Question

Integrate 144(sin(theta)^2)^(3/2) from 0 to 2pi.

I Simplified it to 144sin(theta)*(1-cos(theta)^2, substituting u for cos and then integrating 1-u^2.

That becomes u-(u^3)/3 which then becomes 144 [cos(theta) - (cos(theta)^3)/3].

When I evaluate it my answer comes out to 0 but it's supposed to be 384. I don't understand what I'm doing wrong.

Answer 1

\[\sqrt{a^2}=|a|\]

Answer 2

\[144 \int\limits_{0}^{2 \pi}(\sin^2(\theta))^\frac{3}{2} d \theta \]
\[144 \int\limits_{0}^{2\pi}|\sin(\theta)|^3 d \theta \]

\[144[\int\limits_{0}^{\pi}(\sin(\theta))^3 d \theta+\int\limits_{\pi}^{2 \pi}(-\sin(\theta))^3 d \theta]\]
now we need to find out how to integrate (sin(theta))^3
\[\int\limits_{}^{}\sin^3(\theta) d \theta=\int\limits_{}^{}(1-\cos^2(\theta)) \sin(\theta) d \theta\]
let u=cos(theta)=> du=-sin(theta) d theta
\[-\int\limits_{}^{}(1-u^2) du=-(u-\frac{u^3}{3})+C=-\cos(\theta)+\frac{\cos^3(\theta)}{3}+C\]
so we have
\[\int\limits_{}^{}\sin^3(\theta) d \theta=-\cos(\theta)+\frac{\cos^3(\theta)}{3}+C\]
\[-\int\limits_{}^{}\sin^3(\theta) d \theta=\cos(\theta)-\frac{\cos^3(\theta)}{3}+C\]
so we have
[\[144[(-\cos(\pi)+\frac{\cos^3(\pi)}{3})-(-\cos(0)+\frac{\cos^3(0)}{3})]\]
+
\[144[(\cos(2\pi)-\frac{\cos^3(2\pi)}{3})-(\cos(\pi)-\frac{\cos^3(\pi)}{3})]\]
\[=144[1-\frac{1}{3}+1-\frac{1}{3}]+144[1-\frac{1}{3}+1-\frac{1}{3}]\]

Answer 3

Thank you both. I appreciate it. The absolute value really threw me off.