Integration Using Substitution (College Board AP® Calculus AB)

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Roger B

Expertise

Maths

Determining indefinite integrals using u-substitutions

What is integration by substitution?

Substitution simplifies an integral by defining an alternative variable (usually $u$ ) in terms of the original variable (usually $x$ )
- The integral in $u$ is much easier to solve than the original integral in $x$
- The substitution can be reversed at the end to get the answer in terms of $x$

How do I integrate simple functions using u-substitution?

In a simple integral involving substitution, you will usually be integrating a composite function (i.e., 'function of a function')
- These can also be solved 'by inspection'
  - See the 'Integrals of Composite Functions' study guide
- Substitution can be a safer method when 'by inspection' is awkward or difficult to spot
STEP 1
Identify the substitution to be used – it will be the secondary (or 'inside') function in a composite function
- I.e. if the integral involves $f (g (x))$ , let $u = g (x)$
- E.g. $\int 3 x \cos (x^{2} - 5) d x$
  - Let $u = x^{2} - 5$
STEP 2
Differentiate the substitution and rearrange
- $\frac{d u}{d x}$ can be treated like a fraction (i.e. “multiply by $d x$ " to get rid of fractions)
- E.g. $u = x^{2} - 5 \Rightarrow \frac{d u}{d x} = 2 x$
  - Then $d u = 2 x d x \Rightarrow x d x = \frac{1}{2} d u$
STEP 3
Replace all parts of the integral
- All $x$ terms should be replaced with equivalent $u$ terms, including $d x$
- E.g. $\int 3 x \cos (x^{2} - 5) d x = 3 \int \cos (x^{2} - 5) \cdot x d x$
  - So $\int 3 x \cos (x^{2} - 5) d x = 3 \int \cos u \cdot \frac{1}{2} d u = \frac{3}{2} \int \cos u d u$
STEP 4
Integrate
- E.g. $\frac{3}{2} \int \cos u d u = \frac{3}{2} \sin u + C$
  - Don't forget the constant of integration
STEP 5
Substitute $x$ back in
- Replace $u$ everywhere with the equivalent expression for $x$
- E.g. $\frac{3}{2} \sin u + C = \frac{3}{2} \sin (x^{2} - 5) + C$
  - So $\int 3 x \cos (x^{2} - 5) d x = \frac{3}{2} \sin (x^{2} - 5) + C$

Worked Example

Find the indefinite integral $\int \frac{x^{2} - 1}{2 x^{3} - 6 x + 7} d x$ .

Answer:

Here the 'main' function in the composite function is $\frac{1}{(. . .)}$ , and the 'inside function' is $2 x^{3} - 6 x + 7$

This is an integral that could also be solved by using $\int \frac{f^{'} (x)}{f (x)} d x = \ln |f (x)| + C$ , but here we'll use substitution

Let $u = 2 x^{3} - 6 x + 7$

Differentiate the substitution and rearrange

$\frac{d u}{d x} = 6 x^{2} - 6 d u = 6 (x^{2} - 1) d x \Rightarrow (x^{2} - 1) d x = \frac{1}{6} d u$

Replace all parts of the integral

$\int \frac{x^{2} - 1}{2 x^{3} - 6 x + 7} d x = \int \frac{1}{u} \cdot \frac{1}{6} d u = \frac{1}{6} \int \frac{1}{u} d u$

Integrate

$\frac{1}{6} \int \frac{1}{u} d u = \frac{1}{6} \ln |u| + C$

Substitute $x$ back in

$\int \frac{x^{2} - 1}{2 x^{3} - 6 x + 7} d x = \frac{1}{6} \ln |2 x^{3} - 6 x + 7| + C$

How do I integrate more complicated functions using u-substitution?

The procedure here is exactly the same as for integrating simpler functions
- However the substitution to use may not be as obvious
- Practice questions like this to improve your integration by substitution skills
E.g. $\int x \sqrt{x - 4} d x$
- Note that this is not an integral that can be solved 'by inspection' (i.e. by the 'reverse chain rule')
- Identify the substitution
  - Let $u = x - 4$
- Differentiate the substitution and rearrange
  - $\frac{d u}{d x} = 1 \Rightarrow d u = d x$
- Replace all parts of the integral
  - $u = x - 4 \Rightarrow x = u + 4$
  - So $\int x \sqrt{x - 4} d x = \int (u + 4) \sqrt{u} d u$
- Integrate
  - $\int (u + 4) \sqrt{u} d u = \int (u + 4) u^{\frac{1}{2}} d u = \int (u^{\frac{3}{2}} + 4 u^{\frac{1}{2}}) d u = \frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}} + C$
- Substitute $x$ back in
  - $\int x \sqrt{x - 4} d x = \frac{2}{5} {(x - 4)}^{\frac{5}{2}} + \frac{8}{3} {(x - 4)}^{\frac{3}{2}} + C$

Worked Example

Find the indefinite integral $\int \frac{d x}{\sqrt{16 - 9 x^{2}}}$ .

Answer:

To spot the substitution to use here it helps to recall the standard integral $\int \frac{d x}{\sqrt{1 - x^{2}}} = \arcsin x + C$

First rearrange the integral slightly

$\int \frac{d x}{\sqrt{16 - 9 x^{2}}} = \int \frac{d x}{4 \sqrt{1 - \frac{9}{16} x^{2}}} = \frac{1}{4} \int \frac{d x}{\sqrt{1 - \frac{9}{16} x^{2}}}$

Now the substitution to use is more obvious

Let $u = \frac{3}{4} x \Rightarrow u^{2} = \frac{9}{16} x^{2}$

Differentiate the substitution and rearrange

$\frac{d u}{d x} = \frac{3}{4} \Rightarrow d x = \frac{4}{3} d u$

Replace all parts of the integral

$\frac{1}{4} \int \frac{d x}{\sqrt{1 - \frac{9}{16} x^{2}}} = \frac{1}{4} \int \frac{\frac{4}{3} d u}{\sqrt{1 - u^{2}}} = \frac{1}{3} \int \frac{d u}{\sqrt{1 - u^{2}}}$

Integrate

$\frac{1}{3} \int \frac{d u}{\sqrt{1 - u^{2}}} = \frac{1}{3} \arcsin u + C$

Substitute $x$ back in

$\int \frac{d x}{\sqrt{16 - 9 x^{2}}} = \frac{1}{3} \arcsin (\frac{3}{4} x) + C$

Evaluating definite integrals using u-substitutions

How do I evaluate definite integrals using u-substitution?

Definite integrals can also be solved using u-substitution
- You just need to rewrite the integration limits in terms of u as well
E.g. $\int_{4}^{8} x \sqrt{x - 4} d x$
- We've already seen that this can be integrated using the substitution $u = x - 4$
  - with $d u = d x$ and $x = u + 4$ following from this
- We just need to change the integral limits as well
  - When $x = 4$ , $u = 4 - 4 = 0$
  - When $x = 8$ , $u = 8 - 4 = 4$
- Now when we replace all parts of the integral we get
  - $\int_{4}^{8} x \sqrt{x - 4} d x = \int_{0}^{4} (u + 4) \sqrt{u} d u$
- Integrate, then evaluate the definite integral using the $u$ values
  - $\begin{array}{rcl} \int_{0}^{4} (u + 4) \sqrt{u} d u & = & {[\frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}}]}_{0}^{4} \\ = & \frac{2}{5} \cdot 32 + \frac{8}{3} \cdot 8 - (\frac{2}{5} \cdot 0 + \frac{8}{3} \cdot 0) \\ = & \frac{512}{15} \end{array}$
- Note that there is no need to substitute $x$ back in to evaluate the definite integral!

Worked Example

Evaluate the definite integral $\int_{1}^{2} \frac{2 x + 3}{3 x^{2} + 9 x - 5} d x$ .

Answer:

Choose the substitution

Let $u = 3 x^{2} + 9 x - 5$

Differentiate the substitution and rearrange

$\frac{d u}{d x} = 6 x + 9 d u = 3 (2 x + 3) d x \Rightarrow (2 x + 3) d x = \frac{1}{3} d u$

Find the integration limits in terms of $u$

When $x = 1$ , $u = 3 {(1)}^{2} + 9 (1) - 5 = 7$

When $x = 2$ , $u = 3 {(2)}^{2} + 9 (2) - 5 = 25$

Replace all parts of the integral, including the integration limits

$\int_{1}^{2} \frac{2 x + 3}{3 x^{2} + 9 x - 5} d x = \int_{7}^{25} \frac{1}{u} \cdot \frac{1}{3} d u = \frac{1}{3} \int_{7}^{25} \frac{1}{u} d u$

Integrate and evaluate the definite integral

$\frac{1}{3} \int_{7}^{25} \frac{1}{u} d u = \frac{1}{3} {[\ln u]}_{7}^{25} = \frac{1}{3} (\ln 25 - \ln 7)$

Use laws of logarithms to simplify the final answer

$\int_{1}^{2} \frac{2 x + 3}{3 x^{2} + 9 x - 5} d x = \frac{1}{3} \ln (\frac{25}{7})$

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Integration Using Substitution (College Board AP® Calculus AB)

Revision Note

Determining indefinite integrals using u-substitutions

What is integration by substitution?

How do I integrate simple functions using u-substitution?

Worked Example

How do I integrate more complicated functions using u-substitution?

Worked Example

Evaluating definite integrals using u-substitutions

How do I evaluate definite integrals using u-substitution?

Worked Example

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Author: Roger B