Question

If $\int \frac{\cos (13x)-\cos (14x)}{1+2\cos (9x)}dx=\frac{\sin (4x)}{a}-\frac{\sin (5x)}{b}+c$, then $a^b=$

• A. $4^5$
• B. $5^4$
• C. $4^4$
• D. $5^5$

Answer 1

Answer: (A)

Given, $I=\int \frac{\cos 13x-\cos 14x}{1+2\cos 9x}dx$
Let, $f(x)=\frac{\cos 13x-\cos 14x}{1+2\cos 9x}$
$=\frac{\sin 9x(\cos 13x-\cos 14x)}{\sin 9x+\sin 18x}$
$=\frac{\sin 9x-2\sin \frac{27}{2}x\cdot \sin \left(-\frac{x}{2}\right)}{2\sin \frac{27x}{2}\cdot \cos \left(-\frac{9x}{2}\right)}$
$=\frac{\sin 9x\cdot \sin \frac{x}{2}}{\cos \left(\frac{9x}{2}\right)}=2\sin \frac{9x}{2}\cdot \sin \frac{x}{2}$
$=\cos \left(\frac{9x}{2}-\frac{x}{2}\right)-\cos \left(\frac{9x}{2}+\frac{x}{2}\right)$
$=\cos 4x-\cos 5x$
So, $I=\int (\cos 4x-\cos 5x)dx$
$=\frac{\sin 4x}{4}-\frac{\sin 5x}{5}+C$
Hence, $a^b=4^5$