Question

If $\int \frac{\cos 7x-\cos 8x}{1+2\cos 5x}dx$ = $\frac{1}{a}\sin ax-\frac{1}{b}\sin bx+C$, then $a+b=$

Answer 1

Multiply above and below by $2\sin \frac{5x}{2}.$

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

Now $2\sin A\cos B=\sin (A+B)+\sin (A-B)$

$\therefore I=\frac{1}{2}\int \frac{\sin \frac{19x}{2}+\sin (-\frac{9x}{2})-\{\sin \frac{21x}{2}+\sin (-\frac{11x}{2})\}}{\sin \frac{5x}{2}+\sin \frac{15x}{2}+\sin (-\frac{5x}{2})}$

$=\frac{1}{2}\int \frac{(\sin \frac{19x}{2}+\sin \frac{11x}{2})-(\sin \frac{21x}{2}+\sin \frac{9x}{2})}{\sin \frac{15x}{2}}dx$

$=\frac{1}{2}\int \frac{2\sin \frac{15x}{2}\cos 2x-2\sin \frac{15x}{2}\cos 3x}{\sin \frac{15x}{2}}dx$

$=\int (\cos 2x-\cos 3x)dx=\frac{1}{2}\sin 2x-\frac{1}{3}\sin 3x+C$

$\therefore a+b=2+3=5$