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Math Help - [SOLVED] Trigonometric integration by substitution

  1. #1
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    [SOLVED] Trigonometric integration by substitution

    \int \frac {1-tanx^2}{secx^2} ~dx

    I figure you have to break it up first:

    \int cosx^2 ~dx - \int \frac {tanx^2}{secx^2} ~dx

    But from here, I am somewhat stuck. Would \int \frac {tanx^2}{secx^2} ~dx turn into  \int sinx^2 ~dx?

    If so, then I guess the next part of the question is, how do you take the anti derivative of cosx^2?
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  2. #2
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    Quote Originally Posted by redman223 View Post
    \int \frac {1-tanx^2}{secx^2} ~dx

    I figure you have to break it up first:

    \int cosx^2 ~dx - \int \frac {tanx^2}{secx^2} ~dx

    But from here, I am somewhat stuck. Would \int \frac {tanx^2}{secx^2} ~dx turn into  \int sinx^2 ~dx?

    If so, then I guess the next part of the question is, how do you take the anti derivative of cosx^2?
    Yes.

    \cos (2x) = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 (x) - 1

    Therefore \sin^2 x = \frac{1 - \cos (2x)}{2} and \cos^2 x = \frac{\cos (2x) + 1}{2}\, ......


    But of course, you could always recognise that \frac{1- \tan^2 x}{\text{sec} ^2 x} = \cos^2 x - \sin^2 x = \cos (2x) and save yourself a lot of trouble ......
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  3. #3
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    Quote Originally Posted by redman223 View Post
    \int \frac {1-tanx^2}{secx^2} ~dx

    I figure you have to break it up first:

    \int cosx^2 ~dx - \int \frac {tanx^2}{secx^2} ~dx

    But from here, I am somewhat stuck. Would \int \frac {tanx^2}{secx^2} ~dx turn into  \int sinx^2 ~dx?

    If so, then I guess the next part of the question is, how do you take the anti derivative of cosx^2?
    \int \frac {1-tanx^2}{secx^2} ~dx

    Are you sure its that integral or this?

     <br />
\int \frac {1-(tanx)^2}{(secx)^2} ~dx<br />

    Because if its the first, then its not going to be simple.(I was kindof thrown off by the X^2 part)
    Last edited by Msquared; July 2nd 2008 at 08:03 PM. Reason: rewording
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Msquared View Post
    \int \frac {1-tanx^2}{secx^2} ~dx

    Are you sure its that integral or this?

     <br />
\int \frac {1-(tanx)^2}{(secx)^2} ~dx<br />

    Because if its the first, then its not going to be simple.(I was kindof thrown off by the X^2 part)
    \frac{1-\tan^2(x)}{\sec^2(x)}=\frac{2-(\tan^2(x)+1)}{\tan^2(x)+1}=\frac{2}{\sec^2(x)}-1=2\cos^2(x)-1=\cos(2x)

    EDIT: Oops! Looks as though I am steping on Mr. F's toes again...desole!
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  5. #5
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    Its the second one, not the first one. It is not being evaluated at x^2.

    However, I am confused as to why:
    \cos^2 x - \sin^2 x = \cos (2x)

    I keep getting 2cos^2x -1

    cos^2x - (1-cos^2x)
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by redman223 View Post
    Its the second one, not the first one. It is not being evaluated at x^2.

    However, I am confused as to why:
    \cos^2 x - \sin^2 x = \cos (2x)

    I keep getting 2cos^2x -1

    cos^2x - (1-cos^2x)
    2\cos^2(x)-1=\cos(2x)

    2\cdot\frac{e^{2ix}+2+e^{-2ix}}{4}-1=\frac{e^{2ix}+2+e^{-2ix}}{2}-1 =\frac{e^{2ix}+2+e^{-2ix}-2}{2}=\frac{e^{2ix}+e^{-2ix}}{2}=\cos(2x)\quad\blacksquare
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  7. #7
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    Thanks for the help, my next question is:

    \int cos(2x) ~dx = \frac {sin(2x)}{2} ?
    Last edited by redman223; July 2nd 2008 at 08:47 PM.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by redman223 View Post
    Thanks for the help, my next question is:

    \int cos(2x) = \frac {sin(2x)}{2} ?
    \cdots+\bold{\color{red}C}
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  9. #9
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    Quote Originally Posted by redman223 View Post
    Lol, I always forget that.

    On to the next question. Using x=3sin\theta

    \int x^3 \sqrt {9-x^2} ~dx

    I think I solved the part under the radical correctly ( 3cos\theta)

    If I plug in x=3sin\theta and dx=3cos\theta and use 3cos\theta for the radical I get:

    \int 243sin^3\theta cos^2\theta ~d\theta

    Thats where I get lost.
    New and unrelated questions require new threads. Delete this question (copy it first!) and re-post it as a new thread.


    Edit: Belay that request - I see Mathstud's doing more stepping. But note it for next time.
    Last edited by mr fantastic; July 2nd 2008 at 09:09 PM.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by redman223 View Post
    Lol, I always forget that.

    On to the next question. Using x=3sin\theta

    \int x^3 \sqrt {9-x^2} ~dx

    I think I solved the part under the radical correctly ( 3cos\theta)

    If I plug in x=3sin\theta and dx=3cos\theta and use 3cos\theta for the radical I get:

    \int 243sin^3\theta cos^2\theta ~d\theta

    Thats where I get lost.
    \sin^3(\theta)\cos^2(\theta)=\sin(\theta)(1-\cos^(\theta))(\cos^2(\theta))

    Now expand
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  11. #11
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    Getting off-topic but a legitimate response I think:

    Quote Originally Posted by Mathstud28;163721[snip
    EDIT: Oops! Looks as though I am steping on Mr. F's toes again...desole!
    Like the song says:

    Je suis désolé, mais je n'ai pas le choix
    Je suis désolé, mais la vie me demande ça

    By the way ...... My toes are on my feet. My feet are usually in my mouth. So watch the face, mathstud.


    How's that for setting up the (obscure) double entendre for desole .....?
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    New and unrelated questions require new threads. Delete this question (copy it first!) and re-post it as a new thread.


    Edit: Belay that request - I see Mathstud's doing more stepping. But note it for next time.
    What does stepping mean in this context?
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post

    How's that for setting up the (obscure) double entendre for desole .....?
    Acceptable
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