Evaluate the following integrals:

Solution:

We know $\cos ^{2} x+\sin ^{2} x=1$

Also, $2 \sin x \cos x=\sin 2 x$

$1+\sin 2 x=\cos ^{2} x+\sin ^{2} x+2 \sin x \cos x=(\cos x+\sin x)^{2}$

$1-\sin 2 x=\cos ^{2} x+\sin ^{2} x-2 \sin x \cos x=(\cos x-\sin x)^{2}$

$\therefore$ The equation becomes

$\Rightarrow \int \sqrt{\frac{(\cos x-\sin x)^{2}}{(\cos x+\sin x)^{2}} d x}$

$\Rightarrow \int \frac{(\cos x-\sin x)}{(\cos x+\sin x)} d x$

Assume $\cos x+\sin x=t$

$\therefore d(\cos x+\sin x)=d t$

$=\cos x-\sin x$

$\therefore d t=\cos x-\sin x$

$\Rightarrow \int \frac{\mathrm{d} t}{t}$

$=\ln |t|+c$

But $\mathrm{t}=\cos \mathrm{x}+\sin \mathrm{x}$

$\therefore \ln |\cos x+\sin x|+c$