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    so you have cos^5 x
    separate one cos away so you have cosx(cos^4 x)
    rewrite cos^4 x as (1-sin^2 x)^2
    so you have integral of cosx(1-sin^2 x)^2..

    can you see where to go from there?
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    u cud de moivre dat ho, be like "bam bam *****eZ ima demoivre ur a$$$$" n den integr8 it dat way
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    (Original post by Farhan.Hanif93)
    Would I then use a substitution of u=sinx?
    try it.
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    (Original post by Farhan.Hanif93)
    After using the substitution and integrating I now get:
    (1 - sin^2x)^3
    -------------------
    6sinx

    (the rather shoddy line represents a fraction. )
    try this:

    (integral of) cosx(1-sin^2 x)^2 dx
    (rewriting it) = (1-sin^2 x)^2 cosx dx
    u = sinx
    du = cosx dx


    Now you can replace the cosxdx in the integral with du

    (1-(sin x)^2)^2 du

    and because you made u = sinx, you can sub this in, and now you have everything in terms of u and du (which then you can integrate).
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    (Original post by Farhan.Hanif93)
    Yes, this is what I did but I integrated (1-u^2)^2 using the reversed chain rule and the answer I got was as above.
    Oh

    the 'reserve chain rule' cant really be used here, you are assuming that if you differentiate (1-u^2)^3 / -6u you will get back to (1-u^2)^2 but you dont, because when you differentiate (1-u^2)^3 / -6u you have to use the chain rule AND the quotient rule giving you -36u^2(1-u^2)^2 - 6(1-u^2)^3 / 36u^2 which is quite messy.

    Just expand (1-u^2)^2 and integrate
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    hello there, I manage to get this:

    u-(2u^3)/3 +(u^5)/5 +c

    I guess you then need to plug in u=sinx

    thanks!
 
 
 
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