Evaluate:

Question:

Evaluate: $\int \frac{1}{1-\sin \frac{x}{2}} d x$

Solution:

Let $I=\frac{1}{1-\sin \frac{x}{2}} d x$

$=\frac{1}{1-\sin \frac{x}{2}} \mathrm{dx}$

Now, Multiply with the conjugate we get,

$=\int \frac{1}{1-\sin \frac{x}{2}} \times \frac{1+\sin \frac{x}{2}}{1+\sin \frac{x}{2}} d x$

$=\int \frac{1+\sin ^{\frac{x}{2}}}{1-\sin ^{2} \frac{x}{2}} d x$

$=\int \frac{1+\sin ^{\frac{x}{2}}}{\cos ^{2} \frac{x}{2}} d x$

$=\int \frac{1}{\cos ^{2} \frac{x}{2}} d x+\int \frac{\sin \frac{x}{2}}{\cos ^{2} \frac{x}{2}} d x$

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

$=\frac{\left(\tan \frac{x}{2}\right)}{\frac{1}{2}}+\frac{\left(\sec \frac{x}{2}\right)}{\frac{1}{2}}$

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

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