Evaluate The Integral

[Ryan Rigdon]

I am stuck with this problem i tried working it out twice and got it wrong. can someone help me finish it. thank you. here is my work thus far.

 

[galactus]

One sub you could make is

\(\displaystyle cos^{3}(u)=\frac{1}{4}cos(3u)+\frac{3}{4}cos(u)\)

Now, it is an easy integral:

\(\displaystyle \frac{1}{4}\int cos(3u)du+int\frac{3}{4}cos(u)du\)

 

[Deleted member 4993]

I am stuck with this problem i tried working it out twice and got it wrong. can someone help me finish it. thank you. here is my work thus far.

\(\displaystyle \int_{\frac{-2\pi}{3}}^{\frac{2\pi}{3}}cos^3(\frac{x}{4})dx\)

\(\displaystyle \int_{\frac{-2\pi}{3}}^{\frac{2\pi}{3}}cos(\frac{x}{4})[1-sin^2(\frac{x}{4})]dx\)

let

u = x/4

4 du = dx

\(\displaystyle 4 \int_{\frac{-\pi}{6}}^{\frac{\pi}{6}}cos(u)[1-sin^2(u)]du\)

\(\displaystyle 4 [ \int_{\frac{-\pi}{6}}^{\frac{\pi}{6}}cos(u) du-\int_{\frac{-\pi}{6}}^{\frac{\pi}{6}}cos(u)sin^2(u)]du\)

\(\displaystyle 8 [sin(\frac{\pi}{6}) \ - \ \frac{1}{3}sin^3(\frac{\pi}{6})]\)

\(\displaystyle 8 [\frac{1}{2} \ - \ \frac{1}{3}\cdot \frac{1}{8}]\)

\(\displaystyle \frac{11}{3}\)