Evaluate the following integrals:

Let $x=\cos 2 t$ and $t=\cos ^{-} x \frac{x}{2}$

$=\sqrt{\frac{1+x}{1-x}}=\sqrt{\frac{1+\cos 2 t}{1-\cos 2 t}}$

We know $1+\cos 2 t=2 \cos ^{2} t$ and $1-2 \cos 2 t=2 \sin ^{2} t$

Hence, $\sqrt{\frac{1+\cos 2 t}{1-\cos 2 t}}=\sqrt{\frac{\cos ^{2} t}{\sin ^{2} t}=\sqrt{\cot ^{2} t}}=\cot t$

Therefore, $\int \cos \left\{2 \cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right\} \mathrm{dx}=\int \cos \theta \mathrm{dx}$

Put $\mathrm{t}=\cos ^{-} \mathrm{x} \frac{\mathrm{x}}{2}$

$=\int \cos \theta \mathrm{dx}=\int \cos \frac{\cos ^{-} \mathrm{x}}{2} \mathrm{dx}=\int \frac{\mathrm{x}}{2} \mathrm{dx}=\frac{1}{2} \frac{\mathrm{x}^{2}}{2}+\mathrm{c}=\frac{\mathrm{x}^{2}}{4}+\mathrm{c}$