Evaluate the following integrals:

Given $I=\int \frac{1}{5+7 \cos x+\sin x} d x$

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

$\Rightarrow \int \frac{1}{5+\sin x+7 \cos x} d x=\int \frac{1}{5+\left(\frac{2 \tan \frac{x}{2}}{1+\tan \frac{2}{2}}\right)+7\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$

$=\int \frac{1+\tan ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}+7-7 \tan ^{2} \frac{x}{2}} d x$

Replacing $1+\tan ^{2} x / 2$ in numerator by $\sec ^{2} x / 2$ and putting $\tan x / 2=t$ and $\sec ^{2} x / 2 d x=2 d t$,

$\Rightarrow \int \frac{1+\tan ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}+7-7 \tan ^{2} \frac{x}{2}} \mathrm{dx}=\int \frac{\sec ^{2} \frac{x}{2}}{-2 \tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}+12} \mathrm{dx}$

$=\int \frac{2 d t}{-2 t^{2}+2 t+12}$

$=-\int \frac{1}{t^{2}-t-6} d t$

$=-\int \frac{1}{\left(t-\frac{1}{2}\right)^{2}-\frac{5^{2}}{2}} d t$

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

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

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

$\therefore \mathrm{I}=\int \frac{1}{5+7 \cos \mathrm{x}+\sin \mathrm{x}} \mathrm{dx}=\frac{-1}{5} \log \left|\frac{\tan \frac{\mathrm{x}}{2}-3}{\tan \frac{\mathrm{x}}{2}+2}\right|+\mathrm{c}$