# Antiderivative Help

1. Feb 21, 2016

### Coltjb7

1. The problem statement, all variables and given/known data
Find the antiderivative: ∫[(3x+1)/cos^2(3x^2+2x)]dx

2. Relevant equations

3. The attempt at a solution
I attempted to use "u substitution" but got stuck towards the end:
u=3x^2+2x
du=6x+2dx
=2(3x+1)dx
du/2=(3x+1)dx

After my substitutions it looks like this:
∫(du)/2cos^2(u) = 1/2∫(du)/cos^2(u)

What do I do with the cos^2?

Thanks for any help that can be given

2. Feb 21, 2016

### vela

Staff Emeritus
Rewrite it in terms of $\sec u$.

3. Feb 21, 2016

### Staff: Mentor

Your substitution is the one that will work, but take vela's advice, and write the integrand in terms of sec(u).

4. Feb 21, 2016

### Coltjb7

So I did this two different ways:
A) du/2cos^2u
=1/2 * 1/tanu *du/tanu
=du/2tan^2u
=(3x+1)/2tan^2(3x^2+2x)

OR

B) 1/2 * 1/tanu *du/tanu
=1/2 *cotu *du/tanu
=cot(du)/2tan(u)
=cot(3x+1)/2tan(3x^2+2x)

Which one if wither is correct? Why?

5. Feb 21, 2016

### Coltjb7

Sorry, I just totally avoided the sec somehow....
du/2cos^2u
=1/2 * 1/cosu * du/cosu
=1/2 * sec * sec *du
(1/2) sec^2u du
1/2 tanu du
1/2 tan(3x^2+2x)(3x+1)

6. Feb 21, 2016

### vela

Staff Emeritus
It's easy enough to check. Differentiate the result and see if you get the integrand back.

7. Feb 21, 2016

### Staff: Mentor

I can follow what you're doing, but you don't make any distinction between an integral and the resulting antiderivative. After you have found the antiderivative, the du goes away. If you follow vela's advice, you'll see that you have a mistake in the last line.

The $\Sigma$ sign on the gray menu bar has all sorts of symbols you can use.

8. Feb 21, 2016

### Staff: Mentor

Neither. Remember that $\int sec^2(u)du = tan(u) + C$