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Question Number 142724 by lapache last updated on 04/Jun/21

$${\int }_0^{\frac{\pi }{2}}\frac{sin\left(2t\right)}{1+xsin\left(2t\right)}dt=....$$

Answered by Ar Brandon last updated on 04/Jun/21

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{\sin 2\mathrm{t}}{1+\mathrm{xsin}2\mathrm{t}}\mathrm{dt}=\frac{1}{\mathrm{x}}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{xsin}2\mathrm{t}}{1+\mathrm{xsin}2\mathrm{t}}\mathrm{dt}$$$$=\frac{1}{\mathrm{x}}{\int }_0^{\frac{\pi }{2}}(1-\frac{1}{1+\mathrm{xsin}2\mathrm{t}})\mathrm{dt}=\frac{\pi }{2\mathrm{x}}-\frac{1}{\mathrm{x}}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{1+\mathrm{xsin}2\mathrm{t}}$$$$=\frac{\pi }{2\mathrm{x}}-\frac{1}{\mathrm{x}}{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^2\mathrm{t}}{{\sec }^2\mathrm{t}+\mathrm{xtant}}\mathrm{dt}=\frac{\pi }{2\mathrm{x}}-\frac{1}{\mathrm{x}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{d}j}{j^2+\mathrm{x}j+1}$$$$=\frac{\pi }{2\mathrm{x}}-\frac{1}{\mathrm{x}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{d}j}{{(j+\frac{\mathrm{x}}{2})}^2+1-\frac{{\mathrm{x}}^2}{4}}=\frac{\pi }{2\mathrm{x}}-\frac{2}{\mathrm{x}\sqrt{4-{\mathrm{x}}^2}}{\left[\arctan \left(\frac{2j+\mathrm{x}}{\sqrt{4-{\mathrm{x}}^2}}\right)\right]}_0^{\mathrm{\infty }}$$$$=\frac{\pi }{2\mathrm{x}}-\frac{2}{\mathrm{x}\sqrt{4-{\mathrm{x}}^2}}(\frac{\pi }{2}-\arctan \left(\frac{\mathrm{x}}{\sqrt{4-{\mathrm{x}}^2}}\right))$$

Answered by mathmax by abdo last updated on 06/Jun/21

$$\mathrm{I}\left(\mathrm{x}\right)={\int }_0^{\frac{\pi }{2}}\frac{\sin \left(2\mathrm{t}\right)}{1+\mathrm{xsin}\left(2\mathrm{t}\right)}\mathrm{dt}\Rightarrow \mathrm{I}\left(\mathrm{x}\right){=}_{2\mathrm{t}=\mathrm{u}}\frac{1}{2}{\int }_0^{\pi }\frac{\mathrm{sinu}}{1+\mathrm{xsinu}}\mathrm{du}$$$$=\frac{1}{2\mathrm{x}}{\int }_0^{\pi }\frac{\mathrm{xsinu}+1-1}{1+\mathrm{xsinu}}\mathrm{du}=\frac{\pi }{2\mathrm{x}}-\frac{1}{2\mathrm{x}}{\int }_0^{\pi }\frac{\mathrm{du}}{1+\mathrm{xsinu}}\mathrm{changement}\tan \left(\frac{\mathrm{u}}{2}\right)=\mathrm{y}$$$$\mathrm{give}{\int }_0^{\pi }\frac{\mathrm{du}}{1+\mathrm{xsinu}}={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dy}}{(1+{\mathrm{y}}^2)(1+\mathrm{x}\frac{2\mathrm{y}}{1+{\mathrm{y}}^2})}$$$$\mathrm{H}={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dy}}{1+{\mathrm{y}}^2+2\mathrm{xy}}={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dy}}{{\mathrm{y}}^2+2\mathrm{xy}+1}={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dy}}{{(\mathrm{y}+\mathrm{x})}^2+1-{\mathrm{x}}^2}$$$$\mathrm{case}11-{\mathrm{x}}^2>0\Rightarrow \mid \mathrm{x}\mid <1\Rightarrow \mathrm{H}{=}_{\mathrm{y}+\mathrm{x}=\sqrt{1-{\mathrm{x}}^2}\mathrm{z}}{\int }_{\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}}^{\mathrm{\infty }}\frac{2\sqrt{1-{\mathrm{x}}^2}\mathrm{dz}}{(1-{\mathrm{x}}^2)(1+{\mathrm{z}}^2)}$$$$=\frac{2}{\sqrt{1-{\mathrm{x}}^2}}{\left[\mathrm{arctanz}\right]}_{\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}}^{\mathrm{\infty }}=\frac{2}{\sqrt{1-{\mathrm{x}}^2}}(\frac{\pi }{2}-\arctan \left(\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\right))$$$$=\frac{\pi }{\sqrt{1-{\mathrm{x}}^2}}-\frac{2}{\sqrt{1-{\mathrm{x}}^2}}\arctan \left(\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\right)\Rightarrow$$$$\mathrm{I}\left(\mathrm{x}\right)=\frac{\pi }{2\mathrm{x}}-\frac{\pi }{2\mathrm{x}\sqrt{1-{\mathrm{x}}^2}}-\frac{1}{\mathrm{x}\sqrt{1-{\mathrm{x}}^2}}\arctan \left(\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\right)$$$$\mathrm{case}21-{\mathrm{x}}^2<0\Rightarrow \mathrm{H}={\int }_0^{\mathrm{\infty }}\frac{2\mathrm{dy}}{{(\mathrm{y}+\mathrm{x})}^2-({\mathrm{x}}^2-1)}$$$${=}_{\mathrm{y}+\mathrm{x}=\sqrt{{\mathrm{x}}^2-1}\mathrm{u}}{\int }_{\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2-1}}}^{\mathrm{\infty }}\frac{2\sqrt{{\mathrm{x}}^2-1}\mathrm{du}}{({\mathrm{x}}^2-1)({\mathrm{u}}^2-1)}$$$$=\frac{2}{\sqrt{{\mathrm{x}}^2-1}}{\int }_{\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2-1}}}^{\mathrm{\infty }}\frac{\mathrm{du}}{(\mathrm{u}-1)(\mathrm{u}+1)}$$$$=\frac{1}{\sqrt{{\mathrm{x}}^2-1}}{\int }_{\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2-1}}}^{\mathrm{\infty }}(\frac{1}{\mathrm{u}-1}-\frac{1}{\mathrm{u}+1})$$$$=\frac{1}{\sqrt{{\mathrm{x}}^2-1}}{[\log \mid \frac{\mathrm{u}-1}{\mathrm{u}+1}\mid ]}_{\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2-1}}}^{\mathrm{\infty }}=-\frac{1}{\sqrt{{\mathrm{x}}^2-1}}\log \mid \frac{\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2-1}}-1}{\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2-1}}+1}\mid$$$$=-\frac{1}{\sqrt{{\mathrm{x}}^2-1}}\log \mid \frac{\mathrm{x}-\sqrt{{\mathrm{x}}^2-1}}{\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}}\mid \Rightarrow$$$$\mathrm{I}\left(\mathrm{x}\right)=\frac{\pi }{2\mathrm{x}}+\frac{1}{2\mathrm{x}\sqrt{{\mathrm{x}}^2-1}}\log \mid \frac{\mathrm{x}-\sqrt{{\mathrm{x}}^2-1}}{\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}}\mid$$

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