Question

Integrate :
$I=\int \frac{1+\cos x}{\sin x.\cos x}dx$

Answer 1

$I=\int \frac{1+\cos x}{\sin x.\cos x}dx$

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

$=2\int cosec\left(2x\right)dx+\int cosecxdx$

$=-ln|cosec\left(2x\right)+\cot \left(2x\right)|-ln|cosecx+\cot x|+C$

$I=-ln\left|\left(\frac{1+\cos 2x}{\sin 2x}\right)\right|-ln\left|\left(\frac{1+\cos x}{\sin x}\right)\right|+C$

$I=-[ln\left(\frac{2{\cos }^{2}x}{2\cos x\sin x}\right)+ln\left(\frac{1+\cos x}{\sin x}\right)]+C$

$I=-\left[ln\left(\frac{\cos (1+\cos x)}{{\sin }^{2}x}\right)\right]+C$

$I=ln\left|\frac{{\sin }^{2}x}{\cos x(1+\cos x)}\right|+C$