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Question Number 29043 by yesaditya22@gmail.com last updated on 03/Feb/18

$$\int {\tan }^{-}(1-\mathrm{sinx}/1+\mathrm{sinx})\mathrm{dx}$$

Commented by abdo imad last updated on 03/Feb/18

$$letputI=\int arctan\left(\frac{1-sinx}{1+sinx}\right)dx(arctan={tan}^{-1})wehave$$$$\frac{1-sinx}{1+sinx}=\frac{1-cos(\frac{\pi }{2}-x)}{1+cos(\frac{\pi }{2}-x)}=\frac{2{sin}^2(\frac{\pi }{4}-\frac{x}{2})}{2{cos}^2(\frac{\pi }{4}-\frac{x}{2})}={tan}^2(\frac{\pi }{4}-\frac{x}{2})$$$$I=\int arctan\left({tan}^2(\frac{\pi }{4}-\frac{x}{2})\right)dxletusethech.$$$${tan}^2(\frac{\pi }{4}-\frac{x}{2})=u\iff tan(\frac{\pi }{4}-\frac{x}{2})=\sqrt{u}and$$$$\frac{\pi }{4}-\frac{x}{2}=arctan\left(\sqrt{u}\right)\iff x=-2arctan\left(\sqrt{u}\right)\Rightarrow$$$$dx=-2\frac{\frac{1}{2\sqrt{u}}}{1+u}=\frac{-1}{\sqrt{u}(1+u)}so$$$$I=-\int \frac{artanu}{\sqrt{u}(1+u)}du(\sqrt{u}=t)$$$$I=-\int \frac{arctan\left(t^2\right)}{t(1+t^2)}2tdt=-2\int \frac{arctan\left(t^2\right)}{1+t^2}dt+\lambda andthe$$$$integral\int \frac{arctan\left(t^2\right)}{1+t^2}dtbecalculatedbytheparametric$$$$functionf\left(x\right)=\int \frac{arctan\left(x(1+t^2)\right)}{1+t^2}dtbyderivationunder$$$$\int ....$$$$$$

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