Question

Integrate:

$\int \frac{1}{\cos 3x-\cos x}dx$ ?

Answer 1

$\frac{dx}{{cos}^{3}x}-cosx=\frac{dx}{{4cos}^{3}x}$

by using identity

$cos3x=4co{s}^{3}x-3cosx$

$=\frac{dx}{4}cosx(co{s}^{2}x-1)$

$=\frac{cosxdx}{4{cos}^x({cos}^x-1)}$

$=\frac{-cosxdx}{4{sin}^x({sin}^x-1)}$

now putting $sinx=t,cosxdx=dt$

$=-\frac{dt}{4}(1-t^2)t^2$

$=\frac{1}{4}-\frac{dt}{t^2}-\frac{dt}{1-t^2}$

$=\frac{1}{4}\ast \frac{1}{t}+log1-\frac{t}{1}+t+c$

$=\frac{1}{4}\ast \frac{cosecx+log1-sinx}{1+sinx}$