integrate question:

$\displaystyle

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

$

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

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- Oct 23rd 2008, 09:58 PMjvignaciointegration 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 PMDJMinkus
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 PMjvignacio
- Oct 26th 2008, 03:12 PMDJMinkus
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.