Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I working through Apostol's calculus, and I need to prove integrating by parts that :

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

Now, using the integration by parts formula after first division the integral to parts I arrive at:

$\int (a^2 - x^2)^n \,dx = x(a^2 - x^2)^n + 2n \int x^2 (a^2 - x^2)^{n-1} \,dx $

I could substitute and solve the integral, but I need to do something else. I multiply the first expression on the right by $ \frac{2n + 1}{2n +1}$ which leads to:

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

and I am somewhat close. If I try something else, I end up even closer:

$\int (a^2 - x^2)^n \,dx = \begin{pmatrix} f(x)= a^2 - x^2 | f'(x)=-2x\\ g'(x)=(a^2-x^2)^{n-1} |g(x)=\int(a^2-x^2)^{n-1} \,dx \end{pmatrix} = $ $(a^2-x^2)\int(a^2-x^2)^{n-1} \,dx + 2\int x \Big(\int(a^2-x^2)^{n-1} \,dx \Big)\,dx =$ $(a^2-x^2)\int(a^2-x^2)^{n-1} \,dx +x^2\int(a^2-x^2)^{n-1} \,dx = a^2\int(a^2-x^2)^{n-1} \,dx $

But I think I made a mistake somewhere... could somebody help me out? I'm really stuck!


share|cite|improve this question
up vote 6 down vote accepted

NOTE: You can only pull the constant part out of an integral, but not your variable:

  • This is good: $\int 3x^3 dx = 3 \int x^3 dx$
  • This is BAD: $\int 3x^3 dx = \color{red}x \int 3x^2 dx$

So far, so good:

$\int (a^2 - x^2)^n \,dx = x(a^2 - x^2)^n + 2n \int \color{red}{x^2} (a^2 - x^2)^{n-1} \,dx$

The red part $x^2$ is correct, but should be manipulated somehow to make it disappear. Since it does not show up on the RHS of the original equation.


$\begin{align} \int (a^2 - x^2)^n \,dx &= x(a^2 - x^2)^n + 2n \int \color{red}{x^2} (a^2 - x^2)^{n-1} \,dx \\ &= x(a^2 - x^2)^n + 2n \int \color{red}{(x^2 - a^2 + a^2)} (a^2 - x^2)^{n-1} \,dx\\ &= x(a^2 - x^2)^n + 2na^2 \int (a^2 - x^2)^{n - 1} + 2n \int (x^2 - a^2) (a^2 - x^2)^{n-1} \,dx \\ &= x(a^2 - x^2)^n + 2na^2 \int (a^2 - x^2)^{n - 1} - 2n \int (a^2 - x^2)^{n} \,dx \end{align}$

It should be straight forward from here. :)

share|cite|improve this answer
it is! thank you. finally :) – Sarunas Nov 18 '12 at 16:36

OK. Another problem. Miserably stuck again :/. Integration by parts. Need to show that:

If $I_n(x)=\int_{0} ^{x}t^n(t^2+a^2)^{-\frac{1}{2}}dt$

Then: $nI_n(x) = x^{n-1}\sqrt{x^2+a^2}-(n-1)a^2I_{n-2(x)}$ if $x\geq2$

I can get to the point where:

$nI_{n}=x^{n+1}(x^2+a^2)^{-\frac{1}{2}}-a^2I_{n-2}+a^4\int t^{n-2}(t^2+a^2)^{-\frac{3}{2}}dt$

I get there by dividing it into parts and then using the trick user49685 suggested using.

Now, is this a good start or should I have taken another route? Because I can not find a way out :/

share|cite|improve this answer
You have to look at the final result, to have a good aim. What did you define $u$, and $v$ to be when you tried integrating it by parts, to arrive at: $nI_{n}=x^{n+1}(x^2+a^2)^{\color{red}{-\frac{1}{2}}}-a^2I_{n-2}+a^4\int t^{n-2}(t^2+a^2)^{-\frac{3}{2}}dt$? You should choose another $u$, and $v$, so that the power $-\frac{1}{2}$, should be changed to $\frac{1}{2}$ as required. HINT: $\int \frac{x}{\sqrt{x^2 + a^2}}dx = \sqrt{x^2 + a^2} + C$. Btw, you should create another new question, instead of posting in your old thread. – user49685 Nov 19 '12 at 10:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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