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Question Number 29443 by prof Abdo imad last updated on 08/Feb/18

$$find{\int }_0^{\pi }\frac{sinx}{\sqrt{1+{sin}^{2}x}}dx$$

Answered by sma3l2996 last updated on 09/Feb/18

$$t=cosx\Rightarrow dt=-sinxdx$$$${\int }_0^{\pi }\frac{sinx}{\sqrt{1+{sin}^{2}x}}dx={\int }_{-1}^1\frac{dt}{\sqrt{1+(1-t^2)}}={\int }_{-1}^1\frac{dt}{\sqrt{2}\sqrt{1-{\left(\frac{t}{\sqrt{2}}\right)}^2}}$$$$t=\sqrt{2}u\Rightarrow dt=\sqrt{2}du$$$${\int }_0^{\pi }\frac{sinx}{\sqrt{1+{sin}^{2}x}}dx={\int }_{-\sqrt{2}/2}^{\sqrt{2}/2}\frac{du}{\sqrt{1-u^2}}$$$$u=siny\Rightarrow dy=\frac{du}{\sqrt{1-u^2}}$$$${\int }_0^{\pi }\frac{sinx}{\sqrt{1+{sin}^{2}x}}dx={\int }_{-\pi /4}^{\pi /4}dy=\frac{\pi }{2}$$

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