# integration using trigonmetric functions

• Oct 23rd 2008, 09:58 PM
jvignacio
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?
• Oct 23rd 2008, 10:08 PM
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:
$\displaystyle (-3/4)cos(4x+5) +(3/4)sin(7-8x) + C$
• Oct 23rd 2008, 10:48 PM
jvignacio
Quote:

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?
• Oct 26th 2008, 03:12 PM
DJMinkus
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.