$\int \frac{1}{1+3 \sin ^2 x} d x$ is eq

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

$\int \frac{1}{1+3 \sin ^2 x} d x$ is equal to

Options:

$\frac{1}{3} \tan ^{-1}\left(3 \tan ^2 x\right)+C$

$\frac{1}{2} \tan ^{-1}(2 \tan x)+C$

$\tan ^{-1}(\tan x)+C$

none of these

Correct Answer:

$\frac{1}{2} \tan ^{-1}(2 \tan x)+C$

Explanation:

We have,

$I =\int \frac{1}{1+3 \sin ^2 x} d x=\int \frac{\sec ^2 x}{\sec ^2 x+3 \tan ^2 x} d x$

$\Rightarrow I =\frac{1}{2} \int \frac{1}{(2 \tan x)^2+1^2} d(2 \tan x)$

$\Rightarrow I =\frac{1}{2} \tan ^{-1}(2 \tan x)+C$