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Math Help - definite integral!

  1. #1
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    definite integral!

    evaluate  \int\sin^2x\cos^6xdx having a limit  x=\frac{\pi}{2}, x=0

    pls help on the problem above, thanks.
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  2. #2
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    Re: definite integral!

    Quote Originally Posted by lawochekel View Post
    evaluate  \int\sin^2x\cos^6xdx having a limit  x=\frac{\pi}{2}, x=0

    pls help on the problem above, thanks.
    Maybe try Integration by Parts with \displaystyle \begin{align*} u = \sin{x} \end{align*} and \displaystyle \begin{align*} dv = \sin{x}\cos^6{x}\,dx \end{align*}...
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  3. #3
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    Re: definite integral!

    \text{ Since }: \sin^2 x=1-\cos^2 x \text{ it follows :}

    I= \int\limits_{0}^{\pi /2} \cos^6 x\,dx -\int\limits_{0}^{\pi /2} \cos^8 x\,dx

    Now , use reduction formula for cosine several times .
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  4. #4
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    Re: definite integral!

    Quote Originally Posted by princeps View Post
    \text{ Since }: \sin^2 x=1-\cos^2 x \text{ it follows :}

    I= \int\limits_{0}^{\pi /2} \cos^6 x\,dx -\int\limits_{0}^{\pi /2} \cos^8 x\,dx

    Now , use reduction formula for cosine several times .
    My way should still work...

    \displaystyle \begin{align*} \int{\sin^2{x}\cos^6{x}\,dx} &= -\frac{\sin{x}\cos^7{x}}{7} - \int{-\frac{\cos^8{x}}{7}\,dx} \\ &= -\frac{\sin{x}\cos^7{x}}{7} + \int{\frac{\cos^8{x}}{7}\,dx} \end{align*}

    and now use the double angle formula to simplify \displaystyle \begin{align*} \cos^8{x} \end{align*}.
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