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Question Number 158973 by mnjuly1970 last updated on 11/Nov/21

Answered by qaz last updated on 11/Nov/21

$${\int }_0^{\pi /2}\frac{\mathrm{x}}{{(1+\sin \mathrm{x})}^2}\mathrm{dx}$$$$=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\mathrm{dx}}{{(1+\cos \mathrm{x})}^2}-{\int }_0^{\pi /2}\frac{\mathrm{x}}{{(1+\cos \mathrm{x})}^2}\mathrm{dx}$$$$=\frac{\pi }{4}{\int }_0^{\pi /4}\frac{\mathrm{dx}}{\cos {}^4\mathrm{x}}-{\int }_0^{\pi /4}\frac{\mathrm{x}}{\cos {}^4\mathrm{x}}\mathrm{dx}$$$$=\frac{\pi }{4}{\int }_0^{\pi /4}(1+\tan {}^2\mathrm{x})\mathrm{d}\left(\tan \mathrm{x}\right)-{\int }_0^{\pi /4}\mathrm{xd}(\tan \mathrm{x}+\frac{1}{3}\tan {}^3\mathrm{x})$$$$=\frac{\pi }{4}\cdot {[\tan \mathrm{x}+\frac{1}{3}\tan {}^3\mathrm{x}]}_0^{\pi /4}-\mathrm{x}{[\tan \mathrm{x}+\frac{1}{3}\tan {}^3\mathrm{x}]}_0^{\pi /4}+{\int }_0^{\pi /4}(\tan \mathrm{x}+\frac{1}{3}\tan {}^3\mathrm{x})\mathrm{dx}$$$$=\frac{\pi }{3}-\frac{\pi }{3}+{\int }_0^{\pi /4}\tan \mathrm{x}(\frac{2}{3}+\frac{1}{3}\sec {}^2\mathrm{x})\mathrm{dx}$$$$=\frac{1}{3}\ln 2+\frac{1}{6}$$$$\Rightarrow \mathrm{K}=\frac{\frac{1}{3}\ln 2+\frac{1}{6}}{\frac{\pi }{3}-\frac{1}{3}\ln 2-\frac{1}{6}}=\frac{2\ln 2+1}{2\pi -2\ln 2-1}$$

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