# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{\sin x-\sqrt{3} \cos x} d x$

Solution:

Given $\mathrm{I}=\int \frac{1}{\sin \mathrm{x}-\sqrt{3} \cos \mathrm{x}} \mathrm{dx}$

Let $1=r \cos \theta$ and $\sqrt{3}=r \sin \theta$

$r=\sqrt{3+1}=2$

And $\tan \theta=\sqrt{3} \rightarrow \theta=\pi / 3$

$\Rightarrow \int \frac{1}{\sin x-\sqrt{3} \cos x} d x=\int \frac{1}{r \cos \theta \sin x-r \sin \theta \cos x} d x$

$=\frac{1}{r} \int \frac{1}{\sin (x-\theta)} d x$

$=\frac{1}{r} \int \operatorname{cosec}(x-\theta) d x$

We know that $\int \operatorname{cosec} x d x=\log \left|\tan \frac{x}{2}\right|+c$

$\Rightarrow \frac{1}{r} \int \operatorname{cosec}(x-\theta) d x=\frac{1}{2} \log \left|\tan \left(\frac{x}{2}-\frac{\theta}{2}\right)\right|+c$

$=\frac{1}{2} \log \left|\tan \left(\frac{x}{2}-\frac{\pi}{6}\right)\right|+c$

$\therefore \mathrm{I}=\int \frac{1}{\sin \mathrm{x}-\sqrt{3} \cos \mathrm{x}} \mathrm{dx}=\frac{1}{2} \log \left|\tan \left(\frac{\mathrm{x}}{2}-\frac{\pi}{6}\right)\right|+\mathrm{c}$