Question

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

Answer 1

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

$\Rightarrow \int \frac{2sin\left(\frac{5x}{2}\right)}{2sin\left(\frac{5x}{2}\right)}\left[\frac{cos7x-cos8x}{1+2cos5x}\right]dx$

$=\int \frac{2sin\left(\frac{5x}{2}\right)cos7x-2sin\left(\frac{5x}{2}\right)cos8x}{\left(2sin\frac{5x}{2}\right)+4sin\frac{5x}{2}cos5x}dx$

$2sinAcosB=sin(A+B)-sin(B-A)$

$\Rightarrow \int \frac{[sin\left(\frac{19x}{2}\right)-sin\left(\frac{9x}{2}\right)]-[sin\left(\frac{21x}{2}\right)-sin\left(\frac{11x}{2}\right)]}{2(sin\frac{5x}{2}+2sin\frac{5x}{2}cos5x)}$

$\Rightarrow \int \frac{(sin\frac{19x}{2}+sin\frac{11x}{2})-(sin\frac{9x}{2}+sin\frac{21x}{2})}{2[sin\frac{5x}{2}+(sin\frac{15x}{2}-sin\frac{5x}{2})]}$

$\Rightarrow sinA+sinB=2sin\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)$

$=\int \frac{2sin\frac{15x}{2}cos\frac{4x}{2}-2sin\frac{15x}{2}cos(-3x)}{2sin\frac{15x}{2}}dx$

$=\int cos2x-cos3xdx$

$=\frac{-sin2x}{2}+\frac{sin3x}{3}+c$

$=\frac{sin3x}{3}-\frac{sin2x}{2}+c$