Question

$\int \frac{cos5x+cos4x}{1-2cos3x}dx$

Answer 1

$\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}dx$

To solve this goofy looking integral

$\cos \theta =2{\cos }^2\frac{\theta }{2}-1\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}dx=\int \frac{2\cos \left(\frac{5x+4x}{2}\right)\cos \left(\frac{5x-4x}{2}\right)}{1-2\left[2{\cos }^2\left(\frac{3x}{2}\right)-1\right]}dx=\int \frac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)}{3-4{\cos }^2\left(\frac{3x}{2}\right)}dx$

Multiply and divide by

$\int \frac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)}{3-4{\cos }^2\left(\frac{3x}{2}\right)}\times \frac{\cos \left(\frac{3x}{2}\right)}{\cos \left(\frac{3x}{2}\right)}dx=\int \frac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right)}{3\cos \left(\frac{3x}{2}\right)-4{\cos }^2\left(\frac{3x}{2}\right)}dx=-\int \frac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right)}{\cos (3\times \frac{3x}{2})}dx=-\int \frac{2\cos \left(\frac{9x}{2}\right)\cos \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right)}{\cos \left(\frac{9x}{2}\right)}dx=-\int 2\cos \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right)dx=-\int \left(\cos 2x+\cos x\right)dx=--\int \cos 2xdx-\int \cos xdx=-\frac{\sin 2x}{2}-\sin x+C$