Question

Evaluate: \[ \int \left( \sqrt{\cot x} + \sqrt{\tan x} \right) dx \]

Hint:

When solving integrals involving square roots of trigonometric functions, use substitution to simplify the expression.

Answer 1

The given integral is: \[ \int \left( \sqrt{\cot x} + \sqrt{\tan x} \right) dx \] We can split the integral into two parts: \[ \int \sqrt{\cot x} \, dx + \int \sqrt{\tan x} \, dx \] Let's solve each part separately: First Integral: \[ I_1 = \int \sqrt{\cot x} \, dx \] Using the identity \( \cot x = \frac{\cos x}{\sin x} \), we rewrite the integral: \[ I_1 = \int \sqrt{\frac{\cos x}{\sin x}} \, dx = \int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \, dx \] Now, let's use substitution. Let \( u = \sin x \), so \( du = \cos x \, dx \). The integral becomes: \[ I_1 = \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C_1 = 2\sqrt{\sin x} + C_1 \] Second Integral: \[ I_2 = \int \sqrt{\tan x} \, dx \] Using the identity \( \tan x = \frac{\sin x}{\cos x} \), we rewrite the integral: \[ I_2 = \int \sqrt{\frac{\sin x}{\cos x}} \, dx = \int \frac{\sqrt{\sin x}}{\sqrt{\cos x}} \, dx \] Now, let's use substitution. Let \( v = \cos x \), so \( dv = -\sin x \, dx \). The integral becomes: \[ I_2 = -\int \frac{1}{\sqrt{v}} \, dv = -2\sqrt{v} + C_2 = -2\sqrt{\cos x} + C_2 \] Final Answer: The final solution is: \[ \int \left( \sqrt{\cot x} + \sqrt{\tan x} \right) dx = 2\sqrt{\sin x} - 2\sqrt{\cos x} + C \] Where \( C = C_1 + C_2 \) is the constant of integration.