Integrals of Trig Functions

  • Thread starter scorpa
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  • #1
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Hello everyone,

I am having some trouble finding the integral of tanxsec^2x. I honestly have no idea where to go with this one. I've finished all the others but this one is really screwing me up. I tried doing something with the chain rule but it didn't work out at all. I know that I am supposed to show my work, but at this point I don't have anything to show you guys. Thanks a lot for any help you can give me.
 

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  • #2
OlderDan
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scorpa said:
Hello everyone,

I am having some trouble finding the integral of tanxsec^2x. I honestly have no idea where to go with this one. I've finished all the others but this one is really screwing me up. I tried doing something with the chain rule but it didn't work out at all. I know that I am supposed to show my work, but at this point I don't have anything to show you guys. Thanks a lot for any help you can give me.
Write everything in terms of sines and cosines. Then you can do a simple substitution.
 
  • #3
Or just note that the derivative of tangent is the secant squared... u-substitution, anybody?

--J
 
  • #4
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Ok guys, thanks for the help, but I'm still completely lost. I still don't have a clue of what I should do.
 
  • #5
Are you familiar with u-substitution? You're going to have to make one to evaluate the integral. You have two choices for what to let u equal, both work. What do you think u might equal?

--J
 
  • #6
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I am sort of familiar with it. Could you let u equal tanx?
 
  • #7
OlderDan
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scorpa said:
I am sort of familiar with it. Could you let u equal tanx?
If you did, what would du be?
 
  • #8
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Ok, actually I might let u = sec^2x, then du should equal tanx+c?
 
  • #9
OlderDan
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scorpa said:
Ok, actually I might let u = sec^2x, then du should equal tanx+c?
That is not correct

You should review the derivatives of all the trig functions.
 
  • #10
You're integrating u to get du for some reason. You must differentiate u to get du.

Stick with your original substitution.

--J
 
  • #11
dextercioby
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[tex] \int \sec^{2} x \tan x \ dx=-\int \frac{d {}\cos x}{\cos^{3} x}=\frac{1}{2}\sec^{2}x +\mathcal{C} [/tex]

Daniel.
 
  • #12
shouldnt it be:
[tex]\int sec^2x~tanx~dx=\int\frac{sinx}{cos^3x}dx[/tex]
then, you could do u-sub and set [itex]u=cosx[/itex] and go from there?

*Edit*
dextercioby, you put sinx instead of cosx. i think tanx = sinx/cosx
*Edit*
 
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  • #13
OlderDan
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p53ud0 dr34m5 said:
shouldnt it be:
[tex]\int sec^2x~tanx~dx=\int\frac{sinx}{cos^3x}dx[/tex]
then, you could do u-sub and set [itex]u=cosx[/itex] and go from there?
and du would be

[tex]du = d cosx = -sinx dx[/tex]

giving

[tex] sinx dx = -d cosx[/tex]

which is what Dexter did to get his result, and what earlier hints were pointing to. It could also be done using

[tex] u= tanx [/tex]

[tex]du= d tanx = sec^2x dx[/tex]

[tex]sec^2x dx = d tanx[/tex]

as suggested by Justin early on
 
  • #14
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Oh ok....I think I am starting to get it
 

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