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    Problem solved, no more help needed.

    as title says - i dont know how to start with integrating ((sinh(x))^9)cosh(x)

    am i missing a key formula? (im not very good with hyperbolic trig functions)

    any help appreciated + rep
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    (Original post by Lewk)
    as title says - i dont know how to start with integrating ((sinh(x))^9)cosh(x)

    am i missing a key formula? (im not very good with hyperbolic trig functions)

    any help appreciated + rep
    My initial thoughts would be to use the substitution u=sinhx.
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    (Original post by ttoby)
    My initial thoughts would be to use the substitution u=sinhx.
    ^^ This. If you do this the solution comes out very quickly.
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    (Original post by ttoby)
    My initial thoughts would be to use the substitution u=sinhx.
    This is correct. A dx might be useful to the OP too.
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    Or you might want to consider what happens when you differentiate sinh^{10}(x).
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    (Original post by ttoby)
    My initial thoughts would be to use the substitution u=sinhx.
    lol that was very straight forward thanks, it just 'looks' hard
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    i'd consider the reduction formula \displaystyle I_n=\int sinh^nxcoshx\ dx

    Spoiler:
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    nah not really, i'd just do what marcus said
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    (Original post by Lewk)
    lol that was very straight forward thanks, it just 'looks' hard
    If you see something that looks hard, try to spot tricks. Here, we have a function of a function (that is, (\sinh x)^9 which is multiplied by the derivative of the inner function.

    So for example if you saw \displaystyle \int \dfrac{(2x-5\sin x)\sqrt[3]{\ln (x^2+5\cos x)}}{x^2+5\cos x}\, dx then you'd probably panic, but when you notice that \dfrac{2x-5\sin x}{x^2+5\cos x} is the derivative of \ln (x^2+5\cos x) you can deduce that the substitution u=\ln (x^2+5\cos x) will give you \displaystyle \int \sqrt[3]{u}\, du, which is much easier to get your head around!

    Other such things to look for are:
    - partial fractions
    - things you can use integration by parts on
    - things in the form \frac{f'(x)}{f(x)}, which simply integrate to \ln |f(x)| + C
 
 
 
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