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Question Number 80416 by jagoll last updated on 03/Feb/20

$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{x\cos x}{{(1+\sin x)}^2}dx?$$

Answered by MJS last updated on 03/Feb/20

$$\int \frac{x\cos x}{{(1+\sin x)}^2}dx=$$$$\mathrm{by}\mathrm{parts}$$$$u=x\to u^{\prime }=1$$$$v^{\prime }=\frac{\cos x}{{(1+\sin x)}^2}\to v=-\frac{1}{1+\sin x}$$$$=-\frac{x}{1+\sin x}+\int \frac{dx}{1+\sin x}=$$$$=-\frac{x}{1+\sin x}+\tan x-\frac{1}{\cos x}+C$$$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{x\cos x}{{(1+\sin x)}^2}dx=$$$$[\underset{x\to \frac{\pi }{2}}{\lim }(\tan x-\frac{1}{\cos x})=0]$$$$=1-\frac{\pi }{4}$$

Commented by jagoll last updated on 03/Feb/20

$$\frac{-x}{1+\sin x}{\mid }_0^{\frac{\pi }{2}}=\frac{-\frac{\pi }{2}}{2}=-\frac{\pi }{4}.$$$$howget1-\frac{\pi }{4}sir?$$

Commented by jagoll last updated on 03/Feb/20

$$ooiunderstandsir.get1$$

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