When does substitution not work?

[TylerH]

I noticed that there are some functions that when integrated by substitution, are incorrect. Such as (1-x^2)^(-1/2). The answer is obviously arcsinx, but if you integrate with substitution, set u = 1-x^2, du = -2x dx. Then use anti power rule to go from u^(-.5) to 2u^(.5), then divide by -2x and rewrite u in terms of x you get -(1-x^2)^.5/x. As you'll notice, there's an asymptote at x=0, thus they are not the same(although very similar in terms of slope at x near +-1). Closer analysis shows that it's the quotient rule that leads to extraneous terms when you differentiate the false integral.

So, my question is: whether there is a set of rules to know when substitution will result in something like this? And if so, what are they?

PS I tried using Latex, and for some reason it keeps showing the last equation I posted yesterday. I don't know whether it's a BB bug or a FF bug, but it's annoying.

Thanks,
Tyler

 

[phyzguy]

As far as the LaTex bug, if you hit "Preview Post" and then refresh your browser before hitting "Submit Reply" this will fix it.

I think you are confused about how integration by substitution works. You need to substitute for the integration measure (dx, du, or whatever) before you do the integral, not after. So the substitution you suggested would go as follows:
$$\int\frac{1}{\sqrt{1-x^2}}dx$$
$$u=1-x^2; x=\sqrt{1-u}; du = -2x dx ; dx = \frac{-du}{2\sqrt{1-u}}$$
$$\int\frac{1}{\sqrt{1-x^2}}dx = -\int\frac{du}{2\sqrt{u-u^2}}$$

This hasn't really simplified things, and isn't simply a power that you can apply the anti-power rule.

 

[TylerH]

OH! Well, I guess that's what I get for trying to learn Calculus from Wikipedia. (Not that I blame them, I blame myself for delving intosuch a topic without being properly prepared.) So, to do substitution, you

$$\int f(g(x)) dx$$
$$u=g(x);x=g^{-1}(u);du=dg(x);dx=dg^{-1}(u)$$
$$\int f(g(x)) dx = \int f(g(x))*dg^{-1}(u)$$
Is that right? Seems wrong. Once you have all the info in line 2, what do you do with it?

 

[Mark44]

OH! Well, I guess that's what I get for trying to learn Calculus from Wikipedia. (Not that I blame them, I blame myself for delving intosuch a topic without being properly prepared.) So, to do substitution, you

$$\int f(g(x)) dx$$
$$u=g(x);x=g^{-1}(u);du=dg(x);dx=dg^{-1}(u)$$
$$\int f(g(x)) dx = \int f(g(x))*dg^{-1}(u)$$
Is that right? Seems wrong. Once you have all the info in line 2, what do you do with it?

With the substitution u = g(x), du = g'(x)dx, you get
$$\int f(u) du$$
If you can find an antiderivative for f, then the choice for the substitution was a good one. The substitution method is basically the chain rule in reverse.

For example, suppose you need to evaluate the following:
$$\int xsin(x^2) dx$$

The obvious substitution is u = x², du = 2xdx.
So
$$\int xsin(x^2) dx = (1/2) \int sin(x^2)2xdx = (1/2)\int sin(u)du = -(1/2)cos(u) + C$$
$$=-(1/2)cos(x^2) + C$$

 

[TylerH]

So, the x outside the sine canceled out with the x in du, you divided by 2 because of the 2 in du, and it's negative because the antiderivative of sine is -cosine. Got it, thanks.

 

[Prove It]

So, the x outside the sine canceled out with the x in du, you divided by 2 because of the 2 in du, and it's negative because the antiderivative of sine is -cosine. Got it, thanks.

$$\displaystyle \int{f\left(u(x)\right)\,\frac{du}{dx}\,dx} = \int{f(u)\,du}$$.

Basically your integrand needs to have an "inner" function and also have the inner function's derivative as a factor.

For your example

$$\displaystyle \int{x\,\sin{(x^2)}\,dx}$$

the "inner" function is $$\displaystyle x^2$$, and its derivative is $$\displaystyle 2x$$. Your function almost has $$\displaystyle 2x$$ as a factor.

If you rewrite it as $$\displaystyle \int{x\,\sin{(x^2)}\,dx} = \frac{1}{2}\int{2x\,\sin{(x^2)}\,dx}$$ now the integral is of the required form.

So make the substitution $$\displaystyle u = x^2$$ so that $$\displaystyle \frac{du}{dx} = 2x$$ and the integral becomes

$$\displaystyle \frac{1}{2}\int{\sin{u}\,\frac{du}{dx}\,dx} = \frac{1}{2}\int{\sin{u}\,du}$$

$$\displaystyle = -\frac{1}{2}\cos{u} + C$$

$$\displaystyle = -\frac{1}{2}\cos{(x^2)} + C$$.