Question

$\int \frac{cos7x-cos8x}{1+2cos5x}dx$

Answer 1

$\int \frac{\cos 7x-\cos 8x}{1+2\cos 5x}=-\int \frac{\cos 8x-\cos 7x}{2\cos \left(5x\right)+1}$

Simplify using trigonometric/hyperbolic identies:-

$\to \int \left(\cos \right(3x)-\cos (2x\left)\right)dx$

$\Rightarrow$ Put $3x=u\to dx=\frac{1}{3}du$

Put $2x=v\to dv=\frac{1}{2}du$

$=\frac{1}{3}\int \cos udu-\frac{1}{2}\int \cos vdv$

$=\frac{\sin u}{3}-\frac{1}{2}\sin v+c$

$=\frac{\sin \left(3x\right)}{3}-\frac{1}{2}\sin \left(2x\right)+c$

$\Rightarrow -\int \frac{\cos 8x-\cos 7x}{2\cos \left(5x\right)+1}=\frac{\sin \left(3x\right)}{3}-\frac{\sin \left(2x\right)}{2}$

$\Rightarrow \int \frac{\cos 7x-\cos 8x}{1+2\cos \left(5x\right)}=\frac{\sin \left(2x\right)}{2}-\frac{\sin \left(3x\right)}{3}+c$.