AmTran_Bus
  • AmTran_Bus
Help me integrate tan^3x
Mathematics
schrodinger
  • schrodinger
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AmTran_Bus
  • AmTran_Bus
|dw:1434148868513:dw|
AmTran_Bus
  • AmTran_Bus
|dw:1434148890008:dw|
AmTran_Bus
  • AmTran_Bus
So now u sub?

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anonymous
  • anonymous
sec^2 x - 1 = tan^2 x
AmTran_Bus
  • AmTran_Bus
Didn't I do that in my step above?
anonymous
  • anonymous
Oh! sorry u did
anonymous
  • anonymous
So now distribute. Do you know the antiderivative of \(\tan x\)?
AmTran_Bus
  • AmTran_Bus
Yes. ln sec x +c
anonymous
  • anonymous
Okay, so from distributing you should get \[\int \tan x\sec^2x\,dx-\int\tan x\,dx=\int\tan x\sec^2x\,dx-\ln|\sec x|+C\] You mentioned a substitution, that's a good idea. What do you think might work?
AmTran_Bus
  • AmTran_Bus
I would say u= tanx
anonymous
  • anonymous
Yep, that should work. Another one you can use: write \(\tan x\sec x\sec x\), then \(u=\sec x\) will work too.
anonymous
  • anonymous
\[\int\limits \tan x ~dx=\int\limits \frac{ \sin x }{ \cos x }dx=-\ln \left| \cos x \right|\]
AmTran_Bus
  • AmTran_Bus
|dw:1434149575520:dw|
AmTran_Bus
  • AmTran_Bus
Well, but I need to consider my du
anonymous
  • anonymous
Right, if \(u=\tan x\), then \(du=\sec^2x\,dx\), yes?
AmTran_Bus
  • AmTran_Bus
So it should say integral ( u du - rest) right?
anonymous
  • anonymous
Yeah that's \[\int u\,du-\ln|\sec x|+C\] and so on.
AmTran_Bus
  • AmTran_Bus
Nice. Let me evaluate that and plug in u and see if it is right.
AmTran_Bus
  • AmTran_Bus
Well, guess im stuck
AmTran_Bus
  • AmTran_Bus
I know part of the answer has to be|dw:1434149823843:dw|
anonymous
  • anonymous
If I had to guess what the problem was... Hint: Identities.
AmTran_Bus
  • AmTran_Bus
Hum. I gave up and looked at the book. The answer is |dw:1434149984333:dw|
AmTran_Bus
  • AmTran_Bus
But I just dont see it.
anonymous
  • anonymous
\[\int u\,du=\frac{1}{2}u^2+C\] by the power rule. You set \(u=\tan x\), so this becomes \(\dfrac{\tan ^2x}{2}+C\).
anonymous
  • anonymous
I was thinking: \[\begin{split} \int\tan^3(x)~dx &= \int \frac{\sin^3(x)}{\cos^3(x)}dx \\ &= \int \frac{1-\cos^2(x)}{\cos^3(x)}~d(-\cos(x)) &,&u=\cos(x) \\ &= -\int\left(\frac{1}{u^3}-\frac{1}{u}\right)~du \end{split}\]
anonymous
  • anonymous
Somehow it seems wrong though.
AmTran_Bus
  • AmTran_Bus
AmTran_Bus
  • AmTran_Bus
I forgot a basic step, you were right.
anonymous
  • anonymous
@wio seems right to me. \[-\int\frac{du}{u^3}+\int\frac{du}{u}=\frac{1}{2u^2}+\ln|u|+C=\frac{\sec^2x}{2}+\ln|\sec x|+C\]

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