The value of

Question:

$\int \frac{d x}{1+\cos x}$

Solution:

Let $\mathrm{I}=\int \frac{d x}{1+\cos x}=\int \frac{d x}{2 \cos ^{2} x / 2} \quad\left[\because 1+\cos x=2 \cos ^{2} \frac{x}{2}\right]$

$=\frac{1}{2} \int \sec ^{2} \frac{x}{2} d x=\frac{1}{2} \cdot 2 \tan \frac{x}{2}+C=\tan \frac{x}{2}+C$

Therefore,

$I=\tan \frac{x}{2}+C$

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