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Question Number 90997 by jagoll last updated on 27/Apr/20

$$\int \underset{0}{\stackrel{\pi }{}}\frac{\sin xdx}{1+\sin x}$$

Commented by john santu last updated on 27/Apr/20

Commented by jagoll last updated on 27/Apr/20

$$thankyou$$

Commented by mathmax by abdo last updated on 27/Apr/20

$$I={\int }_0^{\pi }\frac{sinx}{1+sinx}dx\Rightarrow I={\int }_0^{\pi }\frac{1+sinx-1}{1+sinx}dx$$$$=\pi -{\int }_0^{\pi }\frac{dx}{1+sinx}wehave{\int }_0^{\pi }\frac{dx}{1+sinx}{=}_{tan\left(\frac{x}{2}\right)=t}{\int }_0^{\mathrm{\infty }}\frac{2dt}{(1+t^2)(1+\frac{2t}{1+t^2})}$$$${={\int }_0^{\mathrm{\infty }}\frac{2dt}{1+t^2+2t}={\int }_0^{\mathrm{\infty }}\frac{2dt}{{(t+1)}^2}=-\frac{2}{t+1}]}_0^{\mathrm{\infty }}=2\Rightarrow I=\pi -2$$

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