Question

Solve the equation
$\int \frac{\cos 7x-\cos 8x}{1+2\cos 5x}dx$

Answer 1

$I=\int \frac{\cos 7x-\cos 8x}{1+2\cos 5x}dx$

first are multiplying $\sin 5x$ below and above we get

$I=\int \frac{(\cos 7x-\cos 8x)\sin 5x}{\sin 5x+2\sin 5x\cos 5x}dx$

$I=\int \frac{(\cos 7x-\cos 8x)\sin 5x}{\sin 5x+\sin 10x}dx\{\because \sin 2A=2\sin A\cos A\}$

$I=\int \frac{2\sin \left(\frac{7x+8x}{2}\right)\sin \left(\frac{8x-7x}{2}\right)}{2\sin \left(\frac{5x+10x}{2}\right)\left(\frac{10x-5x}{2}\right)}\sin 5xdx$

$I=\int \frac{2\sin 15x/2\sin x/2\sin 5x}{2\sin 15/2\cos \frac{5}{2}x}dx(\because \cos C-\cos D\Rightarrow 2\sin \frac{C+D}{2}\sin \left(\frac{D-C}{2}\right))$

$I=\int \frac{\sin x/2\sin 5x}{\cos \left(\frac{5x}{2}\right)}dx$

Here $\sin \left(5x\right)=\sin (2\times \frac{5x}{2})$

$I=\int \frac{\sin x/2\times 2\sin \frac{5x}{2}\cos \frac{5x}{2}}{\cos 5x/2}dx$

$I=2\int \sin x/2\sin 5x/2dx\left\{2\sin \frac{C+D}{2}\sin \frac{D-C}{2}=\cos C-\cos D\right\}$

$I=\int [\cos \left(2x\right)-\cos 3x]dx$

$I=\frac{\sin 2x}{2}-\frac{\sin 3x}{3}+C$