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Question Number 30180 by abdo imad last updated on 17/Feb/18

$$find{\int }_0^{\frac{\pi }{2}}\frac{xsinxcosx}{{tan}^{2}x+{cotan}^{2}x}dx.(usethech.x=\frac{\pi }{2}-t).$$

Commented by abdo imad last updated on 24/Feb/18

$$ch.x=\frac{\pi }{2}-tgiveI={\int }_0^{\frac{\pi }{2}}\frac{(\frac{\pi }{2}-t)costsint}{{cotan}^{2}t+{tan}^{2}t}dt$$$$=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{dt}{{tan}^{2}t+{cotan}^{2}t}-{\int }_0^{\frac{\pi }{2}}\frac{sint?cost}{{tan}^{2}t+{cotan}^{2}t}dt\Rightarrow$$$$2I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{dt}{{tan}^{2}t+{cotan}^{2}t}dtthech.tant=ugive$$$${\int }_0^{\frac{\pi }{2}}\frac{dt}{{tan}^{2}t+{cotan}^{2}t}={\int }_0^{\mathrm{\infty }}\frac{1}{u^2+\frac{1}{u^2}}\frac{du}{1+u^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{u^2}{(1+u^2)(1+u^4)}du={\int }_0^{\mathrm{\infty }}\frac{1+u^2-1}{(1+u^2)(1+u^4)}du$$$$={\int }_0^{\mathrm{\infty }}\frac{du}{1+u^4}-{\int }_0^{\mathrm{\infty }}\frac{du}{(1+u^2)(1+u^4)}thech.u^4=xgive$$$${\int }_0^{\mathrm{\infty }}\frac{du}{1+u^4}={\int }_0^{\mathrm{\infty }}\frac{\frac{1}{4}{x}^{\frac{1}{4}-1}}{1+x}dx=\frac{1}{4}\frac{\pi }{sin\left(\frac{\pi }{4}\right)}$$$$=\frac{\pi }{4}\frac{1}{\frac{1}{\sqrt{2}}}=\frac{\pi \sqrt{2}}{4}.letdecomposeF\left(x\right)=\frac{1}{(1+u^2)(1+u^4)}$$$$F\left(x\right)=\frac{1}{(1+u^2)({(u^2+1)}^2-2u^2)}$$$$=\frac{1}{(1+u^2)(u^2+1-\sqrt{2}u)(u^2+1+\sqrt{2}u)}$$$$=\frac{au+b}{1+u^2}+\frac{cu+d}{u^2-\sqrt{2}u+1}+\frac{eu+f}{u^2+\sqrt{2}u+1}....becontinued...$$

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