The value of $\int \frac{1+\sin x}{1-\sin x} d x$ is

$2 \tan \left(\frac{x}{2}+\frac{\pi}{4}\right)+C$

$2 \tan ^2\left(\frac{x}{2}+\frac{\pi}{4}\right)-x+C$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

The value of

\int \frac{1+\sin x}{1-\sin x} d x

View answer and explanation

Correct option: C

2 \tan \left(\frac{x}{2}+\frac{\pi}{4}\right)-x+C

Option C (

2 \tan \left(\frac{x}{2}+\frac{\pi}{4}\right)-x+C

) is correct because it is the result of integrating the given expression and simplifying using trigonometric identities.

Practice All Chapter MCQs →

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{C}

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