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Thread: Trig Antiderivative

  1. #1
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    Trig Antiderivative

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

    $\displaystyle \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

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

    this equals

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

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

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

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

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

    this then equals

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

    Then I'm kind of stuck... my next guess would be to take this identity $\displaystyle sin(2x) = 2sin(x)cos(x)$
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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
    $\displaystyle \int sec(x)sin^3(x)dx $

    $\displaystyle \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

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

    this equals

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

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

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

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

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

    this then equals

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

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

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


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

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

    Another approach . . .


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

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

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

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

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

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

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

    Another approach . . .



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

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

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

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

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

    Thank you .
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