Question

The integral \(\int \frac{\csc x}{\cos ^{2}[1+\log (\tan \frac{x}{2})]}\ dx\) is equal to:

• A. \(\tan [1+\log (\tan \frac{x}{2})]+c\)
• B. \(\frac{1}{2}\tan [1+\log (\tan \frac{x}{2})]+c\)
• C. \(2\,\tan [1+\log (\tan \frac{x}{2})]+c\)
• D. \(\frac{1}{4}\tan [1+\log (\tan \frac{x}{2})]+c\)

Hint:

Substitute u = 1 + log tan(x/2); its differential turns out to be cosec x dx.

Answer 1

The Correct Option is A

Step 1: Differentiate the answer
If $F=\tan u$ with $u=1+\log \tan \frac{x}{2}$, then $F'=\sec ^{2}u\cdot u'$.

Step 2: Use u'
$u'=\csc x$, so $F'=\frac{\csc x}{\cos ^{2}u}$, which is the integrand.

Step 3: Conclude
So the antiderivative has coefficient 1, not $\frac12$, 2 or $\frac14$.

Final Answer:
Differentiating $\tan u$ returns the integrand exactly. \[ \boxed{\text{(A) }\tan \left[1+\log \left(\tan \dfrac{x}{2}\right)\right]+c} \]