Question

If \[ \int \frac{dx}{2\sin^2x+\sin2x} = \frac12\log|f(x)|+C \] and \[ f\!\left(\frac{\pi}{4}\right)=\frac12, \] then \(f(x)\) is

• A. \[ \frac{\sin x}{1+\sin x} \]
• B. \[ \frac{\cos x}{1+\cos x} \]
• C. \[ \frac{\tan x}{1+\tan x} \]
• D. \[ \frac{\cot x}{1+\cot x} \]

Hint:

For integrals involving \[ \sin x+\cos x, \] the substitution \[ t=\tan x \] after dividing by \(\cos^2x\) often converts the integral into a simple rational function.

Answer 1

The Correct Option is C

Step 1: Simplify denominator: \(2\sin^2 x + \sin 2x = 2\sin^2 x + 2\sin x\cos x = 2\sin x(\sin x + \cos x)\). Integral: \(I = \int \frac{dx}{2\sin x(\sin x+\cos x)}\). Divide numerator and denominator by \(\cos^2 x\): \(I = \frac12 \int \frac{\sec^2 x\,dx}{\tan x(1+\tan x)}\).

Step 2: Substitute \(t = \tan x\), \(dt = \sec^2 x\,dx\). \(I = \frac12 \int \frac{dt}{t(1+t)}\). Partial fractions: \(\frac1{t(1+t)} = \frac1t - \frac1{1+t}\). So \(I = \frac12(\ln|t| - \ln|1+t|) + C = \frac12 \ln\left|\frac{\tan x}{1+\tan x}\right| + C\). Thus \(f(x) = \frac{\tan x}{1+\tan x}\).

Step 3: Check \(f(\pi/4) = \frac{1}{1+1} = 1/2\), matches condition.

Step 4: Write the final answer. \(\boxed{f(x)=\frac{\tan x}{1+\tan x}}\)