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Math Help - integral reduction

  1. #1
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    integral reduction

    find the reduction formula of the following integral: ∫((xⁿ)/(√(2ax-x)))dx

    =(1/(√(2ax-x)))∫xⁿ.dx-∫[(d/(dx))((1/(√(2ax-x))))∫xⁿ.dx].dx
    =(1/(√(2ax-x))).((xⁿ⁺)/(n+1))-∫[((a-x)/((2ax-x)^{3/2})).((xⁿ⁺)/(n+1))].dx
    couldnot figure out how to proceed further...please help!!!
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  2. #2
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    Quote Originally Posted by Pulock2009 View Post
    find the reduction formula of the following integral: ∫((xⁿ)/(√(2ax-x)))dx

    =(1/(√(2ax-x)))∫xⁿ.dx-∫[(d/(dx))((1/(√(2ax-x))))∫xⁿ.dx].dx
    =(1/(√(2ax-x))).((xⁿ⁺)/(n+1))-∫[((a-x)/((2ax-x)^{3/2})).((xⁿ⁺)/(n+1))].dx
    couldnot figure out how to proceed further...please help!!!
    Are you sure you posted the correct integral?

    Wolfram does not give a reduction formula for this one...

    %29]]integral[(x^n)/(Sqrt[2*a*x - x^2])] - Wolfram|Alpha
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    yes i am sure the integral is corrrect. i am unable to figure out how to do the second integral in the second step. i know some sort of regrouping and then agin by-parts need to be done . but how i dont know????
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    Quote Originally Posted by Pulock2009 View Post
    yes i am sure the integral is corrrect. i am unable to figure out how to do the second integral in the second step. i know some sort of regrouping and then agin by-parts need to be done . but how i dont know????
    Once again, I'll say have a look at the link I gave.

    The integral is clearly NOT a reduction.

    Either you have posted the wrong integral or the original question is wrong.
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    perhaps u misunderstood my question. finding the reduction formulae for an integral means expressing it in terms of an integral which has a lower index. say for example:∫tanⁿx=I(n) where n is the subscript of I actually . Now reduction formulae of this is I(n)=((tanⁿ⁻x)/(n-1))-I(n-2). after finding this we can go on reducing to get the lowest-index integral which would be ∫tan x.dx.and thus get the solution.....something of a similar sort needs to be done in my problem.
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