2
$\begingroup$

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.

$\endgroup$
10
$\begingroup$

Let $u=1-x^2$.

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

$\endgroup$
  • 3
    $\begingroup$ I suggest using >! to hide the answers of homework problems, to give the poster a chance to work it out by themselves. $\endgroup$ – DanielV Jan 1 '15 at 2:50
  • $\begingroup$ @DanielV Thanks; done! $\endgroup$ – Ahaan S. Rungta Jan 1 '15 at 2:52
  • 1
    $\begingroup$ Thanks! Somehow I didn't think to do it this way! $\endgroup$ – Mark Jan 1 '15 at 3:01
  • $\begingroup$ @Mark You're welcome. That happens to me all the time! $\endgroup$ – Ahaan S. Rungta Jan 1 '15 at 3:06
  • 1
    $\begingroup$ 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 math.stackexchange.com/questions/1087205/… $\endgroup$ – Mark Jan 1 '15 at 3:08
3
$\begingroup$

Another way directly, using

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

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

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

$\endgroup$
  • $\begingroup$ This seems to be a restatement of the substitution answer, where we do the replacement $1-x^2 = f$, $df = -1/2 x dx$ $\endgroup$ – Mark Jan 1 '15 at 19:49
  • $\begingroup$ @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. $\endgroup$ – Timbuc Jan 1 '15 at 19:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.