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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