Question

Integrate w.r.to $x$

${\tan }^{-1}\left(\frac{\sin 3x-\cos 3x}{\sin 3x+\cos 3x}\right)$

Answer 1

$\int {\tan }^{-1}\frac{\sqrt{2}(\frac{\sin 3x}{\sqrt{2}}-\frac{\cos 3x}{\sqrt{2}})}{\sqrt{2}(\frac{\sin 3x}{\sqrt{2}}+\frac{\cos 3x}{\sqrt{2}})}dx$

$=\int {\tan }^{-1}\left[\frac{\sin (\frac{\pi }{4}-3x)}{\cos (\frac{\pi }{4}-3x)}\right]dx$

$=\int {\tan }^{-1}\left[\tan (\frac{\pi }{4}-3x)\right]dx$

$=\int (\frac{\pi }{4}-3x)dx$

$=\frac{\pi }{4}\int dx-3\int xdx$

$=\frac{\pi }{4}x-\frac{3x^2}{2}+C.$

Hence, the answer is $\frac{\pi }{4}x-\frac{3x^2}{2}+C.$