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Math Help - Trig Antiderivative

  1. #1
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    Trig Antiderivative

     \int sec(x)sin^3(x)dx

     \int \frac {tan(x)}{sin(x)} sin^3(x)dx

    So then the sin(x)'s cancel out leaving...
    [tex] \int tan(x)sin^2(x)dx

    then

     \int \frac {sin(x)}{cos(x)} sin^2(x)dx

    this equals

     \int \frac {sin^3(x)}{cos(x)}dx

    Then split up the sin^3(x) and then set the sin^2(x) equal to 1 - cos^2(x)

     \int \frac {1 - cos^2(x)}{cos(x)} sin(x)dx

    Then the cos(x)'s can cancel...

     \int (1 - cos(x)) sin(x)dx

    this then equals

     \int sin(x) - cos(x)sin(x)dx

    Then I'm kind of stuck... my next guess would be to take this identity sin(2x) = 2sin(x)cos(x)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by larson View Post
     \int sec(x)sin^3(x)dx

     \int \frac {tan(x)}{sin(x)} sin^3(x)dx

    So then the sin(x)'s cancel out leaving...
    [tex] \int tan(x)sin^2(x)dx

    then

     \int \frac {sin(x)}{cos(x)} sin^2(x)dx

    this equals

     \int \frac {sin^3(x)}{cos(x)}dx

    Then split up the sin^3(x) and then set the sin^2(x) equal to 1 - cos^2(x)

     \int \frac {1 - cos^2(x)}{cos(x)} sin(x)dx

    Then the cos(x)'s can cancel...

     \int (1 - cos(x)) sin(x)dx

    this then equals

     \int sin(x) - cos(x)sin(x)dx

    Then I'm kind of stuck... my next guess would be to take this identity sin(2x) = 2sin(x)cos(x)
    for \int \cos x \sin x~dx, why not just do a simple substitution? u = \sin x or u = \cos x will suffice

    by the way, you could have went from \int \sec x \sin^3 x~dx directly to \int \frac {\sin^3 x}{\cos x}~dx


    but wait...you "canceled" the cosines incorrectly, you can't cancel through a sum like that.

    from \int \frac {\sin^3 x}{\cos x}~dx make the substitution u = \cos x
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  3. #3
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    Hello, larson

    Another approach . . .


     \int \sec x \sin^3\!x \,dx

    We have: . \frac{\sin^3\!x}{\cos x} \;=\;\frac{\sin^2x}{\cos x}(\sin x) \;=\;\frac{1-\cos^2\!x}{\cos x}(\sin x)

    Then we have: . \int \left(\frac{1}{\cos x} - \cos x\right)(\sin x\,dx)

    Let u \,=\,\cos x\quad\Rightarrow\quad du \,=\,-\sin x\,dx\quad\Rightarrow\quad \sin x\,dx \,=\,-du

    Substitute: . \int\left(\frac{1}{u} - u\right)(-du) \;=\;\int\left(u - \frac{1}{u}\right)\,du \;=\;\frac{1}{2}u^2 - \ln|u| + C

    Back-substitute: . \frac{1}{2}\cos^2\!x - \ln|\cos x| + C

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, larson

    Another approach . . .



    We have: . \frac{\sin^3\!x}{\cos x} \;=\;\frac{\sin^2x}{\cos x}(\sin x) \;=\;\frac{1-\cos^2\!x}{\cos x}(\sin x)

    Then we have: . \int \left(\frac{1}{\cos x} - \cos x\right)(\sin x\,dx)

    Let u \,=\,\cos x\quad\Rightarrow\quad du \,=\,-\sin x\,dx\quad\Rightarrow\quad \sin x\,dx \,=\,-du

    Substitute: . \int\left(\frac{1}{u} - u\right)(-du) \;=\;\int\left(u - \frac{1}{u}\right)\,du \;=\;\frac{1}{2}u^2 - \ln|u| + C

    Back-substitute: . \frac{1}{2}\cos^2\!x - \ln|\cos x| + C

    Thank you .
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