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Math Help - 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

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

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


    \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:

    \frac{1}{4}\sin ^4 x + C<br />
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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

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

    I get as far as this, I seem to be having issues with these.
     <br /> <br />
\int {\sin ^2 } x\_\sin x\_\cos x\_dx<br /> <br /> <br />
\int {(1 - \cos ^2 } x)\sin x\cos x\_dx<br /> <br />
u = \cos x du = - \sin x<br /> <br />

    The answer is:

    \frac{1}{4}\sin ^4 x + C<br />
    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 u=\sin(x)\Rightarrow{du=\cos(x)dx}

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

    back subbing we get

    \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

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

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

    RonL
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