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Question Number 118111 by bemath last updated on 15/Oct/20

$$\underset{0}{\stackrel{\pi /4}{\int }}\frac{{\mathrm{x}}^2}{{(\mathrm{x}\sin \mathrm{x}+\cos \mathrm{x})}^2}\mathrm{dx}=?$$

Answered by Lordose last updated on 15/Oct/20

$$$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{4}}\frac{{\mathrm{x}}^2}{{(\mathrm{xsinx}+\mathrm{cosx})}^2}\mathrm{dx}$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{4}}\frac{{\mathrm{x}}^2{\sin }^2\mathrm{x}+{\mathrm{x}}^2{\cos }^2\mathrm{x}}{{(\mathrm{xsinx}+\mathrm{cosx})}^2}\mathrm{dx}$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{4}}\frac{{\mathrm{x}}^2{\sin }^2\mathrm{x}+\mathrm{xsinxcosx}}{{(\mathrm{xsinx}+\mathrm{cosx})}^2}\mathrm{dx}+{\int }_0^{\frac{\pi }{4}}\frac{{\mathrm{x}}^2{\cos }^2\mathrm{x}-\mathrm{xsinxcosx}}{{(\mathrm{xsinx}+\mathrm{cosx})}^2}\mathrm{dx}$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{4}}\frac{\mathrm{xsinx}(\mathrm{xsinx}+\mathrm{cosx})}{{(\mathrm{xsinx}+\mathrm{cosx})}^2}\mathrm{dx}+{\int }_0^{\frac{\pi }{4}}\frac{-\mathrm{xcosx}(-\mathrm{xcosx}+\mathrm{sinx})}{{(\mathrm{xsinx}+\mathrm{cosx})}^2}\mathrm{dx}$$$$\mathrm{\Omega }=\mathrm{\Phi }+\mathrm{\Lambda }$$$$\mathrm{resolving}\mathrm{\Lambda }\mathrm{by}\mathrm{IBP}$$$$\mathrm{u}=\mathrm{sinx}-\mathrm{xcosx}\Rightarrow \mathrm{du}=\mathrm{xsinxdx}$$$$\mathrm{dv}=\frac{-\mathrm{xcosx}}{{(\mathrm{xsinx}+\mathrm{cosx})}^2}\mathrm{dx}\Rightarrow \mathrm{v}=\frac{1}{\mathrm{xsinx}+\mathrm{cosx}}$$$$\mathrm{\Lambda }=\mid \frac{\mathrm{sinx}-\mathrm{xcosx}}{\mathrm{xsinx}+\mathrm{cosx}}{\mid }_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}\frac{\mathrm{xsinx}}{\mathrm{xsinx}+\mathrm{cosx}}\mathrm{dx}$$$$\mathrm{\Lambda }=\mid \frac{\mathrm{sinx}-\mathrm{xcosx}}{\mathrm{xsinx}+\mathrm{cosx}}{\mid }_0^{\frac{\pi }{4}}-\mathrm{\Phi }$$$$\mathrm{\Omega }=\mathrm{\Phi }+\mid \frac{\mathrm{sinx}-\mathrm{xcosx}}{\mathrm{xsinx}+\mathrm{cosx}}{\mid }_0^{\frac{\pi }{4}}-\mathrm{\Phi }$$$$\mathrm{\Omega }=\mid \frac{\mathrm{sinx}-\mathrm{xcosx}}{\mathrm{xsinx}+\mathrm{cosx}}{\mid }_0^{\frac{\pi }{4}}$$$$\mathrm{\Omega }=\frac{4-\pi }{4+\pi }$$$$$$

Commented by bemath last updated on 15/Oct/20

$$gavekudos$$

Answered by Ar Brandon last updated on 15/Oct/20

$$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{xsinx}+\mathrm{cosx})=\mathrm{xcosx}+\mathrm{sinx}-\mathrm{sinx}=\mathrm{xcosx}$$$$\mathcal{I}={\int }_0^{\frac{\pi }{4}}\frac{{\mathrm{x}}^2}{{(\mathrm{xsinx}+\mathrm{cosx})}^2}\mathrm{dx}={\int }_0^{\frac{\pi }{4}}\frac{\mathrm{xcosx}}{{(\mathrm{xsinx}+\mathrm{cosx})}^2}\cdot \frac{\mathrm{x}}{\mathrm{cosx}}\mathrm{dx}$$$$={\{\frac{\mathrm{x}}{\mathrm{cosx}}\int \frac{\mathrm{xcosx}}{{(\mathrm{xsinx}+\mathrm{cosx})}^2}\mathrm{dx}\}}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}}{\mathrm{cosx}}\right)\cdot \int \frac{\mathrm{xcosx}}{{(\mathrm{xsinx}+\mathrm{cosx})}^2}\mathrm{dx}\}$$$$=-{[\frac{\mathrm{x}}{\mathrm{cosx}}\cdot \frac{1}{\mathrm{xsinx}+\mathrm{cosx}}]}_0^{\frac{\pi }{4}}+{\int }_0^{\frac{\pi }{4}}\left\{\left(\frac{\mathrm{cosx}+\mathrm{xsinx}}{{\cos }^2\mathrm{x}}\right)\left(\frac{1}{\mathrm{xsinx}+\mathrm{cosx}}\right)\right\}\mathrm{dx}$$$$=-{[\frac{\mathrm{x}}{\mathrm{cosx}}\cdot \frac{1}{\mathrm{xsinx}+\mathrm{cosx}}]}_0^{\frac{\pi }{4}}+{\int }_0^{\frac{\pi }{4}}\frac{\mathrm{dx}}{{\cos }^2\mathrm{x}}$$$$={[\mathrm{tanx}-\frac{\mathrm{x}}{\mathrm{cosx}}\cdot \frac{1}{\mathrm{xsinx}+\mathrm{cosx}}]}_0^{\frac{\pi }{4}}=[1-\frac{2\pi }{4\sqrt{2}}\cdot \frac{8}{\pi \sqrt{2}+4\sqrt{2}}]$$$$=1-\frac{16\pi }{8\pi +32}=\frac{8\pi +32-16\pi }{8\pi +32}=\frac{32-8\pi }{32+8\pi }=\frac{4-\pi }{4+\pi }$$

Commented by bemath last updated on 15/Oct/20

$$santuyy$$

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