Question

If \[ \int\frac1{1+\cos x}\,dx = \frac1{f\!\left(\frac x2\right)}+C_1, \] then \[ \int f(x)\,dx= \ ? \]

• A. \(\log|\sin x|+C\)
• B. \(\log|\cos x|+C\)
• C. \(-\csc^2x+C\)
• D. \(\tan x+C\)

Hint:

Remember: \[ \int \cot x\,dx=\log|\sin x|+C. \] and \[ \int \tan x\,dx=-\log|\cos x|+C. \]

Answer 1

The Correct Option is A

Concept: The identity \[ 1+\cos x = 2\cos^2\frac x2 \] is extremely useful in integrals involving \(1+\cos x\).

Step 1: Evaluate the given integral.
Using \[ 1+\cos x = 2\cos^2\frac x2, \] we obtain \[ \frac1{1+\cos x} = \frac12\sec^2\frac x2. \] Hence \[ \int\frac1{1+\cos x}\,dx = \int\frac12\sec^2\frac x2\,dx. \] Let \[ t=\frac x2. \] Then \[ dx=2dt. \] Therefore \[ = \int\sec^2 t\,dt = \tan t+C. \] Thus \[ \tan\frac x2 = \frac1{f\!\left(\frac x2\right)}. \]

Step 2: Determine \(f(x)\).
Replacing \[ \frac x2=t, \] we get \[ \frac1{f(t)} = \tan t. \] Hence \[ f(t) = \cot t. \] Therefore \[ f(x)=\cot x. \]

Step 3: Integrate \(f(x)\).
\[ \int f(x)\,dx = \int\cot x\,dx. \] \[ = \log|\sin x|+C. \] Hence \[ \boxed{\log|\sin x|+C}. \]