Question

Evaluate the integral: \[ \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx \]

• A. \( \frac{2}{\cos^2 x} \)
• B. \( \frac{2}{\sin^2 x} \)
• C. \( \frac{2}{\cos x} \)
• D. \( \frac{2}{\sin x} \)

Hint:

Remember to use standard trigonometric identities to simplify complex expressions and recognize common integrals.

Answer 1

The Correct Option is A

We are asked to evaluate the following integral: \[ I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx \] First, notice that we can express \( \tan x \) in terms of \( \sin x \) and \( \cos x \): \[ \tan x = \frac{\sin x}{\cos x} \] Thus, the integral becomes: \[ I = \int \frac{\sqrt{\frac{\sin x}{\cos x}}}{\sin x \cos x} \, dx = \int \frac{\sqrt{\sin x}}{\sqrt{\cos x}} \cdot \frac{1}{\sin x \cos x} \, dx \] Simplifying the expression, we get: \[ I = \int \frac{1}{\cos^2 x} \, dx \] This is a standard integral, which simplifies to: \[ I = 2 \tan x + C \] Thus, the solution to the integral is: \[ \boxed{2 \sec^2 x} \]