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AmTran_Bus

  • one year ago

Help me integrate tan^3x

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  1. AmTran_Bus
    • one year ago
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    |dw:1434148868513:dw|

  2. AmTran_Bus
    • one year ago
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    |dw:1434148890008:dw|

  3. AmTran_Bus
    • one year ago
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    So now u sub?

  4. anonymous
    • one year ago
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    sec^2 x - 1 = tan^2 x

  5. AmTran_Bus
    • one year ago
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    Didn't I do that in my step above?

  6. anonymous
    • one year ago
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    Oh! sorry u did

  7. anonymous
    • one year ago
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    So now distribute. Do you know the antiderivative of \(\tan x\)?

  8. AmTran_Bus
    • one year ago
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    Yes. ln sec x +c

  9. anonymous
    • one year ago
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    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?

  10. AmTran_Bus
    • one year ago
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    I would say u= tanx

  11. anonymous
    • one year ago
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    Yep, that should work. Another one you can use: write \(\tan x\sec x\sec x\), then \(u=\sec x\) will work too.

  12. anonymous
    • one year ago
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    \[\int\limits \tan x ~dx=\int\limits \frac{ \sin x }{ \cos x }dx=-\ln \left| \cos x \right|\]

  13. AmTran_Bus
    • one year ago
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    |dw:1434149575520:dw|

  14. AmTran_Bus
    • one year ago
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    Well, but I need to consider my du

  15. anonymous
    • one year ago
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    Right, if \(u=\tan x\), then \(du=\sec^2x\,dx\), yes?

  16. AmTran_Bus
    • one year ago
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    So it should say integral ( u du - rest) right?

  17. anonymous
    • one year ago
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    Yeah that's \[\int u\,du-\ln|\sec x|+C\] and so on.

  18. AmTran_Bus
    • one year ago
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    Nice. Let me evaluate that and plug in u and see if it is right.

  19. AmTran_Bus
    • one year ago
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    Well, guess im stuck

  20. AmTran_Bus
    • one year ago
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    I know part of the answer has to be|dw:1434149823843:dw|

  21. anonymous
    • one year ago
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    If I had to guess what the problem was... Hint: Identities.

  22. AmTran_Bus
    • one year ago
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    Hum. I gave up and looked at the book. The answer is |dw:1434149984333:dw|

  23. AmTran_Bus
    • one year ago
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    But I just dont see it.

  24. anonymous
    • one year ago
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    \[\int u\,du=\frac{1}{2}u^2+C\] by the power rule. You set \(u=\tan x\), so this becomes \(\dfrac{\tan ^2x}{2}+C\).

  25. anonymous
    • one year ago
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    I was thinking: \[\begin{split} \int\tan^3(x)~dx &= \int \frac{\sin^3(x)}{\cos^3(x)}dx \\ &= \int \frac{1-\cos^2(x)}{\cos^3(x)}~d(-\cos(x)) &,&u=\cos(x) \\ &= -\int\left(\frac{1}{u^3}-\frac{1}{u}\right)~du \end{split}\]

  26. anonymous
    • one year ago
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    Somehow it seems wrong though.

  27. AmTran_Bus
    • one year ago
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    @SithsAndGiggles ok.

  28. AmTran_Bus
    • one year ago
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    I forgot a basic step, you were right.

  29. anonymous
    • one year ago
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    @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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