Question

I = $\int$ ( $\frac{1+tan2x}{1-tan2x}$) dx

Answer 1

$I=\int \frac{1+\tan 2x}{1-\tan 2x}dx=\int \frac{1+\frac{\sin 2x}{\cos 2x}}{1-\frac{\sin 2x}{\cos 2x}}dx$

$=\int \frac{\cos 2x+\sin 2x}{\cos 2x-\sin 2x}dx$

$=\int \frac{{(\cos 2x+\sin 2x)}^2}{{\cos }^{2}2x-{\sin }^{2}2x}dx$ (Multiplying $(cos2x+sin2x)$ in the numerator & deminator)

$=\int \frac{{\cos }^{2}2x+{\sin }^{2}2x+2\cos 2x.\sin 2x}{{\cos }^{2}2x-{\sin }^{2}2x}dx$

$=\int \frac{1+2\cos 2x.\sin 2x}{{\cos }^{2}2x-{\sin }^{2}2x}dx$

$=\int \frac{1+\sin 4x}{\cos 4x}dx$ $[\sin 4x=2\sin 2x.\cos 2x]$

$=\int (\sec 4x+\tan 4x)dx$ $[\cos 4x={\cos }^{2}2x-{\sin }^{2}2x]$

$=\int \sec 4xdx+\int \tan 4xdx$

$=\frac{In(\sec 4x+\tan 4x)}{4}+\frac{In\sec 4x}{4}+C$