$\int \tan ^{-1}\left(\frac{1-\sin x}{1+\sin x}\right) d x=$
$\int \tan ^{-1}\left(\frac{1-\sin x}{1+\sin x}\right) d x=$
- $\frac{\pi}{4} x-x+c$, where $c$ is a constant of integration.
- $\frac{\pi}{4}-\frac{x}{2}+\mathrm{c}$, where c is a constant of integration.
- $\frac{\pi}{4} x-\frac{x^2}{4}+\mathrm{c}$, where c is a constant of integration.
- $\frac{\pi}{4} x+\frac{x^2}{4}+\mathrm{c}$, where c is a constant of integration.
Solution
Let $\mathrm{I}=\int \tan ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) \mathrm{d} x$
$=\int \tan ^{-1}\left(\sqrt{\frac{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}-2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}}\right) \mathrm{d} x$
$=\int \tan ^{-1}\left(\sqrt{\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}}\right) d x$
$\begin{aligned} & =\int \tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) d x \\ & =\int \tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right) d x\end{aligned}$
$\begin{aligned} & =\int \tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right) \mathrm{d} x \\ & =\int\left(\frac{\pi}{4}-\frac{x}{2}\right) \mathrm{d} x \\ & =\frac{\pi}{4} x-\frac{x^2}{4}+\mathrm{c}\end{aligned}$
Asked in: MHT CET 2024 (04 May Shift 2)
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