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Integration Techniques: (lesson 1 of 4)

Integration by Substitution

The substitution method turns an unfamiliar integral into one that can be evaluatet. In other words, substitution gives a simpler integral involving the variable $u$. This lesson shows how the substitution technique works.

Let's review the five steps for integration by substitution.

Step 1: Choose a new variable $u$

Step 2: Determine the value $dx$

Step 3: Make the substitution

Step 4: Integrate resulting integral

Step 5: Return to the initial variable x

Integration by substituting $u = ax + b$

These are typical examples where the method of substitution is used.

Example 1: Solve:

$$ \int {(2x + 3)^4dx} $$

Solution:

Step 1: Choose the substitution function $u$

The substitution function is $\color{blue}{u = 2x + 3}$

Step 2: Determine the value of $dx$

$$ \begin{aligned} 2x + 3 &= u \\ (2x + 3)'dx &= du \\ 2dx &= du \\ \color{red}{dx} &\color{red}{= \frac{1}{2}du} \end{aligned} $$

Step 3: Do the substitution

$$ \int {\color{blue}{(2x + 3)}^4 dx} = \int {\color{blue}{u}^4} \cdot \color{red}{\frac{1}{2} \cdot du} $$

Step 4: Integrate the resulting integral

$$ \int {{{(2x + 3)}^4}} dx = \int {{u^4}} \cdot \frac{1}{2} \cdot du \color{red}{=} \frac{1}{2}\int {{u^4}} du = \frac{1}{2} \cdot \frac{{{u^{4 + 1}}}}{{4 + 1}} + C = \frac{{{u^5}}}{{10}} + C $$

Step 5: Return to the initial variable: $x$

$$ \frac{\color{blue}{u}^5}{10} + C = \frac{\color{blue}{(2x + 3)}^5}{10} + C $$

So, the solution is: $\int (2x + 3) ^ 4 dx = \frac{(2x + 3)^5}{10} + C $

Example 2: Solve:

$$\int {\frac{15}{3 - 2x}dx}$$

Solution:

Step 1: Chose a substitution function $u$

The substitution function is $\color{blue}{u = 3 - 2x}$

Step 2: Determine the value $dx$

$$ \begin{aligned} 3 - 2x &= u \\ (3 - 2x)' dx &= du \\ -2dx &= du \\ \color{red}{dx} &\color{red}{= - \frac{1}{2} du} \end{aligned} $$

Step 3: Do the substitution

$$ \int {\frac{15}{\color{blue}{3 - 2x}} \color{red}{dx}} = \int {\frac{15}{\color{blue}{u}} \cdot \color{red}{\left( { - \frac{1}{2}} \right) \cdot du}} $$

Step 4: Integrate resulting integral

$$ \int {\frac{15}{u} \cdot \left( { - \frac{1}{2}} \right) \cdot du = - \frac{15}{2}\int {\frac{1}{u}du = - \frac{15}{2}\ln \left| u \right|} + C} $$

Step 5: Return to the initial variable: $x$

$$ - \frac{{15}}{2}\ln \left| \color{blue}{u} \right| + C = - \frac{{15}}{2}\ln \left| \color{blue}{3 - 2x} \right| + C $$

The solution is: $\int {\frac{{15}}{{3 - 2x}}dx = - \frac{{15}}{2}\ln \left| {3 - 2x} \right|} + C$

Exercise 1: Solve using substitution $u = ax + b$

Level 1

$$ \color{blue}{\int (2x + 4) dx =} $$ $ \frac{1}{2} \sin{(2x + 4)} + C $
$ - \frac{1}{2} \sin{(2x + 4)} + C $
$ \frac{1}{2} \cos{(2x + 4)} + C $
$ - \frac{1}{2} \cos{(2x + 4)} + C $

Level 2

$$ \color{blue}{\int {\frac{4}{4 - 5x}} dx =} $$ $ - \frac{4}{5} \ln\left| 4 - 5x \right| + C $
$ \frac{4}{5} \ln\left| 4 - 5x \right| + C $
$ - \frac{5}{4} \ln\left| 4 - 5x \right| + C $
$ \frac{5}{4} \ln\left| 4 - 5x \right| + C $

More complicated examples

The steps for integration by substitution in this section are the same as the steps for previous one, but make sure to chose the substitution function wisely.

Example 3: Solve:

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

Solution:

$$ \text{Let} \ \ \color{blue}{u = x^2} \ \ \text{then:} \ \ du = (x^2)' dx = 2xdx \ \Rightarrow \ \color{red}{xdx = \frac{du}{2}} \\ \int {\color{red}{x} \sin (\color{blue}{x^2}) \color{red}{dx} = } \int {\sin \color{blue}{u} \cdot \color{red}{\frac{{du}}{2}}} = \frac{1}{2}\int {\sin udu = - \frac{1}{2}\cos u + C = - \frac{1}{2}\cos {x^2} + C} $$

Example 4: Solve:

$$ \int {{x^2}} \sqrt {1 + {x^3}} dx $$

Solution:

$$ \text{Let} \ \ \color{blue}{u = 1 + x^3} \ \ \text{then:} \ \ du = (1 + x^3)' dx = 3x^2dx \ \Rightarrow \ \color{red}{x^dx = \frac{du}{3}} \\ \int {\color{red}{x^2}} \sqrt {\color{blue}{1 + x^3}} \color{red}{dx} = \int {\color{blue}{\sqrt u} \cdot \color{red}{\frac{du}{3}} = \frac{1}{3}\int {\sqrt u du} = \\ = \frac{1}{3}\int {{u^{\frac{1}{2}}}} } du = \frac{1}{3} \cdot \frac{{{u^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} + C = \frac{1}{3} \cdot \frac{2}{3} \cdot {\color{blue}{u}^{\frac{3}{2}}} + C = \frac{2}{9} \cdot {\color{blue}{(1 + x^3)}^{\frac{3}{2}}} + C $$

Example 5: Solve:

$$ \int {\frac{1}{{{x^2}{{\left( {1 + \frac{1}{x}} \right)}^2}}}} dx $$

Solution:

$$ \text{Let} \ \ \color{blue}{u = 1 + \frac{1}{x}} \ \ \text{then:} \ \color{red}{du} = \left( 1 + \frac{1}{x} \right)' dx = -x^{-2} = \color{red}{- \frac{1}{x^2} dx} \\ \int {\frac{1}{{{x^2}\left( {1 + \frac{1}{x}} \right)}}} dx = - \int {\frac{\color{red}{1}}{{ \color{red}{- {x^2}}{{\left( {\color{blue}{1 + \frac{1}{x}}} \right)}^2}}}} \color{red}{dx} = - \int {\frac{1}{\color{blue}{u^2}}} \color{red}{du} = \\ = - \int {{u^{ - 2}}du = - \frac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}} + C = \color{blue}{u}^{ - 1} + C} = {\left( \color{blue}{1 + \frac{1}{x}} \right)^{ - 1}} + C = \\ = \left( \frac{x + 1}{x} \right)^{-1} + C = \frac{x}{x + 1} + C $$

Exercise 2: Solve using substitution

Level 1

$$ \color{blue}{\int x^2 \cos(x^3) dx =} $$ $ \frac{1}{3} \sin(x^3) + C $
$ 3 \sin(x^3) + C $
$ \frac{1}{3} \cos(x^3) + C $
$ 3 \cos(x^3) + C $

Level 2

$$ \color{blue}{\int {\frac{10x}{5x^2 - 8}} dx =} $$ $ \ln\left| 8 - 5x^2 \right| + C $
$ - \ln\left| 8 - 5x^2 \right| + C $
$ \ln\left| 5x^2 - 8 \right| + C $
$ - \ln\left| 5x^2 - 8 \right| + C $