1. ## Integral u-substitution

*EDIT - Sorry, I was typing on a phone, and missed a couple of key bits of information on the equation. Have fixed them up :/

Hello,

I just have a quick integral question in which I can not figure out how to get the right answer. Or, I know what I have to do to get it, I just don't know WHY I can do it...No answers please, just a tip in the right direction

$\int \frac {x^{1/2}}{(x^{3/2} - 8)^4} dx$

$Let u = x^{3/2}-4, du=\frac {3}{2} x^{1/2}dx$

$\frac {3}{2} \int \frac {1}{u^4} du = u^{-4}$

Now, have I got this set up right? U-substitution still confuses me a little, and I'm not sure if I've done it properly. If I go ahead and finish this equation, I get.

$\frac {1}{-3(x^{3/2}-8)^3}$

I should be getting $\frac {2}{-9(x^{3/2} - 8)^3}$

I can see how I should be getting that answer - by inverting the 3/2, to make it 2/3. But why is the 3/2 being inverted to 2/3 in the first place? The only reason I can think of is that it's in the denominator of a fraction, so it would be inverted, but I'm not sure...

2. ## Re: Integral u-substitution

I'd advise writing the "dx" everywhere. It helps make what's happening during subsitution clearer I think.

I don't think you should be getting what you say you "should" be getting, since the numerator there is, up to a constant multiple, the derivative of the denominator.

3. ## Re: Integral u-substitution

Originally Posted by astuart
Hello,

I just have a quick integral question in which I can not figure out how to get the right answer. Or, I know what I have to do to get it, I just don't know WHY I can do it...No answers please, just a tip in the right direction

$\int \frac {x^{1/2}}{x^{3/2} - 8}$

$Let u = x^{3/2}, du=\frac {3}{2} x^{1/2}$

$\frac {3}{2} \int \frac {1}{u^4} du$

Now, have I got this set up right? U-substitution still confuses me a little, and I'm not sure if I've done it properly. If I go ahead and finish this equation, I get.

$\frac {1}{-3(x^{3/2}-8)^3}$

I should be getting $\frac {2}{-9(x^{3/2} - 8}$

I can see how I should be getting that answer - by inverting the 3/2, to make it 2/3. But why is the 3/2 being inverted to 2/3 in the first place? The only reason I can think of is that it's in the denominator of a fraction, so it would be inverted, but I'm not sure...
The whole point of using a u-substitution is to use the chain rule in reverse. This means, you can simplify an integral if you can see an "inner function" which has this inner function's derivative as a multiple.

You could see that there is an inner function \displaystyle \begin{align*} x^{\frac{3}{2}} \end{align*} and its derivative is \displaystyle \begin{align*} \frac{3}{2}x^{\frac{1}{2}} \end{align*}. You have \displaystyle \begin{align*} x^{\frac{1}{2}} \end{align*} as a multiple, which is ALMOST what you need. That's good.

Some things to think about though. First of all, you need \displaystyle \begin{align*} \frac{3}{2}x^{\frac{1}{2}} \end{align*} as a multiple, not just \displaystyle \begin{align*} x^{\frac{1}{2}} \end{align*}. That's easily fixed, just remember that whatever you multiply by you need to divide by as well, to keep the expression equivalent to what you started with.

So \displaystyle \begin{align*} \int{\frac{x^{\frac{1}{2}}}{x^{\frac{3}{2}} - 8} \, dx} &= \frac{2}{3} \int{\frac{\frac{3}{2}x^{\frac{1}{2}}}{x^{\frac{3} {2}} - 8}\,dx} \end{align*}.

Now for the more pressing issue. If you let \displaystyle \begin{align*} u = x^{\frac{3}{2}} \end{align*}, that means in your denominator you are going to get \displaystyle \begin{align*} u - 8 \end{align*}. It's difficult to integrate \displaystyle \begin{align*} \frac{1}{u - 8} \end{align*}, in fact, you will need to use ANOTHER substitution. So to get around this, we remember we can substitute ANY function we like. So why not substitute \displaystyle \begin{align*} u = x^{\frac{3}{2}} - 8 \end{align*} instead? Its derivative is still \displaystyle \begin{align*} \frac{3}{2}x^{\frac{1}{2}} \end{align*}.

So by making the substitution \displaystyle \begin{align*} u = x^{\frac{3}{2}} - 8 \implies du = \frac{3}{2}x^{\frac{1}{2}}\,dx \end{align*} then the integral becomes

\displaystyle \begin{align*} \int{\frac{x^{\frac{1}{2}}}{x^{\frac{3}{2}} - 8} \, dx} &= \frac{2}{3} \int{\frac{\frac{3}{2}x^{\frac{1}{2}}}{x^{\frac{3} {2}} - 8}\, dx} \\ &= \frac{2}{3} \int{\frac{1}{u} \, du} \end{align*}

I'm sure you can integrate from here.

4. ## Re: Integral u-substitution

I would let:

$u=x^{\frac{3}{2}}-8\,\therefore\,du=\frac{3}{2}x^{\frac{1}{2}}\,dx$

So you now have:

$\frac{2}{3}\int\frac{1}{u}\,du$

You should get a logarithmic anti-derivative. Have you copied the problem correctly?

5. ## Re: Integral u-substitution

Originally Posted by johnsomeone
I'd advise writing the "dx" everywhere. It helps make what's happening during subsitution clearer I think.

I don't think you should be getting what you say you "should" be getting, since the numerator there is, up to a constant multiple, the derivative of the denominator.
Yep, sorry, I should include that - was just typing in a hurry.

I'm not sure to be honest, I thought my answer made sense, but I normally run most questions through Wolfram to confirm I'm getting them right, and according to Wolfram, I should be getting...

$\frac {-2}{9(x^{3/2} - 8)^3}$

6. ## Re: Integral u-substitution

That's what you should get if you are given:

$\int\frac{x^{\frac{1}{2}}}{\left(x^{\frac{3}{2}}-8 \right)^4}\,dx$

7. ## Re: Integral u-substitution

Originally Posted by MarkFL2
That's what you should get if you are given:

$\int\frac{x^{\frac{1}{2}}}{\left(x^{\frac{3}{2}}-8 \right)^4}\,dx$
That's right, I made a bit of a mess on the original post. Really sorry about that. I'm on a phone, and didn't exactly check everything properly :/.

Apologies.

8. ## Re: Integral u-substitution

Originally Posted by astuart
Yep, sorry, I should include that - was just typing in a hurry.

I'm not sure to be honest, I thought my answer made sense, but I normally run most questions through Wolfram to confirm I'm getting them right, and according to Wolfram, I should be getting...

$\frac {-2}{9(x^{3/2} - 8)^3}$
Wolfram gives this, which is not what you are writing...

9. ## Re: Integral u-substitution

No worries, it happens, well at least I can say I have had mistakes in posts before!

I think your error will be corrected when you make the correct substitution for the differential.

10. ## Re: Integral u-substitution

Originally Posted by MarkFL2
No worries, it happens, well at least I can say I have had mistakes in posts before!

I think your error will be corrected when you make the correct substitution for the differential.
Now when you're talking about making the correct substitution for the differential, are you talking about the following that 'Prove It' mentioned?

Some things to think about though. First of all, you need as a multiple, not just . That's easily fixed, just remember that whatever you multiply by you need to divide by as well, to keep the expression equivalent to what you started with.
I wasn't aware that was a required step to be honest. I was aware that it would be set up with the (3/2) constant making the equation equal to the derivative. But that's not the case, is it?

$\frac {3}{2}\int \frac {x^{1/2}}{x^{3/2}-8)^4}dx$ isn't right, it should be $\frac{3}{2}\,\int \frac {\frac{3}{2}x^{1/2}}{(x^{3/2}-8)^4}}$

11. ## Re: Integral u-substitution

Originally Posted by astuart
Now when you're talking about making the correct substitution for the differential, are you talking about the following that 'Prove It' mentioned?

I wasn't aware that was a required step to be honest. I was aware that it would be set up with the (3/2) constant making the equation equal to the derivative. But that's not the case, is it?

$\frac {3}{2}\int \frac {x^{1/2}}{x^{3/2}-8)^4}dx$ isn't right, it should be $\frac{3}{2}\,\int \frac {\frac{3}{2}x^{1/2}}{(x^{3/2}-8)^4}}$
You can NOT fiddle with an integrand so that it is something different, and integrate it, then expect to get the right answer. The only "fiddling" that is allowed is when you get an expression EQUIVALENT to what you started with.

\displaystyle \begin{align*} \frac{3}{2} \int{\frac{x^{\frac{1}{2}}}{\left( x^{\frac{3}{2}} - 8 \right)^4 } \, dx} \end{align*} is not the same as \displaystyle \begin{align*} \int{\frac{x^{\frac{1}{2}}}{\left( x^{\frac{3}{2}} - 8 \right)^4 } \, dx} \end{align*}, and neither is \displaystyle \frac{3}{2}\int{\frac{\frac{3}{2}x^{\frac{1}{2}}}{ \left( x^{\frac{3}{2}} - 8 \right)^4 } \, dx}\begin{align*} \end{align*}. But \displaystyle \begin{align*} \frac{2}{3} \int{\frac{\frac{3}{2}x^{\frac{1}{2}}}{\left( x^{\frac{3}{2}} - 8 \right)^4}\, dx} \end{align*} is.

12. ## Re: Integral u-substitution

No, you should have:

$x^{\frac{1}{2}}\,dx=\frac{2}{3}\,du$ so the integral becomes:

$\frac{2}{3}\int u^{-4}\,du$

13. ## Re: Integral u-substitution

Originally Posted by Prove It
You can NOT fiddle with an integrand so that it is something different, and integrate it, then expect to get the right answer. The only "fiddling" that is allowed is when you get an expression EQUIVALENT to what you started with.

\displaystyle \begin{align*} \frac{3}{2} \int{\frac{x^{\frac{1}{2}}}{\left( x^{\frac{3}{2}} - 8 \right)^4 } \, dx} \end{align*} is not the same as \displaystyle \begin{align*} \int{\frac{x^{\frac{1}{2}}}{\left( x^{\frac{3}{2}} - 8 \right)^4 } \, dx} \end{align*}, and neither is \displaystyle \frac{3}{2}\int{\frac{\frac{3}{2}x^{\frac{1}{2}}}{ \left( x^{\frac{3}{2}} - 8 \right)^4 } \, dx}\begin{align*} \end{align*}. But \displaystyle \begin{align*} \frac{2}{3} \int{\frac{\frac{3}{2}x^{\frac{1}{2}}}{\left( x^{\frac{3}{2}} - 8 \right)^4}\, dx} \end{align*} is.
Right, I understand that, but I don't understand where the $\frac {2}{3}$ came from? Could you explain how that came to be, because that's what's causing me grief at the moment. It's the constant that the numerator was multiplied by, so that it was equal to $du$, but it's been inverted. What part of the u-substitution process calls for it to be inverted?

Is it simply because you have to divide the entire equation by $\frac {3}{2}$ as well, due to multiplying it by $\frac {3}{2}$ or something else?

14. ## Re: Integral u-substitution

Originally Posted by astuart
Is it simply because you have to divide the entire equation by $\frac {3}{2}$ as well, due to multiplying it by $\frac {3}{2}$ ?
Yes, pretty much. Just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

The general drift is...

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

15. ## Re: Integral u-substitution

Originally Posted by astuart
Right, I understand that, but I don't understand where the $\frac {2}{3}$ came from? Could you explain how that came to be, because that's what's causing me grief at the moment. It's the constant that the numerator was multiplied by, so that it was equal to $du$, but it's been inverted. What part of the u-substitution process calls for it to be inverted?

Is it simply because you have to divide the entire equation by $\frac {3}{2}$ as well, due to multiplying it by $\frac {3}{2}$ or something else?
That's exactly right. To keep the expression exactly the same as it was, it needs to be multiplied by 1. Notice that \displaystyle \begin{align*} \frac{2}{3} \cdot \frac{3}{2} = 1 \end{align*}.