Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Indefinite Integration

Question:

The value of $\int \frac{\sin x+\cos x}{3+\sin 2 x} d x$, is

Options:

$\frac{1}{4} \log \left(\frac{2+\sin x-\cos x}{2-\sin x+\cos x}\right)+C$

$\frac{1}{2} \log \left(\frac{2+\sin x}{2-\sin x}\right)+C$

$\frac{1}{4} \log \left(\frac{1+\sin x}{1-\sin x}\right)+C$

none of these

Correct Answer:

$\frac{1}{4} \log \left(\frac{2+\sin x-\cos x}{2-\sin x+\cos x}\right)+C$

Explanation:

We have,

$I =\int \frac{\sin x+\cos x}{3+\sin 2 x} d x$

$\Rightarrow I =\int \frac{1}{4-(1-\sin 2 x)} d(\sin x-\cos x)$

$\Rightarrow I =\int \frac{1}{2^2-(\sin x-\cos x)^2} d(\sin x-\cos x)$

$\Rightarrow I=\frac{1}{4} \log \left(\frac{2+\sin x-\cos x}{2-\sin x+\cos x}\right)+C$