# Integration by substitution

## Summary and examples

In the topic Integration of standard funtions the primitive functions of a restricted number of functions are given.

The primitive functions of most functions cannot easily be calculated.
However, there are also special methods to calculate a primitive function. These methods can be quite complex, but here we focus on a special category, namely substitution.

The functions that we want to discuss now can be written as:

$y=f(ax+b)$ met $a\neq0$

If $F(x)$ is the primitive function of $f(x)$ then:

$\displaystyle\frac{1}{a}F(ax+b)$

is the primitive function of:

$f(ax+b)$

We can verify whether this is correct by differentiating the primitive function, using the chain rule:

$\displaystyle\frac{d}{dx}[\displaystyle\frac{1}{a}F(ax+b)]=\displaystyle\frac{1}{a}\displaystyle\frac{dF(ax+b)}{dx}=\displaystyle\frac{1}{a}f(ax+b)\cdot{a=f(ax+b)}$

And indeed, this result is correct.
The word substitution is used because the result above can also be calculated by the following substitution\:

$u=ax+b$

and then differentiating $F(ax+b)$:

$\displaystyle\frac{dF(ax+b)}{dx}=\displaystyle\frac{dF(u)}{du}\displaystyle\frac{du}{dx}=f(u)\cdot{a}=af(ax+b)$

The following examples show how the method of substitution works.

##### Example 1

Solve:

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

The function:

$f(x)=x^3$

has the primitive function:

$F(x)=\displaystyle\frac{1}{4}x^4$

We have for the integrand of the integral:

$(3x+2)^3=f(3x+2)$

and thus the primitive function of this function is:

$\displaystyle\frac{1}{3}F(3x+2)=\displaystyle\frac{1}{3}\cdot\displaystyle\frac{1}{4}(3x+2)^4=\displaystyle\frac{1}{12}(3x+2)^4$

and thus:

$\displaystyle\int{(3x+2)^3}dx=\displaystyle\frac{1}{12}(3x+2)^4+C$

The second method gives the same result.

Take:

$u=3x+2$

Differentiating this function yields:

$du=3dx$

or:

$dx=\displaystyle\frac{1}{3}du$

If we subtitute these results in the original integral, we get:

$\displaystyle\int{u^3\cdot\displaystyle\frac{1}{3}}du=\displaystyle\frac{1}{3}\cdot\displaystyle\frac{1}{4}u^4+C$

If we 'transform back':

$\displaystyle\frac{1}{12}u^4+C=\displaystyle\frac{1}{12}(3x+2)^4+C$

It is a matter of personal preference which of the methods is chosen. In the following examples we will choose either method.

##### Example 2

Solve:

$\displaystyle\int{2(4x+1)^2}dx$

The primitive of:

$(4x+1)^2$

is equal to:

$\displaystyle\frac{1}{3}(4x+1)^3\cdot\displaystyle\frac{1}{4}=\displaystyle\frac{1}{12}(4x+1)^3$

and thus:

$\displaystyle\int{2(4x+1)^2}dx=\displaystyle\frac{1}{6}(4x+1)^3+C$

##### Example 3

Solve:

$\displaystyle\int{5e^{8x+3}}dx$

We write:

$\displaystyle\int{5e^{8x+3}}dx=5\displaystyle\int{e^{8x+3}}dx$

and the primitive of:

$e^{8x+3}$

is equal to:

$e^{8x+3}\displaystyle\frac{1}{8}$

So we can write:

$\displaystyle\int{5e^{8x+3}}dx=\displaystyle\frac{5}{8}e^{8x+3}+C$

##### Example 4

Solve:

$\displaystyle\int_{1}^{2}{\displaystyle\frac{3}{2x+1}}dx$

We write this integral as:

$\displaystyle\int_{1}^{2}{\displaystyle\frac{3}{2x+1}}dx=3\displaystyle\int_{1}^{2}{\displaystyle\frac{1}{2x+1}}dx$

The primitive of:

$\displaystyle\frac{1}{2x+1}$

is:

$\displaystyle\frac{1}{2}\ln(|2x+1|)$

and thus:

$\displaystyle\int_{1}^{2}{\displaystyle\frac{3}{2x+1}}dx=3[\displaystyle\frac{1}{2}\ln(|2x+1|)+1]_{1}^{2}+C=\displaystyle\frac{3}{2}\ln(\displaystyle\frac{5}{3})+C$

##### Example 5

Solve:

$\displaystyle\int_{0}^{\pi}\sin(2x+\displaystyle\frac{\pi}{2})dx$

The primitive of:

$\sin(2x+\displaystyle\frac{\pi}{2})$

is equal to:

$-\cos(2x+\displaystyle\frac{\pi}{2})\cdot\displaystyle\frac{1}{2}$

and thus:

$\displaystyle\int_{0}^{\pi}\sin(2x+\displaystyle\frac{\pi}{2})dx=[-\cos(2x+\displaystyle\frac{\pi}{2})]_{0}^{\pi}=\cos(\displaystyle{\pi}{2})-\cos(\displaystyle{\pi}{2})=0$

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