Question

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

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

Answer 1

Answer: (A)

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