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 the denominator. \[ 2\sin^2x+\sin2x = 2\sin^2x+2\sin x\cos x. \] \[ = 2\sin x(\sin x+\cos x). \] Hence \[ I = \int \frac{dx} {2\sin x(\sin x+\cos x)}. \] Dividing numerator and denominator by \(\cos^2x\), \[ I = \frac12 \int \frac{\sec^2x\,dx} {\tan x(1+\tan x)}. \] Let \[ t=\tan x. \] Then \[ dt=\sec^2x\,dx. \] Therefore, \[ I = \frac12 \int \frac{dt}{t(1+t)}. \]

Step 2: Use partial fractions. \[ \frac1{t(1+t)} = \frac1t-\frac1{1+t}. \] Thus \[ I = \frac12 \int \left( \frac1t-\frac1{1+t} \right)dt. \] \[ = \frac12 \left[ \log|t| - \log|1+t| \right] +C. \] \[ = \frac12 \log \left| \frac{t}{1+t} \right| +C. \] Substituting \[ t=\tan x, \] \[ I = \frac12 \log \left| \frac{\tan x} {1+\tan x} \right| +C. \] Comparing with \[ \frac12\log|f(x)|+C, \] we get \[ f(x) = \frac{\tan x} {1+\tan x}. \]

Step 3: Verify the condition. \[ f\!\left(\frac{\pi}{4}\right) = \frac{1}{1+1} = \frac12, \] which satisfies the given condition.

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