Why should I use a substitution here?

[thomas49th]

Homework Statement

Area = \frac{2}{\sqrt{3}}\int\sqrt{1-\frac{x^{2}}{4e^{2}}}dx

Apparently I should use a substitution x/2e = sin(u) and I can see why it's those value but I don't understand why they need the substitution in the first place. Isn't the integral simply:

\frac{2}{\sqrt{3}}\cdot \frac{2}{3}\cdot \frac{-4e^{2}}{2x} (1-\frac{x^{2}}{4e^{2}})^{\frac{3}{2}}

Having tried the other method, the answers are different and I cannot see why.

Thanks
Thomas

 

[HallsofIvy]

No, it isn't. It looks to me like you are trying to apply the chain rule "in reverse"- dividing by the derivative of a function of x rather than multiplying. You can't do that. If you had, say, (f(x))^2 and wanted to differentiate it, it would be 2(f(x))f'(x). But to integrate (f(x))^2, you can't just divide by f'(x). You could try dividing and multiplying by that:
\int (f(x))^2 \frac{f'(x)}{f'(x)}dx}
but then since f'(x) is a function of x, you cannot take it outside the integral.
That is NOT the same as
\frac{1}{f'(x)}\int (f(x))^2f'(x)dx= \frac{1}{3f'(x)}(f(x))^3

Did you look at the derivative of what you give?

The derivative of
\frac{-8e^2}{3x\sqrt{3}}\left(1- \frac{x^2}{4e^2}\right)^{3/2}
is
\frac{4e^2}{3x^2\sqrt{3}}\left(1- \frac{x^2}{4e^2}\right)^{3/2}- \frac{8e^2}{3x\sqrt{3}}\left(1- \frac{x^2}{4e^2}\right)^{1/2}\left(-\frac{x}{2e^2}\right)
not at all what you were trying to integrate.

The point of the substitution you mention,
sin(u)= \frac{x}{2e}
is that
\left(1- \frac{x^2}{4e^2}\right)^{1/2}
becomes
\left(1- sin^2(u)\right)^{1/2}= (cos^2(u))^{1/2}= cos(x)
and, of course,
cos(u)du= \frac{1}{2e}dx
so that
2e cos(u)du= dx
and the integral
\frac{2}{\sqrt{3}}\int \sqrt{1- \frac{x^2}{4e^2}}dx
becomes
\frac{2}{\sqrt{3}}\int (cos(u))(2e cos(u)du)= \frac{4e}{\sqrt{3}}\int cos^2(u)du
which can be integrated using the identity cos^2(u)= (1/2)(1+ cos(2u)).

That way

 

[thomas49th]

No, it isn't. It looks to me like you are trying to apply the chain rule "in reverse"- dividing by the derivative of a function of x rather than multiplying. You can't do that. If you had, say, (f(x))^2 and wanted to differentiate it, it would be 2(f(x))f'(x). But to integrate (f(x))^2, you can't just divide by f'(x). You could try dividing and multiplying by that:
\int (f(x))^2 \frac{f'(x)}{f'(x)}dx}
but then since f'(x) is a function of x, you cannot take it outside the integral.
That is NOT the same as
\frac{1}{f'(x)}\int (f(x))^2f'(x)dx= \frac{1}{3f'(x)}(f(x))^3

Did you look at the derivative of what you give?

The derivative of
\frac{-8e^2}{3x\sqrt{3}}\left(1- \frac{x^2}{4e^2}\right)^{3/2}
is
\frac{4e^2}{3x^2\sqrt{3}}\left(1- \frac{x^2}{4e^2}\right)^{3/2}- \frac{8e^2}{3x\sqrt{3}}\left(1- \frac{x^2}{4e^2}\right)^{1/2}\left(-\frac{x}{2e^2}\right)
not at all what you were trying to integrate.

The point of the substitution you mention,
sin(u)= \frac{x}{2e}
is that
\left(1- \frac{x^2}{4e^2}\right)^{1/2}
becomes
\left(1- sin^2(u)\right)^{1/2}= (cos^2(u))^{1/2}= cos(x)
and, of course,
cos(u)du= \frac{1}{2e}dx
so that
2e cos(u)du= dx
and the integral
\frac{2}{\sqrt{3}}\int \sqrt{1- \frac{x^2}{4e^2}}dx
becomes
\frac{2}{\sqrt{3}}\int (cos(u))(2e cos(u)du)= \frac{4e}{\sqrt{3}}\int cos^2(u)du
which can be integrated using the identity cos^2(u)= (1/2)(1+ cos(2u)).

That way

I was doing the chain rule in reverse. Is that method never never valid. I spent a year out doing little integration, have I just made the method up all together or is it allowed (just not in this circumstance). I'm sure that's how I was taught. Hmm

 

[Mark44]

You are using an ordinary substitution, u = 1 - x²/(4e²), so du = -2x/(4e²) dx.

In your original integral you have the dx, but not the remaining part needed to make du, so you are tacking on a factor of -2x/(4e²) and accounting for it by multiplying by the reciprocal, -4e²/(2x). You can't do that.

You can bring in factors that are constants, but you can't bring in factors that involve your variable x.

The quickest way to see that your substitution is incorrect is to differentiate the answer you got.

Here's an example that shows why your technique doesn't work.

\int cos(x^2) dx

Let u = x², so du = 2x dx.

The integrand is missing a factor of 2x, so let's just stick it in. To account for this, we'll also bring in a factor of 1/(2x), which we'll move outside the integral (this is what HallsOfIvy was saying was invalid).

\frac{1}{2x}\int cos(x^2) 2x ~dx = \frac{1}{2x} \int cos(u) du = \frac{1}{2x} sin(u) + C = \frac{1}{2x} sin(x^2) + C

By differentiating the answer above, you should see that you don't get cos(x²).

 

[LCKurtz]

I was doing the chain rule in reverse. Is that method never never valid?

The "reverse chain rule" is the u-substitution method.

Suppose, for example, you differentiate

y = (2 + x³)⁴ to get
y' = 4(2 + x³)³(3x²) = 12x²(2 + x³)³

Now suppose you are asked to integrate f(x) =(2 + x³)⁴. Notice that this didn't come from the above example because it is missing the "derivative of the inside" 3x². As a consequence, trying the u-substitution u =2 + x³ won't help integrate f(x).

That's why u-substitutions are not appropriate for some integrals. But suppose you have a problem that did come from the chain rule like:

\int x^2(2+x^3)^4\, dx

Here the "derivative of the inside" is there, to within a constant. That's the x² term. Now if you try the u-substitution u =2 + x³ it will work with you only having to adjust for the missing constant. That reverses the chain rule. The u-substitution method is the "reverse chain rule".

 

[thomas49th]

Thanks, I see what you mean