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Trouble with trig identities in solving integral.

  1. Aug 8, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\int[/itex][itex]\frac{sec^{2}x}{tan^{4}x}[/itex]dx


    2. Relevant equations



    3. The attempt at a solution

    I have the answer as -1/3 cot^3(x) + C listed.

    All the intermediate steps are given, but the first one is they have converted the sec/tan integral in to:

    [itex]\int[/itex]{cot^{2}x}{csc^{2}x} dx

    I am a little confused by the trig identities used and manipulated to get this.
     
  2. jcsd
  3. Aug 8, 2011 #2
    you can just write it all in terms of sines and cosines and then reduce it to get that.
    without using any identities
     
  4. Aug 8, 2011 #3
    Hint: what's the derivative of [itex]tan(x)[/itex]?
     
  5. Aug 8, 2011 #4

    SammyS

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    [tex]\frac{\sec^{2}(x)}{\tan^{4}(x)}=\frac{1}{\cos^2(x)} \frac {\cos^4(x)}{\sin^4(x)}[/tex]

    Try it from there.
     
  6. Aug 8, 2011 #5

    Mark44

    Staff: Mentor

    Following TheoMcCloskey's hint, the problem can be done without converting to sines and cosines. After a fairly simple substitution, the integral looks like
    [tex]\int u^{-4}du[/tex]
     
  7. Aug 8, 2011 #6
    Thanks for the responses guys, yes the d/dx (tan x) substitution can be done to get a result of

    -1/3*1/tan^3 x + C

    But there's that other result which is bugging me, I will take a closer look at the post with the other identities as I want to see how I can get the cot^2csc^2 result.
     
  8. Aug 9, 2011 #7

    SammyS

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    Taking it one step further ...

    [itex]\displaystyle \frac{\sec^{2}(x)}{\tan^{4}(x)}=\frac{1}{\cos^2(x) } \frac {\cos^4(x)}{\sin^4(x)}[/itex]
    [itex]\displaystyle =\frac{1}{\cos^2(x) }\frac {\cos^2(x)}{\sin^2(x)}\frac {\cos^2(x)}{\sin^2(x)}[/itex]​
     
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