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: Choose the substitution
Let \(u=1+\log \tan \frac{x}{2}\).

Step 2: Find du
\(du=\frac{1}{\tan (x/2)}\cdot \sec ^{2}\frac{x}{2}\cdot \frac{1}{2}dx=\frac{dx}{2\sin \frac{x}{2}\cos \frac{x}{2}}=\csc x\,dx\).

Step 3: Integrate
The integral becomes \(\int \frac{du}{\cos ^{2}u}=\int \sec ^{2}u\,du=\tan u+c\).

Step 4: Substitute back
\(\tan \left[1+\log \left(\tan \frac{x}{2}\right)\right]+c\).

Final Answer:
The integral equals \(\tan \left[1+\log \tan \frac{x}{2}\right]+c\). \[ \boxed{\text{(A) }\tan \left[1+\log \left(\tan \dfrac{x}{2}\right)\right]+c} \]