1. ## Evaluate the Integral

Is there a quick way to do this one?

$\int sin(8x)cos(5x)dx$

2. $\int\sin (c_1x)\cos (c_2x)\;dx = -\frac{\cos(c_1+c_2)x}{2(c_1+c_2)}-\frac{\cos(c_1-c_2)x}{2(c_1-c_2)} \qquad\mbox{(for }|c_1|\neq|c_2|\mbox{)}\,\!$

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.

3. Originally Posted by xifentoozlerix
$\int\sin (c_1x)\cos (c_2x)\;dx = -\frac{\cos(c_1+c_2)x}{2(c_1+c_2)}-\frac{\cos(c_1-c_2)x}{2(c_1-c_2)} \qquad\mbox{(for }|c_1|\neq|c_2|\mbox{)}\,\!$

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.
Interesting. I haven't actually seen this method before.

4. From $\sin(a+b)=\sin a\cos b+\cos a\sin b,$ combine this one with $\sin(a-b)$ and you'll get a formula to split the original product into a sum.

5. I have another one that I tried but couldn't get the answer.

$\int_{0}^{\frac{pi}{3}} tan^{5}x sec^{6}x dx$

Here's my work

$u = sec(x) du = sec(x)tan(x)dx$

$\int_{0}^{\frac{pi}{3}} (tan^{2}x)^{2} sec^{5}x tan(x)sec(x)dx$

$\int_{0}^{\frac{pi}{3}} (1-sec^{2}x)^{2} sec^{5}x tan(x)sec(x)dx$

$\int_{0}^{\frac{pi}{3}} (1- u^{2})^{2} u^{5} du$

$\int_{0}^{\frac{pi}{3}} u^{5} - 2u^{7} + u^{9} du$

$\frac{sec^{6}x}{6} - \frac{sec^{8}x}{8} + \frac{sec^{10}x}{10}$
from 0 to $\frac{Pi}{3}$

I plug those in and I get a huge number that isn't correct.

6. Originally Posted by FalconPUNCH!
I have another one that I tried but couldn't get the answer.

$\int_{0}^{\frac{pi}{3}} tan^{5}x sec^{6}x dx$
Do the make-up first:

$\tan ^5 x\sec ^6 x = \tan ^5 x\left( {\sec ^2 x} \right)^2 \sec ^2 x = \tan ^5 x\left( {1 + \tan ^2 x} \right)^2 \sec ^2 x.$

Now just make the obvious substitution. (Don't forget to change integration limits for $u,$ or whatever.)

7. Originally Posted by Krizalid
Do the make-up first:

$\tan ^5 x\sec ^6 x = \tan ^5 x\left( {\sec ^2 x} \right)^2 \sec ^2 x = \tan ^5 x\left( {1 + \tan ^2 x} \right)^2 \sec ^2 x.$

Now just make the obvious substitution. (Don't forget to change integration limits for $u,$ or whatever.)
To change the limits do I just plug them into whatever u equals?

8. Yes.

So by setting $u=\tan x$ get the lower integration limit by makin' $x=0$ and the upper integration limit by makin' $x=\frac\pi3.$

9. Originally Posted by Krizalid
Yes.

So by setting $u=\tan x$ get the lower integration limit by makin' $x=0$ and the upper integration limit by makin' $x=\frac\pi3.$
Thanks if I have anymore questions I'll just post them here.

10. Does this look right?

$\frac{1}{6}tan^{6}(\frac{1}{\sqrt{3}}) + \frac{2}{8}tan^{8}(\frac{1}{\sqrt{3}}) + \frac{1}{10}tan^{10}(\frac{1}{\sqrt{3}})$

it just doesn't look right to me

11. I left it in terms of u so I ended up getting

$\frac{1}{6}\sqrt{3}^{6} - \frac{2}{8}\sqrt{3}^{8} + \frac{1}{10}\sqrt{3}^{10}$

I get $\frac{171}{10}$

Can anyone check if it's right?

12. Can't see your TeX, but I plugged the integral into my TI-89 and got $\frac{981}{20}$. So, uhh, yeah.

13. Originally Posted by xifentoozlerix
Can't see your TeX, but I plugged the integral into my TI-89 and got $\frac{981}{20}$. So, uhh, yeah.
I edited it

14. $\frac{1}{6}\sqrt{3}^{6} - \frac{2}{8}\sqrt{3}^{8} + \frac{1}{10}\sqrt{3}^{10}$
There is a typo here. You should have $\frac{1}{6}\sqrt{3}^{6} + \frac{2}{8}\sqrt{3}^{8} + \frac{1}{10}\sqrt{3}^{10}$.

There is a typo here. You should have $\frac{1}{6}\sqrt{3}^{6} + \frac{2}{8}\sqrt{3}^{8} + \frac{1}{10}\sqrt{3}^{10}$.