Evaluate the following integrals:

Given I $=\int \frac{1}{4 \cos x-1} d x$

We know that $\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

$\Rightarrow \int \frac{1}{-1+4 \cos x} d x=\int \frac{1}{-1+4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$

$=\int \frac{1+\tan ^{2} \frac{x}{2}}{-1\left(1+\tan ^{2} \frac{x}{2}\right)+4\left(1-\tan ^{2} \frac{x}{2}\right)} d x$

Replacing $1+\tan ^{2} x / 2$ in numerator by $\sec ^{2} x / 2$

$\Rightarrow \int \frac{1+\tan ^{2} \frac{x}{2}}{-1\left(1+\tan ^{2} \frac{x}{2}\right)+4\left(1-\tan ^{2} \frac{x}{2}\right)} d x=\int \frac{\sec ^{2} \frac{x}{2}}{-5 \tan ^{2} \frac{x}{2}+3} d x$

Puttingtan $\frac{x}{2}=\operatorname{tand} \frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right) d x=d t$

$\Rightarrow \int \frac{\sec ^{2} \frac{x}{2}}{-5 \tan ^{2} \frac{x}{2}+3} d x=\int \frac{d t}{3-5 t^{2}}$

$=\frac{1}{5} \int \frac{1}{\frac{3}{5}-t^{2}} d t$

We know that $\int \frac{1}{\mathrm{a}^{2}-\mathrm{x}^{2}} \mathrm{dx}=\frac{1}{2 \mathrm{a}} \log \left|\frac{\mathrm{a}+\mathrm{x}}{\mathrm{a}-\mathrm{x}}\right|+\mathrm{c}$

$\Rightarrow \frac{1}{5} \int \frac{1}{\frac{3}{5}-t^{2}} d t=\frac{1}{5}\left(\frac{1}{\sqrt{\frac{3}{5}}}\right) \log \left|\frac{\sqrt{\frac{3}{5}}+t}{\sqrt{\frac{3}{5}}-t}\right|+c$

$=\frac{1}{\sqrt{15}} \log \left|\frac{\sqrt{3}+\sqrt{5} \tan \frac{x}{2}}{\sqrt{3}-\sqrt{5} \tan \frac{x}{2}}\right|+c$

$\therefore \mathrm{I}=\int \frac{1}{4 \cos \mathrm{x}-1} \mathrm{dx}=\frac{1}{\sqrt{15}} \log \left|\frac{\sqrt{3}+\sqrt{5} \tan \frac{\mathrm{x}}{2}}{\sqrt{3}-\sqrt{5} \tan \frac{\mathrm{x}}{2}}\right|+\mathrm{c}$