Is there a quick way to do this one?
$\displaystyle \int sin(8x)cos(5x)dx $
$\displaystyle \int\sin (c_1x)\cos (c_2x)\;dx = -\frac{\cos(c_1+c_2)x}{2(c_1+c_2)}-\frac{\cos(c_1-c_2)x}{2(c_1-c_2)} \qquad\mbox{(for }|c_1|\neq|c_2|\mbox{)}\,\! $
Thats the really quick way though. How to derive it, I don't know off hand, and don't have the time to work through it. Someone else here who is good with integrals probably can though.
I have another one that I tried but couldn't get the answer.
$\displaystyle \int_{0}^{\frac{pi}{3}} tan^{5}x sec^{6}x dx $
Here's my work
$\displaystyle u = sec(x) du = sec(x)tan(x)dx$
$\displaystyle \int_{0}^{\frac{pi}{3}} (tan^{2}x)^{2} sec^{5}x tan(x)sec(x)dx $
$\displaystyle \int_{0}^{\frac{pi}{3}} (1-sec^{2}x)^{2} sec^{5}x tan(x)sec(x)dx $
$\displaystyle \int_{0}^{\frac{pi}{3}} (1- u^{2})^{2} u^{5} du $
$\displaystyle \int_{0}^{\frac{pi}{3}} u^{5} - 2u^{7} + u^{9} du $
$\displaystyle \frac{sec^{6}x}{6} - \frac{sec^{8}x}{8} + \frac{sec^{10}x}{10}$
from 0 to $\displaystyle \frac{Pi}{3} $
I plug those in and I get a huge number that isn't correct.
Do the make-up first:
$\displaystyle \tan ^5 x\sec ^6 x = \tan ^5 x\left( {\sec ^2 x} \right)^2 \sec ^2 x = \tan ^5 x\left( {1 + \tan ^2 x} \right)^2 \sec ^2 x.$
Now just make the obvious substitution. (Don't forget to change integration limits for $\displaystyle u,$ or whatever.)
I left it in terms of u so I ended up getting
$\displaystyle \frac{1}{6}\sqrt{3}^{6} - \frac{2}{8}\sqrt{3}^{8} + \frac{1}{10}\sqrt{3}^{10} $
I get $\displaystyle \frac{171}{10}$
Can anyone check if it's right?