Question

$\int \frac{\text{cosec}x}{{\cos }^2(1+\log \tan \frac{x}{2})}dx$ is equal to

• A. ${\sin }^2[1+\log \tan \frac{x}{2}]+C$
• B. $\tan [1+\log \tan \frac{x}{2}]+C$
• C. ${\sec }^2[1+\log \tan \frac{x}{2}]+C$
• D. $-\tan [1+\log \tan \frac{x}{2}]+C$

Answer 1

The correct option is B $\tan [1+\log \tan \frac{x}{2}]+C$

Let $I=\int \frac{\text{cosec}x}{{\cos }^2(1+\log \tan \frac{x}{2})}dx$
Put $(1+\log \tan \frac{x}{2})=t$

$\Rightarrow \frac{1}{\tan \frac{x}{2}}\cdot {\sec }^2\frac{x}{2}\cdot \frac{1}{2}dx=dt$
$\Rightarrow \text{cosec}xdx=dt$
Therefore, $I=\int \frac{1}{{\cos }^{2}t}dt$

$=\int {\sec }^{2}tdt=\tan t+C$
$=\tan [1+\log \tan \frac{x}{2}]+C$