# What is the evaluation of the following of Integral?: int1/(cos3x-cosx)dx

## ans - $\frac{1}{4} | \cos e c x - \log | \sec x + \tan x | | + C$

Mar 17, 2018

$I = \frac{1}{4} \left[\cos e c x - \log | \sec x + \tan x |\right] + c$

#### Explanation:

We have, color(red)(cos3x=4cos^3x-3cosx
$\implies \cos 3 x - \cos x = 4 {\cos}^{3} x - 3 \cos x - \cos x$$\implies \cos 3 x - \cos x = 4 {\cos}^{3} x - 4 \cos x = 4 \cos x \left({\cos}^{2} x - 1\right)$
$\implies \cos 3 x - \cos x = - 4 \cos x \left(1 - {\cos}^{2} x\right) = - 4 \cos x {\sin}^{2} x$
Hence,
$I = \int \frac{1}{\cos 3 x - \cos x} \mathrm{dx} = \int \frac{\textcolor{red}{1}}{- 4 \cos x {\sin}^{2} x} \mathrm{dx}$
$I = - \frac{1}{4} \int \frac{\textcolor{red}{\left({\cos}^{2} x + {\sin}^{2} x\right)}}{\cos x {\sin}^{2} x} \mathrm{dx}$
$I = - \frac{1}{4} \int \left[{\cos}^{2} \frac{x}{\cos x {\sin}^{2} x} + {\sin}^{2} \frac{x}{\cos x {\sin}^{2} x}\right] \mathrm{dx}$
$I = - \frac{1}{4} \int \left[\cos \frac{x}{\sin} ^ 2 x + \frac{1}{\cos} x\right] \mathrm{dx}$
$I = - \frac{1}{4} \int \left[\cos e c x \cot x \mathrm{dx} + \sec x\right] \mathrm{dx}$
$I = - \frac{1}{4} \left[- \cos e c x + \log | \sec x + \tan x |\right] + c$
$I = \frac{1}{4} \left[\cos e c x - \log | \sec x + \tan x |\right] + c$
Hint:,
$\left(1\right) \int \cos e c x \cot x \mathrm{dx} = - \cos e c x + c$
$\left(2\right) \int \sec x \mathrm{dx} = \log | \sec x + \tan x | + c$