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Question Number 50415 by Abdo msup. last updated on 16/Dec/18

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

Commented by Abdo msup. last updated on 18/Dec/18

$$letputcos\theta =\lambda andfindA\left(\lambda \right)={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+\lambda cost}$$$$with\mid \lambda \mid ⩽1changementtan\left(\frac{t}{2}\right)=ugive$$$$A\left(\lambda \right)={\int }_0^1\frac{1}{1+\lambda \frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}$$$$={\int }_0^1\frac{2du}{1+u^2+\lambda -\lambda u^2}={\int }_0^1\frac{2du}{(1-\lambda )u^2+1+\lambda }$$$$=\frac{2}{1-\lambda }{\int }_0^1\frac{du}{u^2+\frac{1+\lambda }{1-\lambda }}$$$${=}_{u=\sqrt{\frac{1+\lambda }{1-\lambda }}t}\frac{2}{1-\lambda }{\int }_0^{\sqrt{\frac{1-\lambda }{1+\lambda }}}\frac{\sqrt{\frac{1+\lambda }{1-\lambda }}dt}{\frac{1+\lambda }{1-\lambda }(1+t^2)}$$$$=\frac{2}{1+\lambda }\frac{\sqrt{1+\lambda }}{\sqrt{1-\lambda }}arctan\left(\sqrt{\frac{1-\lambda }{1+\lambda }}\right)$$$$=\frac{2}{\sqrt{1-{\lambda }^2}}arctan\left(\sqrt{\frac{1-\lambda }{1+\lambda }}\right)\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{dt}{1+co\theta cost}=A\left(cos\theta \right)$$$$=\frac{2}{\sqrt{1-{cos}^2\theta }}arctan\left(\sqrt{\frac{1-cos\theta }{1+cos\theta }}\right)$$$$=\frac{2}{\mid sin\theta \mid }arctan\mid tan\left(\frac{\theta }{2}\right)\mid .$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

$$\int \frac{dt}{1+cos\theta cost}$$$$\frac{1}{cos\theta }\int \frac{dt}{sec\theta +cost}$$$$\frac{1}{cos\theta }\int \frac{dt}{sec\theta +\frac{1-{tan}^2\frac{t}{2}}{1+{tan}^2\frac{t}{2}}}$$$$\frac{1}{cos\theta }\int \frac{{sec}^2\frac{t}{2}dt}{sec\theta +sec\theta {tan}^2\frac{t}{2}+1-{tan}^2\frac{t}{2}}$$$$=sec\theta \int \frac{{sec}^2\frac{t}{2}\times dt}{(sec\theta +1)+{tan}^2\frac{t}{2}(sec\theta -1)}$$$$=sec\theta \int \frac{{sec}^2\frac{t}{2}}{(sec\theta -1)\{\frac{sec\theta +1}{sec\theta -1}+{tan}^2\frac{t}{2}\}}dt$$$$=\frac{sec\theta }{sec\theta -1}\int \frac{{sec}^2\frac{t}{2}}{\frac{1+cos\theta }{1-cos\theta }+{tan}^2\frac{t}{2}}dt$$$$=\frac{1}{1-cos\theta }\int \frac{{sec}^2\frac{t}{2}dt}{{cot}^2\frac{\theta }{2}+{tan}^2\frac{t}{2}}$$$$k=tan\frac{t}{2}dk=\frac{1}{2}{sec}^2\frac{t}{2}dt$$$$=\frac{1}{2{sin}^2\frac{\theta }{2}}\int \frac{2dk}{{cot}^2\frac{\theta }{2}+k^2}$$$$=\frac{1}{{sin}^2\frac{\theta }{2}}\int \frac{dk}{{cot}^2\frac{\theta }{2}+k^2}$$$$=\frac{1}{{sin}^2\frac{\theta }{2}}\times \frac{1}{cot\frac{\theta }{2}}{tan}^{-1}\left(\frac{k}{cot\frac{\theta }{2}}\right)$$$$=\frac{2}{sin\theta }{tan}^{-1}\left(\frac{tan\frac{t}{2}}{cot\frac{\theta }{2}}\right)+c$$$$requiredansweriz$$$$\frac{2}{sin\theta }\mid {tan}^{-1}\left(\frac{tan\frac{t}{2}}{cot\frac{\theta }{2}}\right){\mid }_0^{\frac{\pi }{2}}$$$$=\frac{2}{\mathit{s}\mathit{i}\mathit{n}\theta }[{tan}^{-1}\left(\frac{tan\frac{\pi }{4}}{cot\frac{\theta }{2}}\right)-{tan}^{-1}\left(\frac{tan0}{cot\frac{\theta }{2}}\right)]$$$$=\frac{2}{sin\theta }\left[{tan}^{-1}\left(tan\frac{\theta }{2}\right)\right]$$$$=\frac{2}{sin\theta }\times \frac{\theta }{2}=\frac{\theta }{sin\theta }$$$$$$$$$$

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