# Definite Integrals with Substitution

## Key Questions

• This is a classic case if what's called u-substitution. Meaning that you have to find a function ( u) and it's derivative (du) in the expression. Both the function and it's derivative may be hidden behind coefficients and confusing notation.

In this case, I might start by using exponents to rewrite the integral without fractions.

${\int}_{0}^{2} \frac{10 x}{\sqrt{3 - {x}^{2}}} \mathrm{dx} = {\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(10 x\right) \mathrm{dx}$

Now, I'm looking for a function and it's derivative. Here's where I notice that I have a second degree polynomial ( $3 - {x}^{2}$) inside the exponent part, and I have a first degree polynomial ($10 x$) elsewhere in the integral. In general a polynomial that starts with an x^2 will have a derivative that starts with an x. So:

If I decide that

$u = 3 - {x}^{2}$

then

$\mathrm{du} = - 2 x \mathrm{dx}$

but I don't have a $- 2 x$ anywhere on the integral. I do have a $10 x$.

I'm allowed to manipulate coefficients in an integral (manipulating variables is trickier, and sometimes not possible). So I need to manipulate the $10$ in front of the x into being a $- 2$ in front of an x.

${\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(10 x\right) \mathrm{dx}$

We can pull a coefficient factor out of the integral entirely. We don't have to, but I find it makes for a more clear picture.

$10 {\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(x\right) \mathrm{dx}$

Now we would really like to put a $- 2$ in front of the x, so that we can match the $- 2 x$ above. And we can put in in there. As long as we put it's reciprocal out in front to balance things out.

$10 \left(\frac{1}{- 2}\right) {\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right) \mathrm{dx}$

=$- 5 {\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right) \mathrm{dx}$

Now we have our u and our du. But there's one other thing we need to be aware of.

This is a definite integral, and the 0 and the 2 represent x values, not u values. But we have a way to convert them,

$u = 3 - {x}^{2}$

So for $x = 0$, we get $u = 3 - {0}^{2}$, or $u = 3$.

For $x = 2$, we get $u = 3 - {2}^{2}$, or $u = - 1$.

Now we substitute all our x's for u's.

$- 5 {\int}_{0}^{2} {\left(3 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right) \mathrm{dx} = - 5 {\int}_{3}^{- 1} {u}^{- \frac{1}{2}} \mathrm{du}$

$= - 5 \frac{{u}^{\frac{1}{2}}}{\frac{1}{2}} {|}_{3}^{- 1} = - 10 {u}^{\frac{1}{2}} {|}_{3}^{- 1}$

$= - 10 \sqrt{- 1} - \left(- 10 \sqrt{3}\right)$

$= - 10 i + 10 \sqrt{3}$, or

$10 \left(\sqrt{3} - i\right)$

Hope this helps.

• Since dx has been placed twice, I'm going to assume that the equation should read ${\int}_{e}^{{e}^{4}} \frac{\mathrm{dx}}{x \cdot \sqrt{\ln \left(x\right)}}$. In this case, a good way to find the integral is by substitution, letting $u = \ln \left(x\right)$.

To integrate something by substitution (also known as the change-of-variable rule), we need to select a function $u$ so that its derivative also forms part of the original equation. (For example, when we try to antidifferentiate $\tan \left(x\right)$ we can say that $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$ and select $u = \cos \left(x\right)$. The derivative of $u$ is also within the original equation.)

To find the integral of your function, we do the following:

1. Let $u = \ln \left(x\right)$, then $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$ - this is a standard derivative.
2. Substitute these two new functions into the equation:
${\int}_{e}^{{e}^{4}} \frac{\mathrm{dx}}{x \cdot \sqrt{\ln \left(x\right)}} = {\int}_{e}^{{e}^{4}} \frac{1}{x \cdot u} \mathrm{dx} = {\int}_{e}^{{e}^{4}} \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$
3. Find new terminals - this is a crucial step, because we're changing the variable!
$\text{When "x = e, u = ln(e) = 1 " and when } x = {e}^{4} , u = \ln \left({e}^{4}\right) = 4$
4. Substitute these new terminals in, and "cancel out" the two $\mathrm{dx}$ terms:
${\int}_{e}^{{e}^{4}} \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx} = {\int}_{1}^{4} \frac{\mathrm{du}}{u} = {\int}_{1}^{4} {u}^{- 1} \mathrm{du}$
5. Integrate normally. The answer you get for this "new" integral will be exactly the same answer as the original integral.
• The only thing different from indefinite integral is that you will have to convert the original lower and upper limits into the new ones.

Let us evaluate the definite integral below.

${\int}_{0}^{2} 2 x {e}^{{x}^{2} + 1} \mathrm{dx}$

Let $u = {x}^{2} + 1$.
Rightarrow {du}/{dx}=2x Rightarrow du=2x du

When $x = 0$, $u = {\left(0\right)}^{2} + 1 = 1$
When $x = 2$, $u = {\left(2\right)}^{2} + 1 = 5$

If x goes from 0 to 2, then u goes from 1 to 5.

By Substitution,

${\int}_{0}^{2} 2 x {e}^{{x}^{2} + 1} \mathrm{dx} = {\int}_{1}^{5} {e}^{u} \mathrm{du} = {\left[{e}^{u}\right]}_{1}^{5} = {e}^{5} - e$