Deriving the Formula for Integration by Parts

[threeder]

integration by parts

I'm working through Apostol's Calculus. I have attached the problem. I need to derive the formula integrating by parts.

It is not a hard problem, but I can't seem to understand how on Earth the author came up with that expression.

I take f(x) = (a^2 - x^2)^n, so f'(x)=-2nx(a^2 - x^2)^(n-1).
Take g'(x)=1, so g(x)=x.

Plug the expressions in. We get another integral which could be easily solved with integration by substitution and I get the final expression of x(a^2 - x^2)^n - (a^2 - x^2)^n which does make sense if we check it by taking derivatives of both sides.

On the other hand, if I try to take the derivative of the given expression, I don't really get the initial result. So, am I missing something or what?

EDIT: OK, I've managed to see why his expression makes sense after all. but still, do not see how could I get there. Would my initial picks of f's and g' for integration by parts allow me to get there or should I pick something else? thanks!

 

[Bonaparte]

Its very hard to understand your question threeder. Please say
given expression: show your work on it
explain problem.
Or I might just have reading problems (its probably this), however I can't help you either way, so your going to be forced to explain for me to understand.

Thanks and good luck, Bonaparte

 

[threeder]

I suppose I could have framed it more clearly, sorry. What I have done is:

∫(a^2 - x^2)^n dx = [f(x) = (a^2 - x^2)^n -> f'(x)=-2nx(a^2 - x^2)^(n-1), g'(x)=1 ->g(x)=x.]

(in order to integrate by parts, I have chosen particular f and g)

After using integration by parts ( g(x)f(x) - ∫g'(x)f'(x) )we get:

x(a^2 - x^2)^n - 2n∫x(a^2 - x^2)^(n-1) dx = [ let u=a^2 - x^2, so dx=-1/2 du ]

After substituting we get

x(a^2 - x^2)^n - n∫u^(n-1) du = x(a^2 - x^2)^n - (a^2 - x^2)^n + C

So while this is correct, it does not lead to the result which is I am required to get.

What should I do differently to get the required result?

 

[cjc0117]

From your first post, it looks like you're not performing integration by parts correctly.

\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx

 

[threeder]

Yes, indeed I made the mistake, but it was more of a typo, as I still get the final result using the metod I described which does not lead to the final expression of:

\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 correct 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

Now, as I mentioned before, I could substitute, but I need to do something else. I multiply the first expression on the right by \frac{2n + 1}{2n +1} which leads to:

(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. could anybody help me out? I tried multiple things like dividing expressions into integrals by parts but nothing really helped me to get where I need to. Thanks!

 

[threeder]

I tried different approach, and I think this looks more promising, but I might be making a mistake somewhere. Could anybody give me some hints?

What I tried is:

\int (a^2 - x^2)^n \,dx = \begin{pmatrix}<br /> f(x)= a^2 - x^2 | f&#039;(x)=-2x\\<br /> g&#039;(x)=(a^2-x^2)^{n-1} |g(x)=\int(a^2-x^2)^{n-1} \,dx<br /> \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<br /> =
(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

what is wrong here?

 

[cjc0117]

threeder,

I've found a way to solve this that starts with trig substitution. Use the substitution x=asinθ to begin with. When you do this, you should get the following:

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

Now try to solve it from here. Notice that the problem has become an issue of finding the reduction formula for cosine, for which you can find many online guides. it involves a trig identity and integration by parts. Once you get the reduction formula, back substitute to get the answer in terms of x again, and then simplify. You should arrive at the expression given in your book.

 

[SammyS]

Yes, indeed I made the mistake, but it was more of a typo, as I still get the final result using the metod I described which does not lead to the final expression of:
\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 correct 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
...

Now, look at the integral on the right hand side.

Rewrite it as:\displaystyle \int x^2 (a^2 - x^2)^{n-1} \,dx

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

\displaystyle =\int \left(a^2 (a^2 - x^2)^{n-1}-(a^2 - x^2)^n\right) \,dx​

Split that into two integrals, & substitute back into \displaystyle \ \ \int (a^2 - x^2)^n \,dx = x(a^2 - x^2)^n + 2n \int x^2 (a^2 - x^2)^{n-1} \,dx \ \ and solve for \displaystyle \ \ \int (a^2 - x^2)^n \,dx\ .

 

[cjc0117]

That's a nice and simple way to do it, SammyS. Very clever.

 

[threeder]

simply - elegant. thank you!