How Do You Integrate \(\sin 6x + 3\cos 5x\) Using Substitution?

[duki]

Homework Statement

Integrate \int\sin 6x + 3\cos 5x dx

Homework Equations

The Attempt at a Solution

The way I was taught was the 'u' substitution method.
I know that \int\sin u du = -\cos u + C and I know that \int\cos u du = \sin u + C

Here's what I've done so far

\int\sin 6x + 3\cos 5x dx
\int\sin 6x dx + 3\int\cos 5x dx
u = 6x; u = 5x
du = 6dx; du = 5dx
du/6 = dx; du/5 = dx

\frac{1}{6}\int\sin u du + \frac{3}{5}\int\cos u du

Am I on the right track?
Thanks!

 

[Alienjoey]

Yep, everything you've done looks good so far. Just evaluate the integrals and sub the u values back in and you should be finished.

 

[duki]

Answer: \frac{-1}{6}\cos 6x + \frac{3}{5}\sin 6x + C ?

 

[dynamicsolo]

Your're OK so far...

Such is the glacial pace of my ancient home computer... Yes, your antiderivative is correct.

 

[Alienjoey]

Answer: \frac{-1}{6}\cos 6x + \frac{3}{5}\sin 6x + C ?

Close, not quite. Keep in mind that you have two separate u values for the two different integrals.

(Sometimes it's easier to use both u's and v's to avoid confusion)

 

[duki]

<br /> \frac{-1}{6}\cos 6x + \frac{3}{5}\sin 5x + C<br />
?

 

[Alienjoey]

Yep, that's the answer I got. It seems like you've got the calculus down!

Sometimes the algebra is all that will mess you up, lol.

 

[duki]

Yeah, I don't have as strong of an Algebra background as I should.
Thanks for the help