Question

evaluate $int_{0}^{pi/18}5cos^{5}9x dx$.

a. $u = \frac{1}{9}cos9x$
b. $u=cos9x$
c. $u = 9x$
d. $u=sin9x$
e. $u=\frac{1}{9}sin9x$

rewrite the integral found above using this change $int_{0}^{square}square du$

Explanation:

Step1: Recall derivative of sine

The derivative of $\sin(ax)$ with respect to $x$ is $a\cos(ax)$. Here $a = 9$, so the derivative of $\sin(9x)$ with respect to $x$ is $9\cos(9x)$. We want to use substitution to simplify the integral $\int_{0}^{\pi/18}5\cos^{5}(9x)dx$. If we let $u=\sin(9x)$, then $du = 9\cos(9x)dx$ and $\cos(9x)dx=\frac{1}{9}du$. Also, we can rewrite $\cos^{5}(9x)=(1 - \sin^{2}(9x))^{2}\cos(9x)$.

Step2: Change the limits of integration

When $x = 0$, $u=\sin(9\times0)=0$. When $x=\frac{\pi}{18}$, $u=\sin(9\times\frac{\pi}{18})=\sin(\frac{\pi}{2}) = 1$.

Step3: Rewrite the integral

The original integral $\int_{0}^{\pi/18}5\cos^{5}(9x)dx$ can be rewritten. Since $\cos^{5}(9x)=(1 - \sin^{2}(9x))^{2}\cos(9x)$ and $u = \sin(9x)$, $\cos(9x)dx=\frac{1}{9}du$, the integral becomes $\int_{0}^{1}5(1 - u^{2})^{2}\times\frac{1}{9}du=\frac{5}{9}\int_{0}^{1}(1 - 2u^{2}+u^{4})du$.

Answer:

D. $u = \sin(9x)$; $\frac{5}{9}\int_{0}^{1}(1 - 2u^{2}+u^{4})du$