Evaluate the following integrals:

Solution:

Let $I=\int \frac{\cos x}{1+\cos x} d x$

On multiplying and dividing $(1-\cos x)$, we can write the integral as

$I=\int \frac{\cos x}{1+\cos x}\left(\frac{1-\cos x}{1-\cos x}\right) d x$

$\Rightarrow I=\int \frac{\cos x(1-\cos x)}{(1+\cos x)(1-\cos x)} d x$

$\Rightarrow I=\int \frac{\cos x-\cos ^{2} x}{1-\cos ^{2} x} d x$

$\Rightarrow I=\int \frac{\cos x-\cos ^{2} x}{\sin ^{2} x} d x\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$\Rightarrow I=\int\left(\frac{\cos x}{\sin ^{2} x}-\frac{\cos ^{2} x}{\sin ^{2} x}\right) d x$

$\Rightarrow I=\int\left(\frac{1}{\sin x} \times \frac{\cos x}{\sin x}-\frac{\cos ^{2} x}{\sin ^{2} x}\right) d x$

$\Rightarrow I=\int\left(\operatorname{cosecx} \cot x-\cot ^{2} x\right) d x$

$\Rightarrow \mathrm{I}=\int\left(\operatorname{cosec} \mathrm{x} \cot \mathrm{x}-\operatorname{cosec}^{2} \mathrm{x}+1\right) \mathrm{dx}\left[\because \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\right]$

$\Rightarrow \mathrm{I}=\int \operatorname{cosec} \mathrm{x} \cot \mathrm{x} \mathrm{dx}-\int \operatorname{cosec}^{2} \mathrm{x} \mathrm{dx}+\int \mathrm{dx}$

Recall $\int \operatorname{cosec}^{2} x d x=-\cot x+c$ and $\int d x=x+c$

We also have $\int \operatorname{cosecx} \cot x d x=-\operatorname{cosec} x+c$

$\Rightarrow 1=-\operatorname{cosec} x-(-\cot x)+x+c$

$\Rightarrow 1=-\operatorname{cosec} x+\cot x+x+c$

Thus, $\int \frac{\cos x}{1+\cos x} d x=-\operatorname{cosec} x+\cot x+x+c$