# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{\cos x+\cos e c x} d x$

Solution:

let $I=\frac{1}{\cos x+\operatorname{cosec} x} d x$

Multiply and divide by $\sin x$

$I=\frac{\frac{1}{\sin x}}{\frac{\cos x}{\sin x}+\frac{\operatorname{cosecx}}{\sin x}} d x$

$=\frac{\operatorname{cosec} x}{\cot x+\operatorname{cosec}^{2} x} d x$

$=\frac{\operatorname{cosec} x}{\cot x+1+\cot ^{2} x} d x$

$=\frac{\operatorname{cosec} x}{\cot ^{2} x+\cot x+1} d x$

Let $\cot x=t$

$-\operatorname{cosec} x d x=d t$

So, I $=-\int \frac{d t}{t^{2}+t+1}$

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

$=\int \frac{d t}{\left(t+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}$

$=\frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1} \frac{\mathrm{t}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}+\mathrm{c}$

$=\frac{2}{\sqrt{3}} \tan ^{-1} \frac{2 \mathrm{t}+1}{\sqrt{3}}+\mathrm{c}$

$=\frac{2}{\sqrt{3}} \tan ^{-1} \frac{2 \cot \mathrm{x}+1}{\sqrt{3}}+\mathrm{c}$