Evaluate the following integrals:

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

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

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

$\therefore$ we can write the given equation as

$\Rightarrow \int \frac{\cos 2 x}{1+\sin 2 x} d x$

Assume $1+\sin 2 x=t$

$\Rightarrow \frac{\mathrm{d}(1+\sin 2 \mathrm{x})}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}$

$\Rightarrow 2 \cos 2 \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{dt}$

$\therefore \cos 2 \mathrm{xd} \mathrm{x}=\frac{\mathrm{dt}}{2}$

Substituting these values in the above equation we get

$\Rightarrow \int \frac{1}{2 t} d t$

$\Rightarrow \frac{1}{2} \ln t+c$

substituting $t=1+2 \sin x$ in above equation

$\Rightarrow \frac{1}{2} \ln (1+2 \sin x)+c$