Evaluation of a reduction formula

[NewtonianAlch]

Homework Statement

\int (1 - x^2)^n dx = \frac{2n}{2n+1} \int (1 - x^2)^n-1 dx

for n greater or equal to 1, find \int (1 - x^2)^4 dx

The integrals go from 0 to 1

Homework Equations

The Attempt at a Solution

Well what I did was to keep doing n - 1 whilst pulling a fraction outside the integral and multiplying each subsequent fraction.

So I ended up with 8/9 * 6/7 * 4/5 and the integral of (1 - x) which became x - x^2/2

Substituting in 1 and 0, subtracting and the multiplying by the fractions I got 192/305*1/2 which simplifies to 96/305.

Is this the correct procedure for evaluating reduction formulae, and was my answer correct?

 

[upsidedowntop]

I get:

\int_0^1 (1 - x^2)^4 dx = \frac{8}{9}\frac{6}{7}\frac{4}{5}\frac{2}{3}\int_0^1 1 dx

So the answer is: \frac{(8)(6)(4)(2)}{(9)(7)(5)(3)} = \frac{2^7}{(3^2)(5)(7)} = \frac{128}{315}

Wolframe Alpha agrees with the above answer.

 

[NewtonianAlch]

Ahh, I see my mistake now, for n greater or equal to 1, I needed to have reduced it one more time to get the 2/3.

Thanks

 

[NewtonianAlch]

I'd like to know if possible what page you used to cross check that, I'm looking at Wolfram Alpha and can't really find anything to evaluate reduction formulae, though I can see the step-by-step integrator.

 

[upsidedowntop]

You don't need to give WA the correct formulas with which to evaluate an integral...
"int from 0 to 1 (1-x^2)^4"
returns 128/315
See: http://www.wolframalpha.com/input/?i=int+from+0+to+1+(1-x^2)^4
It will do infinite sums, limits and derivatives too. I find it really useful for checking work.

 

[NewtonianAlch]

Awesome, thanks for that.