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

Concept: Whenever an integral contains powers or functions of \(\sin x\) and \(\cos x\), a useful substitution is \[ t=\cos x. \] This transforms the trigonometric integral into a rational function which can then be decomposed into partial fractions.

Step 1: Use the substitution \(t=\cos x\). Let \[ t=\cos x. \] Then \[ dt=-\sin x\,dx. \] Therefore \[ dx=-\frac{dt}{\sin x}. \] Substituting into the integral, \[ I = \int \frac{1}{\sin x(2\cos^2x-1)} \left(-\frac{dt}{\sin x}\right). \] Using \[ \sin^2x=1-\cos^2x=1-t^2, \] we obtain \[ I = -\int \frac{dt} {(1-t^2)(2t^2-1)}. \]

Step 2: Perform partial fraction decomposition. Decompose \[ \frac{-1} {(1-t^2)(2t^2-1)} = \frac{A}{1-t} +\frac{B}{1+t} +\frac{Ct+D}{2t^2-1}. \] After comparing coefficients and simplifying, we obtain \[ I = \frac12\int\frac{dt}{t-1} -\frac12\int\frac{dt}{t+1} -\sqrt2\int\frac{dt}{2t^2-1}. \]

Step 3: Integrate each term. The first two terms give \[ \frac12\log|t-1| -\frac12\log|t+1|. \] Combining, \[ \frac12 \log\left| \frac{t-1}{t+1} \right|. \] For the third integral, \[ \int\frac{dt}{2t^2-1} = \frac1{2\sqrt2} \log \left| \frac{\sqrt2t-1} {\sqrt2t+1} \right|. \] Hence \[ I = \frac12 \log \left| \frac{t-1}{t+1} \right| -\frac1{\sqrt2} \log \left| \frac{\sqrt2t-1} {\sqrt2t+1} \right| +C. \]

Step 4: Substitute back \(t=\cos x\). \[ I = \frac12 \log \left| \frac{\cos x-1} {\cos x+1} \right| -\frac1{\sqrt2} \log \left| \frac{\sqrt2\cos x-1} {\sqrt2\cos x+1} \right| +C. \]

Step 5: Final Answer. \[ \boxed{ \frac12 \log \left| \frac{\cos x-1} {\cos x+1} \right| -\frac1{\sqrt2} \log \left| \frac{\sqrt2\cos x-1} {\sqrt2\cos x+1} \right| +C } \] Hence option (C) is correct.