$\int \tan ^{-1}(\sec x+\tan x) d x=$

$\int \tan ^{-1}(\sec x+\tan x) d x=$
  1. $\frac{\pi x}{4}+\frac{x^2}{4}+c$
  2. $\sin x \cos x+c$
  3. $\frac{\pi \mathrm{x}}{2}+\frac{\mathrm{x}^2}{2}+\mathrm{c}$
  4. $\sin x+\cos x+c$

Solution

$\begin{aligned} & \text { Let } I=\tan ^{-1}(\sec x+\tan x) d x \\ & =\int \tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right) d x \\ & =\int \tan ^{-1} x\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}\right]_{d x} \\ & =\int \tan ^{-1}\left(\frac{\left.\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right) d x \quad=\int \tan ^{-1}\left(\frac{1+\sin \frac{x}{2}}{1-\tan \frac{x}{2}}\right) d x \\ & =\int \tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right] d x=\int\left(\frac{\pi}{4}+\frac{x}{2}\right) d x \\ & =\frac{\pi x}{4}+\frac{x^2}{4}+c\end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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