Integral sorting

Integral sorting

Excerpt: Students are asked to pair up cards containing integrals and their answers, as well as find a suitable substitution that could be used to solve the integral.

Problem

Below are 8 integrals and their solutions. Can you:

• match up the cards and fill in the blanks?

• write down a substitution that could have been used to solve each integral?

• sort the pairs into groups. What features could be used to define the groups?

$\displaystyle{\int 9x\sqrt{1-9x^2} \, dx}$

$\displaystyle{\int \dfrac{x^2-1}{(x^3 - 3x)^2} \, dx}$

$\displaystyle{\int \dfrac{1}{\sqrt{1-9x^2}} \, dx}$

$\displaystyle{\int \dfrac{9x}{\sqrt{1-9x^2}} \, dx}$

$\displaystyle{\int \dfrac{6-6x^2}{x^3 - 3x} \, dx}$

$\displaystyle{\int \dfrac{\cos x}{\sin^3 x} \, dx}$

$\displaystyle{\int \dfrac{\cos x}{\sin x} \, dx}$

$\dfrac{-1}{2\sin^2x}+c$

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

$\ln{|\sin x|}+c$

$\frac{1}{3}\arcsin 3x +c$

$\dfrac{-1}{3(x^3-3x)}+c$

$-\frac{1}{3}\left(1-9x^2\right)^{\frac{3}{2}} +c$

$-2\ln{|x^3-3x|}+c$

Solution

In some of the integrals there are similar expressions we could use for sorting. B and F both contain $x^3 - 3x$, as do the solutions 2, 6 and 8. To match them we could try to integrate B and F, or differentiate 2, 6 and 8. We find that we can pair them up, and that card 2 differentiates to $\frac{1}{3}(x^3-3x)^{1/2}(3x^2-3)$ which must match the blank card C.

$\displaystyle{\int \dfrac{x^2-1}{(x^3 - 3x)^2} \, dx}$

$\dfrac{-1}{3(x^3-3x)}+c$

$\displaystyle{\int \dfrac{6-6x^2}{x^3 - 3x} \, dx}$

$-2\ln{|x^3-3x|}+c$

$\displaystyle{\int {(x^2-1)}\sqrt{x^3 - 3x} \, dx}$

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

To help us find the correct substitutions let us think about the chain rule. We should be familiar with how we differentiate a composite function.

$$\dfrac{d}{dx} \Big(g(f(x))\Big) = g'(f(x))\times f'(x)$$

If we integrate both sides we can write

$$g(f(x)) = \int g'(f(x))\times f'(x) \, dx.$$

This integral is made up of a composite function multiplied by the differential of the inner function.

If you spot this pattern in an integral, you may be able to integrate it by inspection without needing to substitute. However, as integrals get more complex, it will help to find a substitution that will reduce the integral to $g(u) = \int g'(u) \, du.$

Taking a closer look at the integrals above, we find that B, C and F all include a function of $x^3 - 3x$, as well as a multiple of $3x^2 - 3$, which is the derivative of $x^3 - 3x$. This suggests that the substitution $u = x^3 - 3x$ will work for all three integrals.

Try using this substitution to check that it works. Are there any other options for the substitution?

Now let’s look at the integrals which involve trigonometric functions.

$\displaystyle{\int \dfrac{\cos x}{\sin^3 x} \, dx}$

$\dfrac{-1}{2\sin^2x}+c$

$\displaystyle{\int \dfrac{\cos x}{\sin x} \, dx}$

$\ln{|\sin x|}+c$

For G we see that $\dfrac{1}{\sin^3 x}$ is the composite function. As $\sin^3 x$ is more clearly written as $(\sin x)^3$ we see that $\sin x$ is the inner function and so $u = \sin x$ is a suitable substitution.

For H it is less obvious, but remember $\dfrac{1}{\sin x}$ is a composite function of $\dfrac{1}{x}$ and $\sin x$, so we can use the same substitution of $u = \sin x$.

Why can’t we use the substitution $u = \cos x$? What happens if we do?

Our final three cards match up like this:

$\displaystyle{\int 9x\sqrt{1-9x^2} \, dx}$

$-\frac{1}{3}\left(1-9x^2\right)^{\frac{3}{2}} +c$

$\displaystyle{\int \dfrac{1}{\sqrt{1-9x^2}} \, dx}$

$\frac{1}{3}\arcsin 3x +c$

$\displaystyle{\int \dfrac{9x}{\sqrt{1-9x^2}} \, dx}$

$-\sqrt{1-9x^2} +c$

A and E are similar problems with composite functions $\sqrt{1-9x^2}$ and $\dfrac{1}{\sqrt{1-9x^2}}$. For both, a suitable substitution is $u = 1-9x^2$. This allows us to complete the blank card 5 to match with E.

D also includes $\dfrac{1}{\sqrt{1-9x^2}}$, but not the derivative of $1-9x^2$. What happens if you try the substitution $u = 1-9x^2$?

For this type of integral a trigonometric substitution is useful since $\sin^2 x + \cos^2 x =1$. One possible substitution is $3x = \sin u.$

• How is this different from the other substitutions that have been used?

• Would it matter if you used $x = \sin u$ or $3x = \cos u$ instead?

Rather than focusing on the expressions the integrals contained we could think about the structure of each integral.

$\displaystyle{\int \dfrac{6-6x^2}{x^3 - 3x} \, dx}$

$\displaystyle{\int \dfrac{\cos x}{\sin x} \, dx}$

$\displaystyle{\int \dfrac{x^2-1}{(x^3 - 3x)^2} \, dx}$

$\displaystyle{\int \dfrac{9x}{\sqrt{1-9x^2}} \, dx}$

$\displaystyle{\int \dfrac{\cos x}{\sin^3 x} \, dx}$

$\displaystyle{\int 9x\sqrt{1-9x^2} \, dx}$

$\displaystyle{\int {(x^2-1)}\sqrt{x^3 - 3x} \, dx}$

$\displaystyle{\int \dfrac{1}{\sqrt{1-9x^2}} \, dx}$

What are the features of each group?

Can you write down a different integral that would fit into each group?