Question

Solve: $\int \frac{\cos 2x}{\sin x}dx$

• A. $\log \left|\tan \frac{x}{2}\right|-2\cos x+C$
• B. $\log \left|\cot \frac{x}{2}\right|-2\cos x+C$
• C. $-\log \left|\tan \frac{x}{2}\right|+2\cos x+C$
• D. $\log \left|\tan \frac{x}{2}\right|+2\cos x+C$

Answer 1

The correct option is D $\log \left|\tan \frac{x}{2}\right|+2\cos x+C$

Consider, $I=\int \frac{\cos 2x}{\sin x}dx$

$=\int \frac{1-2{\sin }^{2}x}{\sin x}dx$ $[\because \cos 2x=1-2{\sin }^{2}x]$

$=\int cosecxdx$$-\int 2\sin xdx$

$=\log \left|\tan \frac{x}{2}\right|+2\cos x+c$ [ Where $c$ is integrating constant]