Question

integration of tan inverse (secX + tanx) dx

Answer 1

∫tan⁻1(secx+tanx)

=∫∫tan⁻1(1/cosx+sinx/cosx)

Answer 2

Answer:

The integration is explained step by step below :

Step-by-step explanation:

$\text{To Find : }\int \tan^{-1}(\sec x+\tan x)dx\\\\\text{Solution :}\int \tan^{-1}(\sec x+\tan x)dx\\\\=\int \tan^{-1}(\frac{1}{\cos x}+\frac{\sin x}{\cos x})dx\\\\=\int \tan^{-1}(\frac{1+\sin x}{\cos x})dx\\\\=\int \tan^{-1}(\frac{1+\cos(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x)})dx\\\\=\int \tan^{-1}(\frac{2\cos^2(\frac{\pi}{4}-\frac{x}{2})}{2\sin(\frac{\pi}{4}-\frac{x}{2})\cos(\frac{\pi}{4}-\frac{x}{2})})dx$

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

Hence the result.