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Antiderivative Help

  • Thread starter Coltjb7
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
vela
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Rewrite it in terms of ##\sec u##.
 
  • #3
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Your substitution is the one that will work, but take vela's advice, and write the integrand in terms of sec(u).
 
  • #4
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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?
 
  • #5
15
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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)
 
  • #6
vela
Staff Emeritus
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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.
 
  • #7
33,072
4,772
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.
 
  • #8
33,072
4,772
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##
 

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