Evaluation of a reduction formula

1. Aug 14, 2011

NewtonianAlch

1. The problem statement, all variables and given/known data
$\int$ (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

2. Relevant equations

3. 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?

2. Aug 14, 2011

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.

3. Aug 15, 2011

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

4. Aug 15, 2011

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.

5. Aug 15, 2011

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.

6. Aug 15, 2011

NewtonianAlch

Awesome, thanks for that.