Question

Evaluate: \[ \int \frac{1}{\sin x \cos 2x}\,dx \]

• A. \[ \frac12\log\left|\frac{\cos x+1}{\cos x-1}\right| -\frac1{\sqrt2} \log\left| \frac{\sqrt2\cos x+1} {\sqrt2\cos x-1} \right| +C \]
• B. \[ \frac12\log\left|\frac{\cos x+1}{\cos x-1}\right| +\frac1{\sqrt2} \log\left| \frac{\sqrt2\cos x+1} {\sqrt2\cos x-1} \right| +C \]
• C. \[ \frac12\log\left|\frac{\cos x-1}{\cos x+1}\right| -\frac1{\sqrt2} \log\left| \frac{\sqrt2\cos x-1} {\sqrt2\cos x+1} \right| +C \]
• D. \[ \frac12\log\left|\frac{\cos x-1}{\cos x+1}\right| +\frac1{\sqrt2} \log\left| \frac{\sqrt2\cos x-1} {\sqrt2\cos x+1} \right| +C \]

Hint:

For integrals involving \(\cos 2x\), always rewrite \[ \cos 2x=2\cos^2x-1 \] before applying substitutions. This frequently converts the integral into a rational function.

Answer 1

The Correct Option is C