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Thread: integral of sin^3 cos x dx

  1. #1
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    integral of sin^3 cos x dx

    Definately having some problems with these.
    Question is evalueate the indefinate integral

    $\displaystyle \int {\sin ^3 } x\_\cos x\_dx$

    I get as far as this, I seem to be having issues with these.


    $\displaystyle \int {\sin ^2 } x\_\sin x\_\cos x\_dx$


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

    $\displaystyle u = \cos x $ $\displaystyle du = - \sin x$

    The answer is:

    $\displaystyle \frac{1}{4}\sin ^4 x + C
    $
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Craka View Post
    Definately having some problems with these.
    Question is evalueate the indefinate integral

    $\displaystyle \int {\sin ^3 } x\_\cos x\_dx$

    I get as far as this, I seem to be having issues with these.
    $\displaystyle

    \int {\sin ^2 } x\_\sin x\_\cos x\_dx


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

    u = \cos x du = - \sin x

    $

    The answer is:

    $\displaystyle \frac{1}{4}\sin ^4 x + C
    $
    No need for a use of trig identities either see that cos is the derivative of sin or do it the old fashioned way

    Let $\displaystyle u=\sin(x)\Rightarrow{du=\cos(x)dx}$

    giving $\displaystyle \int{u^3du}=\frac{u^4}{4}+C$

    back subbing we get

    $\displaystyle \frac{\sin^4(x)}{4}+C$
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Craka View Post
    Definately having some problems with these.
    Question is evalueate the indefinate integral

    $\displaystyle \int {\sin ^3 } x\_\cos x\_dx$
    Since $\displaystyle \frac{d}{dx} \sin(x)=\cos(x)$ the integrand is of the form $\displaystyle [f(x)]^3f'(x)$, so the integral is $\displaystyle [f(x)]^4/4+C$, or:

    $\displaystyle \int (\sin(x))^3 \cos(x)~dx = \frac{(\sin(x))^4}{4}+C$

    RonL
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