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    What would be the quickest way to integrate something like sin^3 or cos^3? I'm always faced with these and I can never be bothered to compute them!
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    (Original post by gangsta316)
    What would be the quickest way to integrate something like sin^3 or cos^3? I'm always faced with these and I can never be bothered to compute them!
    First getting them in the form involving sin3x, cos3x, etc. Can't remember exactly what they are but you don't want any cubes or squares. Remember the pythagorean identity. :wink2:

    Please correct if I'm worng here someone cause I'm not 100% sure right now.
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    (Original post by gangsta316)
    What would be the quickest way to integrate something like sin^3 or cos^3? I'm always faced with these and I can never be bothered to compute them!
    The quickest way would be to remember them! They're not that farfetched. In terms of method, write them as \sin x (1-\cos ^2x) and \cos x (1-\sin ^2x) respectively and integrate by recognition noting that you have something of the form g'(x)f'(g(x)). I know you're at university so I probably sound a little patronising here but that is probably the easiest ways to the evaluate those integrals.
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    if you have something in the form of dy/dx= sin^a x where a is an odd number like dy/dx=sin^3 x
    use substitution, let u=cosx
    u'=sinx so
    dy/du=sin^3 x/sinx
    dy/du=sin^2x=1-cos^2x=1-u^2
    and thus you can integrate it from there, if dy/dx=cos^a x where a is an odd number, let u=sinx and you'll get an answer pretty easily
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    Do as Farhan suggests. More generally, to integrate \cos^{2n+1} x, rewrite it as cos x (1-sin^2 x)^n expand by the binomial theorem, and you can write down the integral of each term

    (\int cos x sin^k x = \frac{sin^{k+1}x}{k+1}).

    Even powers are considerably harder.
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    You could evaluate cos^2x by integration by parts then use that value to work out cos^3x by saying u = cosx and dv/dx = cos^2x

    then memorise the results so you don't have to do it again!
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    (Original post by RFextra)
    You could evaluate cos^2x by integration by parts then use that value to work out cos^3x by saying u = cosx and dv/dx = cos^2x

    then memorise the results so you don't have to do it again!
    Try it. It will not be fun.
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    (Original post by RFextra)
    You could evaluate cos^2x by integration by parts then use that value to work out cos^3x by saying u = cosx and dv/dx = cos^2x

    then memorise the results so you don't have to do it again!
    As DFranklin said above, it's considerably more hassle to deal with the even powers so it probably isn't sensible to do that in order to deal with the more simple case of the odd power. Besides, using integration by parts in that way will just complicate matters!
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    (Original post by Farhan.Hanif93)
    It probably wouldn't help, if I'm honest. I can't see a way to use the compound angle expansion in reverse to deal with the integral of sin or cos to an odd power.
    sin^3x = (3 sin x - sin 3x)/4
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    (Original post by DFranklin)
    sin^3x = (3 sin x - sin 3x)/4
    That's why I should really know my expansion of sin(kx) and cos(kx) for at least k=2 and k=3. :sigh:
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    (Original post by Farhan.Hanif93)
    It probably wouldn't help, if I'm honest. I can't see a way to use the compound angle expansion in reverse to deal with the integral of sin or cos to an odd power.
    Oh God. I feel like such a fool making basic mistakes nowadays. I was thinking something along the lines of this:
    \sin 3\theta=3\sin \theta - 4\sin^3 \theta
    So \sin^3 \theta=\frac{1}{4}(3\sin \theta  - \sin 3\theta)

    And using this:
    \int \sin^3 \theta\ dx
    = \frac{1}{4} \int 3\sin \theta  - \sin 3\theta\ dx
    = \frac{1}{4} (-3\cos \theta  + \frac{1}{3}\cos 3\theta) +C
    = \frac{1}{12}\cos 3\theta -\frac{3}{4}\cos \theta +C

    Could you point out the mistake please because I genuinely can't see it. Thanks for the help.
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    (Original post by ElMoro)
    Oh God. I feel like such a fool making basic mistakes nowadays. I was thinking something along the lines of this:
    \sin 3\theta=3\sin \theta - 4\sin^3 \theta
    So \sin^3 \theta=\frac{1}{4}(3\sin \theta  - \sin 3\theta)

    And using this:
    \int \sin^3 \theta\ dx
    = \frac{1}{4} \int 3\sin \theta  - \sin 3\theta\ dx
    = \frac{1}{4} (-3\cos \theta  + \frac{1}{3}\cos 3\theta) +C
    = \frac{1}{12}\cos 3\theta -\frac{3}{4}\cos \theta +C

    Could you point out the mistake please because I genuinely can't see it. Thanks for the help.
    That's fine! It was me that was making the stupid mistake, I apologise. :o:
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    (Original post by DFranklin)
    Try it. It will not be fun.
    Oh yeah, I only managed to prove that the intergral of cos²x is equal to the intergral of cos²x...

    brilliant.
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    (Original post by Farhan.Hanif93)
    That's fine! It was me that was making the stupid mistake, I apologise. :o:
    Haha, well it happens to the best of us.
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    You can always write it in complex exponential form. Expanding and integrating is then easy. (The tough bit is then simplifying again).
 
 
 
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