Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

Evaluate $\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx$.

Options:

$\sin x - \cos x + C$

$x + C$

$\log|\sin x + \cos x| + C$

$2x + C$

Correct Answer:

$x + C$

Explanation:

The correct answer is Option (2) → $x + C$

Let $I = \int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx = \int \frac{(\sin x + \cos x)}{\sqrt{\sin^2 x + \cos^2 x + 2\sin x \cos x}} dx$

$[∵\sin^2 x + \cos^2 x = 1]$

$= \int \frac{\sin x + \cos x}{\sqrt{(\sin x + \cos x)^2}} dx = \int 1 dx = x + C \quad [∵(a+b)^2 = a^2 + b^2 + 2ab]$