Integration Help (Quick if Possible!)

  • Thread starter Chewy0087
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  • #1
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Homework Statement



[tex]\int sec^2 x tan x dx[/tex]

Homework Equations





The Attempt at a Solution


hey guys, sorry about this but i realised i can't do this question (in for tommorow)

I said, let u = tan x
so du/dx = sec^2 x
so dx/du =1 / sec^x

so putting that in I get [tex]\int tan x dx or \int u du[/tex]

which of course is [tex]\frac{u^2}{2}[/tex] = [tex]\frac{tan^2 x}{2}[/tex]

however according to wolfram alpha it's sec^2 x / 2 :O

http://www.wolframalpha.com/input/?i=int+sec^2+x+tan+x

thanks again in advance, would save me :O
 

Answers and Replies

  • #2
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Your answer is correct. Another substitution you can use is u = secx, du = secx tanx dx. With that substitution you have
[tex]\int u du~=~(1/2)u^2 + C~=~(1/2)sec^2x + C[/tex]
The two answers seem to be different, and indeed they are. But they differ only by a constant, since (1/2)sec^2(x) = (1/2)tan^(x) + 1/2.

The upshot is that you are correct and wolfram alpha is also correct.
 
  • #3
370
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i see now, much obliged sir!
 

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