Find Antiderivative Help: ∫[(3x+1)/cos^2(3x^2+2x)]dx

[Coltjb7]

Homework Statement

Find the antiderivative: ∫[(3x+1)/cos^2(3x^2+2x)]dx

Homework Equations

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

 

[vela]

Rewrite it in terms of $\sec u$.

 

[Mark44]

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

 

[Coltjb7]

Rewrite it in terms of $\sec u$.

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?

 

[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?

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)

 

[vela]

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?

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

 

[Mark44]

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)

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.

 

[Mark44]

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?

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