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Question Number 99168 by bemath last updated on 19/Jun/20

Answered by Kunal12588 last updated on 19/Jun/20

$$I={\int }_0^{\pi /2}\frac{x\cos x}{1+{\sin }^{2}x}dx$$$$\Rightarrow I={\int }_0^{\pi /2}\frac{\left(\pi /2\right)\cos x}{1+{\sin }^{2}x}dx-{\int }_0^{\pi /2}\frac{x\cos x}{1+{\sin }^{2}x}dx$$$$\Rightarrow 2I=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\cos x}{1+{\sin }^{2}x}dx$$$$\Rightarrow I=-\frac{\pi }{4}{\int }_0^{\pi /2}\frac{d\left(\sin x\right)}{1+{\sin }^{2}x}$$$$\Rightarrow I=-\frac{\pi }{4}{\tan }^{-1}\left(\sin \frac{\pi }{2}\right)=-\frac{\pi }{4}{\tan }^{-1}\left(1\right)$$$$\Rightarrow I=-\frac{{\pi }^2}{16}$$

Commented by mathmax by abdo last updated on 19/Jun/20

$$\mathrm{miss}\mathrm{kunal}\mathrm{if}\mathrm{yiu}\mathrm{have}\mathrm{done}\mathrm{the}\mathrm{changement}\mathrm{x}=\frac{\pi }{2}-\mathrm{t}\mathrm{we}\mathrm{get}$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{(\frac{\pi }{2}-\mathrm{t})\mathrm{sint}}{1+{\cos }^2\mathrm{t}}\mathrm{dt}=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{sint}}{1+{\cos }^2\mathrm{t}}\mathrm{dt}-{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{tsint}}{1+{\cos }^2\mathrm{t}}\mathrm{dt}$$$$\mathrm{how}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{xcosx}}{1+{\sin }^2\mathrm{x}}\mathrm{dx}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{xsinx}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}?$$

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