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Question Number 38122 by maxmathsup by imad last updated on 22/Jun/18

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

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18

$$t=cosxdt=-sinxdx$$$${\int }_1^{-1}\frac{-dt}{\sqrt{1+t^2}}$$$${\int }_{-1}^1\frac{dt}{\sqrt{1+t^2}}$$$$l$$$$=\mid ln(t+\sqrt{1+t^2}){\mid }_{-1}^1$$$$=ln(1+\sqrt{2})-ln(-1+\sqrt{2})$$

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