Question

Evaluate the integral \( \int \frac{1 - k \cos^2 x}{\sin^k x \cos^2 x} dx \).

• A. \( \frac{\tan x}{\sin^{k+1} x} + C \)
• B. \( \frac{\tan x}{\sin^k x} + C \)
• C. \( \sin^k x \sec^2 x + C \)
• D. \( k \sin^{k-1} x \cos x + C \)

Hint:

Trigonometric Derivative Patterns}
For complex trig integrals, test differentiation of likely answers
\(\frac{d}{dx}\left( \frac{\tan x}{\sin^k x} \right)\) gives the required pattern
Try reverse-engineering instead of full integration

Answer 1

The Correct Option is B

We analyze: \[ \int \frac{1 - k \cos^2 x}{\sin^k x \cos^2 x} dx = \int \left( \frac{1}{\sin^k x \cos^2 x} - \frac{k}{\sin^k x} \right) dx \] Let’s guess: \[ \frac{d}{dx} \left( \frac{\tan x}{\sin^k x} \right) = \frac{1 - k \cos^2 x}{\sin^k x \cos^2 x} \Rightarrow \text{So integral is } \frac{\tan x}{\sin^k x} + C \]