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Question Number 32351 by abdo imad last updated on 23/Mar/18

$$calculate{\int }_0^{\frac{\pi }{2}}\frac{dt}{1+cos\theta sint}.$$

Answered by sma3l2996 last updated on 25/Mar/18

$$letx=tan\left(t/2\right)\Rightarrow dt=\frac{2dx}{1+x^2}$$$$sint=\frac{2x}{1+x^2}$$$${\int }_0^{\pi /2}\frac{dt}{1+cos\theta sint}=2{\int }_0^1\frac{dx}{1+x^2+2xcos\theta }=2{\int }_0^1\frac{dx}{{(x+cos\theta )}^2+{sin}^2\theta }$$$$=\frac{2}{{sin}^2\theta }{\int }_0^1\frac{dx}{{\left(\frac{x+cos\theta }{sin\theta }\right)}^2+1}$$$$letu=\frac{x+cos\theta }{sin\theta }\Rightarrow dx=sin\theta du$$$${\int }_0^{\pi /2}\frac{dt}{1+cos\theta sint}=\frac{2}{sin\theta }{\int }_{cot\theta }^{\frac{1+cos\theta }{sin\theta }}\frac{du}{u^2+1}=\frac{2}{sin\theta }{\left[arctan\left(\frac{x+cos\theta }{sin\theta }\right)\right]}_0^1$$$$=\frac{2}{sin\theta }(arctan\left(\frac{1+cos\theta }{sin\theta }\right)-arctan\left(cot\theta \right))$$$$$$

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