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Math Help - Integration Reduction; Formulae

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    Integration Reduction; Formulae

    Hi , I am stuck on one line in the solution attached . I am OK to the third line in the mark scheme but
    do not understand how ....n a ^(n-1) has been evaluated in the * line .
    Integration Reduction; Formulae-untitled-2.pngIntegration Reduction; Formulae-untitled.png

    Thanks
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    Re: Integration Reduction; Formulae

    Quote Originally Posted by minicooper58 View Post
    Hi , I am stuck on one line in the solution attached . I am OK to the third line in the mark scheme but
    do not understand how ....n a ^(n-1) has been evaluated in the * line .
    Click image for larger version. 

Name:	Untitled 2.png 
Views:	8 
Size:	28.9 KB 
ID:	23860Click image for larger version. 

Name:	Untitled.png 
Views:	4 
Size:	17.9 KB 
ID:	23859
    \left[-n(a-x)^{n-1}\cos x\right]_0^a

    =\left[-n(a-a)^{n-1}\cos a\right] - \left[-n(a-0)^{n-1}\cos 0\right]

    =0 + na^{n-1}\cos 0

    =na^{n-1}
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  3. #3
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    Re: Integration Reduction; Formulae

    Quote Originally Posted by minicooper58 View Post
    Hi , I am stuck on one line in the solution attached . I am OK to the third line in the mark scheme but
    do not understand how ....n a ^(n-1) has been evaluated in the * line .
    Click image for larger version. 

Name:	Untitled 2.png 
Views:	8 
Size:	28.9 KB 
ID:	23860Click image for larger version. 

Name:	Untitled.png 
Views:	4 
Size:	17.9 KB 
ID:	23859

    Thanks
    \displaystyle \begin{align*} I_n &= \int_0^a{(a - x)^n\cos{x}\,dx} \end{align*}

    Using integration by parts with \displaystyle \begin{align*} u = (a - x)^n \implies du = -n(a-x)^{n-1}\,dx \end{align*} and \displaystyle \begin{align*} dv = \cos{x}\,dx \implies v = \sin{x} \end{align*} and the integral becomes

    \displaystyle \begin{align*} \int{(a - x)^n\cos{x}\,dx} &= (a - x)^n\sin{x} - \int{-n(a - x)^{n - 1}\sin{x}\,dx} \\ &= (a - x)^n\sin{x} + n\int{(a - x)^{n-1}\sin{x}\,dx} \end{align*}

    Using Integration by parts again with \displaystyle \begin{align*} u = (a - x)^{n - 1} \implies du = -(n - 1)(a - x)^{n - 2}\,dx \end{align*} and \displaystyle \begin{align*} dv = \sin{x}\,dx \implies v = -\cos{x} \end{align*} and the integral becomes

    \displaystyle \begin{align*} \int{(a - x)^n\cos{x}\,dx} &= (a - x)^n\sin{x} + n\left[-(a-x)^{n-1}\cos{x} - \int{(n - 1)(a - x)^{n-2}\cos{x}\,dx}\right] \\ &= (a - x)^n\sin{x} - n(a - x)^{n - 1}\cos{x} - n(n - 1)\int{(a - x)^{n-2}\cos{x}} \end{align*}

    and evaluating between 0 and a will give you the required result
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