1. ## integral reduction

find the reduction formula of the following integral: ∫((xⁿ)/(√(2ax-x²)))dx

=(1/(√(2ax-x²)))∫xⁿ.dx-∫[(d/(dx))((1/(√(2ax-x²))))∫xⁿ.dx].dx
=(1/(√(2ax-x²))).((xⁿ⁺¹)/(n+1))-∫[((a-x)/((2ax-x²)^{3/2})).((xⁿ⁺¹)/(n+1))].dx

2. Originally Posted by Pulock2009
find the reduction formula of the following integral: ∫((xⁿ)/(√(2ax-x²)))dx

=(1/(√(2ax-x²)))∫xⁿ.dx-∫[(d/(dx))((1/(√(2ax-x²))))∫xⁿ.dx].dx
=(1/(√(2ax-x²))).((xⁿ⁺¹)/(n+1))-∫[((a-x)/((2ax-x²)^{3/2})).((xⁿ⁺¹)/(n+1))].dx
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

3. 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????

4. Originally Posted by Pulock2009
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.

5. 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⁻¹x)/(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.