Question

evaluate $int_{0}^{pi/18}5cos^{5}9x dx$.
e. $int_{0}^{pi/18}5(1 - sin9x)^{4}cos9x dx$
f. $int_{0}^{pi/18}5(1 - sin^{2}9x)^{2}cos9x dx$
use a change of variables to evaluate the integral. what is the best choice for u in the change of variables?
a. $u = \frac{1}{9}cos9x$
b. $u=cos9x$
c. $u = 9x$
d. $u=sin9x$
e. $u=\frac{1}{9}sin9x$

Explanation:

Step1: Rewrite the integrand

We know that $\cos^{2}\theta = 1-\sin^{2}\theta$. So, $\cos^{5}9x=\cos^{4}9x\cdot\cos9x=( \cos^{2}9x)^{2}\cos9x=(1 - \sin^{2}9x)^{2}\cos9x$. The integral $\int_{0}^{\frac{\pi}{18}}5\cos^{5}9x dx=\int_{0}^{\frac{\pi}{18}}5(1 - \sin^{2}9x)^{2}\cos9x dx$.

Step2: Choose substitution variable

We use the substitution method for integration. If we let $u = \sin9x$, then $du=9\cos9x dx$, or $\cos9x dx=\frac{1}{9}du$. When $x = 0$, $u=\sin(0)=0$; when $x=\frac{\pi}{18}$, $u=\sin(\frac{\pi}{2}) = 1$.

Answer:

D. $u = \sin9x$