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Question Number 110551 by shahria14 last updated on 29/Aug/20

Answered by Dwaipayan Shikari last updated on 29/Aug/20

$${\int }_0^{\pi }\frac{x}{1+sinx}dx={\int }_0^{\pi }\frac{\pi -x}{1+sinx}dx=I$$$$2I={\int }_0^{\pi }\frac{\pi }{1+sinx}dx=2\pi {\int }_0^{\mathrm{\infty }}\frac{1}{1+\frac{2t}{1+t^2}}.\frac{1}{1+t^2}dt(tan\frac{x}{2}=t)$$$$2I=2\pi {\int }_0^{\mathrm{\infty }}\frac{1}{{(1+t)}^2}dt$$$$I=-\pi {\left[\frac{1}{1+t}\right]}_0^{\mathrm{\infty }}=\pi$$

Answered by mathmax by abdo last updated on 30/Aug/20

$$\mathrm{I}={\int }_0^{\pi }\frac{\mathrm{xdx}}{1+\mathrm{sinx}}\mathrm{changement}\mathrm{x}=\pi -\mathrm{t}\mathrm{give}$$$$\mathrm{I}={\int }_0^{\pi }\frac{\pi -\mathrm{t}}{1+\mathrm{sint}}\mathrm{dt}=\pi {\int }_0^{\pi }\frac{\mathrm{dt}}{1+\mathrm{sint}}-\mathrm{I}\Rightarrow 2\mathrm{I}=\pi {\int }_0^{\pi }\frac{\mathrm{dt}}{1+\mathrm{sint}}$$$${=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{u}}{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{du}}{(1+{\mathrm{u}}^2)(1+\frac{2\mathrm{u}}{1+{\mathrm{u}}^2})}={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{du}}{1+{\mathrm{u}}^2+2\mathrm{u}}=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{du}}{{(1+\mathrm{u})}^2}$$$$={\left[\frac{-2}{1+\mathrm{u}}\right]}_0^{\mathrm{\infty }}=1\Rightarrow 2\mathrm{I}=\pi \Rightarrow \mathrm{I}=\frac{\pi }{2}$$

Commented by mathmax by abdo last updated on 30/Aug/20

$$\mathrm{sorry}2\mathrm{I}=2\pi \Rightarrow \mathrm{I}=\pi$$

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