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Basic Integration win sin and cos

  • Thread starter duki
  • Start date
  • #1
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Homework Statement



Integrate [tex]\int\sin 6x + 3\cos 5x dx[/tex]

Homework Equations




The Attempt at a Solution



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

Here's what I've done so far

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

[tex]\frac{1}{6}\int\sin u du + \frac{3}{5}\int\cos u du[/tex]

Am I on the right track?
Thanks!
 
Last edited:

Answers and Replies

  • #2
41
0
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
264
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Answer: [tex]\frac{-1}{6}\cos 6x + \frac{3}{5}\sin 6x + C[/tex] ?
 
  • #4
dynamicsolo
Homework Helper
1,648
4
Your're OK so far...

Such is the glacial pace of my ancient home computer... Yes, your antiderivative is correct.
 
Last edited:
  • #5
41
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Answer: [tex]\frac{-1}{6}\cos 6x + \frac{3}{5}\sin 6x + C[/tex] ?
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
264
0
[tex]
\frac{-1}{6}\cos 6x + \frac{3}{5}\sin 5x + C
[/tex]
?
 
  • #7
41
0
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
264
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Yeah, I don't have as strong of an Algebra background as I should.
Thanks for the help
 

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