Question

Evaluate: $\int {\tan }^{-1}(\sec x+\tan x)dx$ where $-\frac{\pi }{2}<x<\frac{\pi }{2}$

Answer 1

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

$=\int {\tan }^{-1}(\frac{1}{\cos x}+\frac{\sin x}{\cos x})dx$

$=\int {\tan }^{-1}\left(\frac{1+\sin x}{\cos x}\right)dx$

$=\int {\tan }^{-1}\left(\frac{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}\right)dx$

$=\int {\tan }^{-1}\left(\frac{(\sin \frac{x}{2}+\cos \frac{x}{2}{)}^2}{(\cos \frac{x}{2}+\sin \frac{x}{2})(\cos \frac{x}{2}-\sin \frac{x}{2})}\right)dx$

Dividing by $\cos \frac{x}{2}$ in numerator & denominator, we get

$=\int {\tan }^{-1}\left(\frac{\tan \frac{x}{2}+1}{1-\tan \frac{x}{2}}\right)dx$

$=\int {\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi }{4}\cdot \tan \frac{x}{2}}\right)dx$

$=\int {\tan }^{-1}\left[\tan (\frac{x}{2}+\frac{\pi }{4})\right]dx$

$=\int (\frac{x}{2}+\frac{\pi }{4})dx=\frac{x^2}{4}+\frac{\pi x}{4}+C=\frac{1}{4}x(x+\pi )+c$

Since, $-\frac{\pi }{2}<x<\frac{\pi }{2}$

$I=\frac{1}{4}[\frac{{\pi }^2}{4}+\frac{{\pi }^2}{2}-\frac{{\pi }^2}{4}+\frac{{\pi }^2}{2}]=\frac{{\pi }^2}{4}$