\:\mathrm{t}\right)\:\left(−\mathrm{dt}\right) \\ $$$$=−\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \mathrm{f}\left(\mathrm{sin}\:\mathrm{t}\right)\:\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{f}\left(\mathrm{sin}\:\mathrm{t}\right)\:\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{f}\left(\mathrm{sin}\:\mathrm{x}\right)\:\mathrm{dx} \\ $$$$ \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\pi} \mathrm{f}\left(\mathrm{sin}\:\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{f}\left(\mathrm{sin}\:\mathrm{x}\right)\mathrm{dx}+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \mathrm{f}\left(\mathrm{sin}\:\mathrm{x}\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{f}\left(\mathrm{sin}\:\mathrm{x}\right)\mathrm{dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{f}\left(\mathrm{sin}\:\mathrm{x}\right)\mathrm{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{f}\left(\mathrm{sin}\:\mathrm{x}\right)\mathrm{dx} \\ $$
Commented by Arnab Maiti last updated on 02/Jul/17
Thank u sir. I really appriciate.
$$\mathrm{Thank}\:\mathrm{u}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appriciate}. \\ $$
Commented by Arnab Maiti last updated on 02/Jul/17
Sir can you give me some more questions like that ? Please do.
$$\mathrm{Sir}\:\mathrm{can}\:\mathrm{you}\:\mathrm{give}\:\mathrm{me}\:\mathrm{some}\:\mathrm{more}\:\mathrm{questions}\: \\ $$$$\mathrm{like}\:\mathrm{that}\:?\:\:\mathrm{Please}\:\mathrm{do}. \\ $$
Commented by mrW1 last updated on 02/Jul/17
But I don′t have any such questions.
$$\mathrm{But}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{any}\:\mathrm{such}\:\mathrm{questions}. \\ $$

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