# Homework Help: Basic Integration win sin and cos

1. Apr 28, 2008

### duki

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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!

Last edited: Apr 28, 2008
2. Apr 28, 2008

### 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.

3. Apr 28, 2008

### duki

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

4. Apr 28, 2008

### dynamicsolo

Your're OK so far...

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

Last edited: Apr 28, 2008
5. Apr 28, 2008

### Alienjoey

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)

6. Apr 28, 2008

### duki

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

7. Apr 28, 2008

### 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.

8. Apr 28, 2008

### duki

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