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Math Help - Integration!!

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    Integration!!

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    Moo
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    Hello,

    Write : \sin^n x=\sin^{n-2} x \cdot \sin^2 x=\sin^{n-2}x \cdot (1-\cos^2 x)=\sin^{n-2}x-\cos^2x \cdot \sin^{n-2}x

    Integrate by parts the latter term
    Last edited by Moo; May 16th 2008 at 10:42 AM. Reason: Checked
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    Thank you(Further Qn)

    Hey Moo could you give me a further tip on integrating especially
    sin^(n-2)x plz!!!!
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    Moo
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    Quote Originally Posted by matty888 View Post
    Hey Moo could you give me a further tip on integrating especially
    sin^(n-2)x plz!!!!
    Actually, you can't integrate it ^^
    I'll give it a try :

    As I told you, \int \sin^n x dx=\int \sin^{n-2}x dx -\int \cos^2 x \sin^{n-2} x dx

    ~~~~~~~~~~~~
    Let's consider \int \cos^2 x \sin^{n-2} x dx

    \int \cos^2 x \sin^{n-2} x dx =\int \cos x (\cos x \cdot \sin^{n-2} x) dx \ \leftarrow \text{IBP}
    v'(x)=\cos x \cdot \sin^{n-2} x \longrightarrow v(x)=\frac{1}{n-1} \cdot \sin^{n-1} x
    u(x)=\cos x \longrightarrow u'(x)=-\sin x


    Therefore :
    \begin{aligned} \int \cos x (\cos x \cdot \sin^{n-2} x) dx &= \frac{1}{n-1} \cdot \sin^{n-1} x \cdot \cos x - \frac{1}{n-1} \int (-\sin x) \cdot \sin^{n-1} x dx \\<br />
&=\boxed{\frac{1}{n-1} \cdot \sin^{n-1} x \cdot \cos x+\frac{1}{n-1} \int \sin^n x dx} \end{aligned}
    ~~~~~~~~~~~~

    Getting back to the original integral :

    \int \sin^n x dx=\int \sin^{n-2}x dx -\frac{1}{n-1} \cdot \sin^{n-1} x \cdot \cos x-\frac{1}{n-1} \int \sin^n x dx


    Does it help you seeing through the problem ?
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