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Question Number 116903 by mnjuly1970 last updated on 07/Oct/20

Answered by mathmax by abdo last updated on 07/Oct/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\pi }\frac{\ln (1+\mathrm{acosx})}{\mathrm{cosx}}\mathrm{dx}(\mathrm{here}\mathrm{a}=\sin \alpha )\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\pi }\frac{\mathrm{dx}}{1+\mathrm{acosx}}{=}_{\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(1+\mathrm{a}.\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{1+{\mathrm{t}}^2+\mathrm{a}-{\mathrm{at}}^2}=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{(1-\mathrm{a}){\mathrm{t}}^2+1+\mathrm{a}}$$$$=\frac{2}{1-\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+\frac{1+\mathrm{a}}{1-\mathrm{a}}}{=}_{\mathrm{t}=\sqrt{\frac{1+\mathrm{a}}{1-\mathrm{a}}}\mathrm{z}}\frac{2}{1-\mathrm{a}}\times \frac{1-\mathrm{a}}{1+\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{1}{1+{\mathrm{z}}^2}\times \frac{\sqrt{1+\mathrm{a}}}{\sqrt{1-\mathrm{a}}}\mathrm{dz}$$$$=\frac{2}{\sqrt{1-{\mathrm{a}}^2}}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{1-{\mathrm{a}}^2}}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\pi \arcsin \left(\mathrm{a}\right)+\mathrm{c}$$$$\mathrm{c}=\mathrm{f}\left(0\right)=0\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\pi \mathrm{arcsina}$$$$\mathrm{take}\mathrm{a}=\sin \alpha \Rightarrow {\int }_0^{\pi }\frac{\ln (1+\sin \alpha \mathrm{cosx})}{\mathrm{cosx}}\mathrm{dx}=\pi \arcsin \left(\sin \alpha \right)=\alpha \pi$$

Commented by mnjuly1970 last updated on 08/Oct/20

$$thankyousomuchsir..$$

Answered by Dwaipayan Shikari last updated on 08/Oct/20

$${\int }_0^{\pi }\frac{log(1+sinacosx)}{cosx}dx$$$$I\left(\mathrm{\Psi }\right)={\int }_0^{\pi }\frac{log(1+\mathrm{\Psi }cosx)}{cosx}dx\mathrm{\Psi }=sina$$$$I^{\prime }\left(\mathrm{\Psi }\right)={\int }_0^{\pi }\frac{1}{(1+\mathrm{\Psi }cosx)}dx$$$$I^{\prime }\left(\mathrm{\Psi }\right)=2{\int }_0^{\mathrm{\infty }}\frac{1}{(1+\mathrm{\Psi }\frac{1-t^2}{1+t^2})}.\frac{1}{1+t^2}dtt=tan\frac{x}{2}$$$$I^{\prime }\left(\mathrm{\Psi }\right)=2{\int }_0^{\mathrm{\infty }}\frac{1}{1+t^2+\mathrm{\Psi }t-\mathrm{\Psi }{t}^2}dt$$$$I^{\prime }\left(\mathrm{\Psi }\right)=\frac{2}{1-\mathrm{\Psi }}{\int }_0^{\mathrm{\infty }}\frac{1}{t^2+{\left(\sqrt{\frac{1+\mathrm{\Psi }}{1-\mathrm{\Psi }}}\right)}^2}dt$$$$I^{\prime }\left(\mathrm{\Psi }\right)=\frac{2}{\sqrt{1-{\mathrm{\Psi }}^2}}{\left[{tan}^{-1}\frac{t\sqrt{1-\mathrm{\Psi }}}{\sqrt{1+\mathrm{\Psi }}}\right]}_0^{\mathrm{\infty }}$$$$I^{\prime }\left(\mathrm{\Psi }\right)=\frac{\pi }{\sqrt{1-{\mathrm{\Psi }}^2}}$$$$I\left(\mathrm{\Psi }\right)=\pi \int \frac{1}{\sqrt{1-{\mathrm{\Psi }}^2}}d\mathrm{\Psi }$$$$I\left(\mathrm{\Psi }\right)=\pi {sin}^{-1}\mathrm{\Psi }+C\mathrm{\Psi }=sina$$$$I\left(\mathrm{\Psi }\right)=\pi a+C$$$$a=0thenC=0$$$$I\left(\mathrm{\Psi }\right)=\pi a$$

Commented by mnjuly1970 last updated on 08/Oct/20

$$thankyoumrdwaipayn...$$

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