# Thread: integration using trigonmetric functions

1. ## integration using trigonmetric functions

integrate question:

$\displaystyle = \int (3sin(4x+5) - 6cos(7-8x)) dx$

any help, im thinking it has to do with identities?

2. It looks kind of evil, but it's actually not that difficult. You can split it up into two different integrals. $\displaystyle 3\int (sin(4x+5))dx - 6\int (cos(7-8x))dx$

Now use u-subs, taking the 4x+5 and the 7-8x to be the u's for the two different integrals. I think it yields:
$\displaystyle (-3/4)cos(4x+5) +(3/4)sin(7-8x) + C$

3. Originally Posted by DJMinkus
It looks kind of evil, but it's actually not that difficult. You can split it up into two different integrals. $\displaystyle 3\int (sin(4x+5))dx - 6\int (cos(7-8x))dx$

Now use u-subs, taking the 4x+5 and the 7-8x to be the u's for the two different integrals. I think it yields:
$\displaystyle (-3/4)cos(4x+5) +(3/4)sin(7-8x) + C$
shouldnt it be $\displaystyle (-3/4)cos(4x+5)-(3/4)sin(7-8x) + C$ ? if theres a minus there already and u have a plus from the sin coming in for the cos?

4. But don't forget, the coefficient for x in the sin function is -8, making du = -8 and thus causing the two minuses (from the original equation and then from the integration) to cancel out, yielding a positive.