1. ## trigonometric integral

integral of cos(x)sin(2x)

2. Originally Posted by xslim12
integral of cos(x)sin(2x)
$\int cos(x)sin(2x)$

$sin (2x) = 2sinxcosx$
so,
$\int cos(x)sin(2x) = \int 2cos^2(x)sinx$

let u = cos x
so, du =-sinx dx

from here, u can continue it.. Ü
should the answer be like that?

3. Originally Posted by xslim12
integral of cos(x)sin(2x)
$sin(a + b) = sin(a)cos(b) + sin(b)cos(a)$
$sin(a - b) = sin(a)cos(b) - sin(b)cos(a)$

Thus
$sin(a + b) + sin(a - b) = 2sin(a)cos(b)$

So
$sin(a)cos(b) = \frac{sin(a + b) + sin(a - b)}{2}$

$
\int sin(2x)cos(x)~dx = \frac{1}{2} \int (sin(3x) + sin(x) )~dx$

-Dan

4. Originally Posted by topsquark
$sin(a + b) = sin(a)cos(b) + sin(b)cos(a)$
$sin(a - b) = sin(a)cos(b) - sin(b)cos(a)$

Thus
$sin(a + b) + sin(a - b) = 2sin(a)cos(b)$

So
$sin(a)cos(b) = \frac{sin(a + b) + sin(a - b)}{2}$

$
\int sin(2x)cos(x)~dx = \frac{1}{2} \int (sin(3x) + sin(x) )~dx$

-Dan
thanks man. blahh i need to start realizing these things.
Kalgota, i havn't tried your method yet, but i found this much simpler. thanks for the help both.

5. Originally Posted by xslim12
thanks man. blahh i need to start realizing these things.
Kalgota, i havn't tried your method yet, but i found this much simpler. thanks for the help both.
Yeah, but what you don't know is how many problems I ran into that I couldn't integrate before I finally started remembering this trick!

You should try kalagota's method also. It's always good to practice trig integrals (since they are, in my opinion, the nastiest kind.)

-Dan