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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!

Thanks!

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2 Answers 2

up vote 5 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.

HINT

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

It should be straight forward from here. :)

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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 :/

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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

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