Tough Integration Problem

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pepperonibread

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Oct 30, 2010
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hey guys, i've got one integration problem on my calc homework where i'm completely stuck. though the full question has a bunch of additional constants and this is only one step of the problem, i've been unable to get past the integration of the function f(x)=(1+x^2)^(-3/2). i've tried both u-substitution and integration by parts, though i can't seem to get to a simpler integral with either of these methods.
specifically, i've substituted for u=x^2, u=x^2+1, and u=(1+x^2)^(3/2), none of these yielding anything useful...
and i've integrated by parts by splitting the function into (1+x^2)^(-1) times (1+x^2)^(-1/2), but solving this just led to a more complex integral.
anyways, i hope this is enough background info. if anyone could help that would be awesome.
 
pepperonibread said:
i've been unable to get past the integration of the function f(x)=(1+x^2)^(-3/2).
Click to expand...

pepperonibread,

\(\displaystyle let \ x \ = \ tan(u).\)

\(\displaystyle And \ (1 + x^2)^{(-3/2)} \ =\)

\(\displaystyle [sec^2(u)]^{(-3/2)} \ = \\)

\(\displaystyle [sec(u)]^{(-3)} \ =\\)

\(\displaystyle \frac{1}{sec^3(u)}\)


\(\displaystyle Because \ dx \ = \ sec^2(u)du, \ then \ the \ integral \ becomes:\)


\(\displaystyle \int\frac{sec^2(u)}{sec^3(u)}du \ = \\)


\(\displaystyle \int\frac{1}{sec(u)}du \ = \\)


\(\displaystyle \int cos(u)du \ = \\)


\(\displaystyle sin(u) \ + \ C \ = \\)


\(\displaystyle sin[tan(x)] \ + \ C\)


\(\displaystyle Can \ you \ continue \ from \ this \ and \ simplify \ the \ trigonometry \ portion \ here?\)
 
ah ok that makes sense... i guess this is trig substitution, we haven't covered that fully yet, but the concept makes sense to me.
and yeah i think i can simplify it from there. thanks!
 
\(\displaystyle \int\frac{1}{(x^{2}+1)^{\frac{3}{2}}}dx\)

A sub that whittles this down to practically nothing is let:

\(\displaystyle u=\frac{x}{\sqrt{x^{2}+1}}, \;\ du=\frac{1}{(x^{2}+1)^{\frac{3}{2}}}dx\)

See?. du is your integral itself.

So, you get:

\(\displaystyle \int du\)

Integrating gives u. See what u is from above?. That's it. Of course, you can add the +C (constant of integration) for thoroughness if you wish
 
hey galactus... i understand what you did there, but isn't that kind of circular? you make u=x/(1+^2)^(1/2), which is also the final answer to the problem, but how would i know to make that substitution unless i already knew the answer?
i guess what i'm saying is, if you're integrating a function f(x), and its antiderivative is F(x), then you could set u=F(x), and so du=f(x)dx. then you're left just integrating du, which gives you the solution F(x)+C. so you're using F(x) to find F(x)? that's where i'm confused.
 
pepperonibread said:
hey galactus... i understand what you did there, but isn't that kind of circular? you make u=x/(1+^2)^(1/2), which is also the final answer to the problem, but how would i know to make that substitution unless i already knew the answer?
i guess what i'm saying is, if you're integrating a function f(x), and its antiderivative is F(x), then you could set u=F(x), and so du=f(x)dx. then you're left just integrating du, which gives you the solution F(x)+C. so you're using F(x) to find F(x)? that's where i'm confused.
Click to expand...

Yes and no.

The substituted function was chosen "judiciously" - that type of judgement will be natural after you have done about 10[sup:2hfvhpxc]100[/sup:2hfvhpxc] antiderivatives - which Galactus has done. There is nothing wrong with that - you just have to be damn good.....
 
Yeah, I can see how it apears circular.

After doing integration, one can 'see' various subsitutions to make. It is likely best for you to go with lookagains's trig sub method.

I just knew that the derivative of \(\displaystyle \frac{x}{\sqrt{x^{2}+1}}\) is \(\displaystyle \frac{1}{(x^{2}+1)^{\frac{3}{2}}}dx\)

The latter being your integral.

You could think of your integral as: \(\displaystyle \frac{x}{x(\sqrt{x^{2}+1})(x^{2}+1)}\) and perhaps see the sub in there.

You can often rewrite integrands in an equivalent form to aid in seeing the subs to make.

For get I posted. I was half asleep this morning. :)
 
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