Question

Evaluate $\int \frac{\left(\cos 5x+\cos 4x\right)}{1-2\cos 3x}dx$

Answer 1

Given:
$\int \frac{(\cos 5x+\cos 4x)}{1-2\cos 3x}dxsolution;\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)\therefore \cos 5x+\cos 4x=2cos\left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)1-2\cos 3x=1-2\left(2{\cos }^2\frac{3x}{2}-1\right)=3-4{\cos }^2\frac{3x}{2}=\frac{3\cos \frac{3x}{2}-4{\cos }^3\frac{3x}{2}}{\cos \frac{3x}{2}}=\frac{-(4{\cos }^3\frac{3x}{2}-3\cos \frac{3x}{2})}{\cos \frac{3x}{2}}=\frac{-\cos (3\times \frac{3x}{2})}{\cos \frac{3x}{2}}\therefore \frac{\cos 5x+\cos 4x}{1-2\cos 3x}=\frac{2\cos \frac{9x}{2}\cos \frac{x}{2}}{-\cos \left(\frac{9x}{2}\right)/\cos \left(\frac{3x}{2}\right)}=-2\cos \frac{x}{2}\cos \frac{3x}{2}=-(\cos (\frac{3x}{2}-\frac{x}{2})+\cos (\frac{3x}{2}+\frac{x}{2}))=-(\cos x+\cos 2x)\therefore \int -2(\cos +\cos 2x)dx=-(2\sin x+\sin 2x)+C$

Hence the correct answer is$=-(2\sin x+\sin 2x)+C$