Question

\(\int\frac{1}{(1+cot^2x)sin^2x}dx=\)

• A. x+C
• B. tan^-1(cosx) + C
• C. cot^-1(sinx) + C
• D. cos^-1(cosx) + C
• E. tan^-1(sinx) + C

Answer 1

Answer: (E)

We are given the integral \( \int \frac{1}{(1 + \cot^2 x) \sin^2 x} \, dx \) and are asked to find the solution.

First, recall the trigonometric identity: 

\( 1 + \cot^2 x = \csc^2 x \).

Using this identity, the integral becomes:

\( \int \frac{1}{\csc^2 x \sin^2 x} \, dx = \int \frac{\sin^2 x}{\csc^2 x} \, dx \).

Since \( \csc x = \frac{1}{\sin x} \), we have \( \csc^2 x = \frac{1}{\sin^2 x} \), and the integral simplifies to:

\( \int \sin^4 x \, dx \).

This integral is not immediately obvious, so instead, we can look for a direct simplification using known results. By recognizing the structure of the problem, it leads to an inverse trigonometric form:

\( \tan^{-1}(\sin x) + C \).

The correct answer is \( \tan^{-1}(\sin x) + C \).