Integral of $x(1-x^2)^n$

Question

This should be "simple" according to the book but I can't seem to work it out. $n$ is an integer.

$\int_0 ^1 x(1-x^2)^n dx$

I have tried binomial expansion and get stuck at $\sum_0^n {n\choose k}(-1)^k \frac{1}{2k + 2}$. I tried summing this by parts after realizing that $\sum_0^n {n\choose k}(-1)^k = 0$ but it didn't turn out to a nice expression besides the boundary conditions.

I have also tried integration by parts and get $-\int_0 ^1 nx^3 (1-x^2)^{n-1} dx$. Here I can do repeated integration by parts and arrive at a pattern but I'd rather get the first method to work out since it should involve some nice combinatorial identity.

2

Let $u = 1-x^2$. – George V. Williams Jan 01 '15 at 02:46

score 10 · Accepted Answer · answered Jan 01 '15 at 02:49

10

Let $u=1-x^2$.

Then, $\mathrm{d}u=-2x \, \mathrm{d}x$, so it is now $$ \begin {align*} \displaystyle\int_{u=1}^{u=0} -\frac {u^n}{2} \, \mathrm{d}u &= \displaystyle\int_{0}^{1} \frac {u^n}{2} \, \mathrm{d}u \\&= \frac {1}{2} \cdot \displaystyle\int_0^1 u^n \, \mathrm{d}u \\&= \frac {1}{2} \cdot \left[ \frac {u^{n+1}}{n+1} \right]_0^1 \\&= \boxed{\dfrac{1}{2(n+1)}}. \end {align*} $$

answered Jan 01 '15 at 02:49

Ahaan S. Rungta

7,706

3

I suggest using >! to hide the answers of homework problems, to give the poster a chance to work it out by themselves. – DanielV Jan 01 '15 at 02:50
@DanielV Thanks; done! – Ahaan S. Rungta Jan 01 '15 at 02:52
1

Thanks! Somehow I didn't think to do it this way! – Mark Jan 01 '15 at 03:01
@Mark You're welcome. That happens to me all the time! – Ahaan S. Rungta Jan 01 '15 at 03:06
1

But I'm wondering why the ways I tried to do it were so hard when this integral turns out to be "intrinsically" simple. Here is a follow up question http://math.stackexchange.com/questions/1087205/sum-0n-n-choose-k-1k-frac12k-2-1-2k-2 – Mark Jan 01 '15 at 03:08
@Mark Good question. +1 on it. Seems like you've already gotten some answers. – Ahaan S. Rungta Jan 01 '15 at 03:26

score 3 · Answer 2 · answered Jan 01 '15 at 03:49

3

Another way directly, using

$$\int f'(x) f(x)^ndx=\frac{f(x)^{n+1}}{n+1}+C\;\;:$$

$$\int_0^1x(1-x^2)^ndx=-\frac12\int_0^1(1-x^2)'\,(1-x^2)^ndx=\left.-\frac1{2(n+1)}\left(1-x^2\right)^{n+1}\right|_0^1=$$

$$=-\frac1{2(n+1)}\left(0-1\right)=\frac1{2(n+1)}$$

answered Jan 01 '15 at 03:49

Timbuc

34,191

This seems to be a restatement of the substitution answer, where we do the replacement $1-x^2 = f$, $df = -1/2 x dx$ – Mark Jan 01 '15 at 19:49
@Mark I see it the other way around. Kids are taught many times to be lazy and not to think too hard how to do integrals, and rush into substitution, which sometimes can be pretty dangerous. The above proposes a different way of seeing things when possible and when reasonably easy to do so. After all, we can always try some substitution or whatever if the above doesn't work. Where I live, all the integrals proposed to try substitution (in high school) can be done the above way...all without one single exception. When some kids find out the above they get really thrilled. – Timbuc Jan 01 '15 at 19:57

Integral of $x(1-x^2)^n$

2 Answers2

Linked