1 Making a substitution

The technique described here involves making a substitution in order to simplify an integral. We let a new variable, $u$ say, equal a more complicated part of the function we are trying to integrate. The choice of which substitution to make often relies upon experience: don’t worry if at first you cannot see an appropriate substitution. This skill develops with practice. However, it is not simply a matter of changing the variable - care must be taken with the differential form $dx$ as we shall see. The technique is illustrated in the following Example.

Example 19

Find $\int \; <!--nolimits\; -->{\left(3x+5\right)}^{6}dx$ .

Solution

First look at the function we are trying to integrate: ${\left(3x+5\right)}^6$ . It looks quite complicated to integrate. Suppose we introduce a new variable, $u$ , such that $u=3x+5$ . Doing this means that the function we must integrate becomes $u^6$ . Would you not agree that this looks a much simpler function to integrate than ${\left(3x+5\right)}^6$ ? There is a slight complication however. The new function of $u$ must be integrated with respect to $u$ and not with respect to $x$ . This means that we must take care of the term $dx$ correctly.

Long Method $u=3x+5\text{so}\frac{du}{dx}=3,\text{or}\frac{dx}{du}=\frac{1}{3}$

Short Method $u=3x+5\text{so}\frac{du}{dx}=3,\text{so}dx=\frac{1}{3}du$

$\text{Let}I=\int \; <!--nolimits\; -->{\left(3x+5\right)}^{6}dx=\int \; <!--nolimits\; -->u^{6}dx=\int \; <!--nolimits\; -->u^6.\frac{1}{3}.du=\frac{1}{3}\int \; <!--nolimits\; -->u^{6}du=\frac{u^7}{21}+\text{constant}$

To finish off we must rewrite this answer in terms of the original variable $x$ and replace $u$ by $3x+5$ :

$\int \; <!--nolimits\; -->{\left(3x+5\right)}^{6}dx=\frac{{\left(3x+5\right)}^7}{21}+c$

In practice the short method is generally used but mathematicians don’t like to separate the ‘ $dx$ ’ from the ‘ $du$ ’ as in the statement ‘ $dx=\frac{1}{3}du$ ’ as it is meaningless mathematically (but it works!). In the future we will use the short method, with apologies to the mathematicians!

Task!

By making the substitution $u=sinx$ find $\int \; <!--nolimits\; -->cosx{sin}^{2}xdx$

You are given the substitution $u=sinx$ . Find $\frac{du}{dx}$ :

Answer

Now make the substitution, simplify the result, and finally perform the integration:

Answer

Exercise

Use suitable substitutions to find

1. $\int \; <!--nolimits\; -->{\left(4x+1\right)}^{7}dx$
2. $\int \; <!--nolimits\; -->t^{2}sin\left(t^3+1\right)dt$ (Hint: you need to simplify $sin\left(t^3+1\right)$ )

Answer

1. $\frac{{\left(4x+1\right)}^8}{32}+c$
2. $-\frac{cos\left(t^3+1\right)}{3}+c$