# Evaluate the following integrals:

Question:

Evaluate the following integrals:

Let $2 \cos x=t$

Solution:

Then $d t=-2 \sin x d x$

Or, $\sin \mathrm{x} \mathrm{dx}=-\frac{\mathrm{dt}}{2}$

Therefore, $\int \frac{\sin x}{\sqrt{4 \cos ^{2} x-1}} d x=\int-\frac{d t}{2 \sqrt{\left(t^{2}-1^{2}\right)}}$

Since, $\int \frac{1}{\sqrt{\left(x^{2}-a^{2}\right)}} d x=\log \left[x+\sqrt{\left(x^{2}-a^{2}\right)}\right]+c$

Therefore, $\int-\frac{d t}{2 \sqrt{\left(t^{2}-1^{2}\right)}}=-\frac{1}{2} \operatorname{lod}\left[t+\sqrt{t^{2}-1}\right]+c$

$=-\frac{1}{2} \log \left[2 \cos x+\sqrt{4 \cos ^{2} x-1}\right]+c$