Integrals with changes of variable

We are going to learn to do integrals by doing changes of variable. The change of variable is a very useful method for solving integrals, but it might be difficult since it needs some "imagination". In fact we have to invent the change of variable.

We will start by indefinite integrals with change of variable and then move to defined integrals.

The change of variable for doing an integral consists of equaling a part of the integrand to a new variable, (we can call it $t$ , $u$ , or as we want), called auxiliary variable.

Then, it is necessary to calculate the derivative of the auxiliary variable and to do the necessary operations, so that the original variable does not appear in the integrand or the differential part of the integration. This is named a change of variable. Namely $\int_{a}^{b} f (x) d x = \int_{φ (a)}^{φ (b)} f (φ (t)) \cdot φ^{'} (t) d t$ where the change of variable has been done $φ (t) = x$ .

After doing the change of variable, in general, we obtain simpler integrals.

Formulary

Typical changes of variable:

$\int F (a x + b) d x = \frac{1}{a} \int F (u) d u$ , where $u = a x + b$

$\int F (\sqrt{a x + b} d x = \frac{2}{a} \int u \cdot F (u) d u$ , where $u = \sqrt{a x + b}$

$\int F (\sqrt[n]{a x + b}) d x = \frac{n}{a} \int u^{n - 1} F (u) d u$ , where $u = \sqrt[n]{a x + b}$

$\int F (\sqrt{a^{2} + b^{2}}) d x = a \int F (a \cdot \cos u) d u$ , where $u = a \cdot \sin u$

$\int F (e^{a x}) d x = \frac{1}{a} \int \frac{F (u)}{u} d u$ , where $u = e^{a x}$

$\int F (\ln x) d x = \int F (u) e^{u} d u$ , where $u = \ln x$

How to proceed

To decide the change of variable to be used ( $t$ as a function of $x$ ).
To calculate $d t$ according to $x$ and $d x$ .
To substitute $t$ and $d t$ in the integral, so that we eliminate $x$ .
To calculate the indefinite integral according to $t$ . If we do not know how to calculate it, try another change of variable or another method of integration.
To replace $t$ by $x$ so that the result is according to $x$ .

Example

To calculate the following integral by the method of the change of variable $\int \frac{x}{\sqrt[5]{x^{2} + 2}} d x$

We will change the variable $t = x^{2} + 2$ .
$d t$ is calculated by deriving the expression of $t$ according to $x$ . But bearing in mind that, once having derived $x$ , we still have to find $d x$ (of the differential part): $d t = 2 x \cdot d x + 0 = 2 x \cdot d x$ and therefore $x \cdot d x = \frac{d t}{2}$
$\int \frac{x}{\sqrt[5]{x^{2} + 2}} d x = \frac{1}{2} \int \frac{1}{\sqrt[5]{t}} d t$
$\frac{1}{2} \int \frac{1}{\sqrt[5]{t}} d t = \frac{1}{2} \frac{t^{\frac{4}{5}}}{\frac{4}{5}} + K = \frac{5}{8} \sqrt[5]{t^{4}}$
$\frac{5}{8} \sqrt[5]{t^{4}} + K = \frac{5}{8} \sqrt[5]{(x^{2} + 2)^{4}} + K$

So we obtain: $\int \frac{x}{\sqrt[5]{x^{2} + 2}} d x = \frac{5}{8} \sqrt[5]{(x^{2} + 2)^{4}} + K$

It is necessary to bear in mind that this integral could be done almost as an immediate integral. Many integrals can be solved by many different methods.

Example

To calculate the following integral by the method of the change of variable: $\int \frac{1}{x^{2} \cdot \sqrt{4 - x^{2}}} d x$

We see that the root is somehow similar to the integral of the arc cosine, so we will use this. We will first do the change of variable $t = \frac{x}{2}$ to eliminate the $4$ from the integral.
$d t = \frac{d x}{2}$
$\int \frac{1}{x^{2} \sqrt{4 - x^{2}}} d x = \int \frac{2}{4 t^{2} \sqrt{4 - 4 t^{2}}} d t = \frac{1}{4} \int \frac{1}{t^{2} \sqrt{1 - t^{2}}} d t$
To calculate the resultant integral, we will do another change of variable. We will take now $z = \arccos t$ , so that $t = \cos z$ and $d t = - \sin z \cdot d z$ , and we obtain: $\int \frac{1}{t^{2} \sqrt{1 - t^{2}}} d t = \frac{1}{4} \int \frac{- \sin z}{\cos^{2} z \cdot \sqrt{1 - \cos^{2} z}} d z = \frac{1}{4} \int \frac{- \sin z}{\cos^{2} z \cdot \sin z} d z =$ $= \frac{- 1}{4} \int \frac{1}{\cos^{2} z} d z = \frac{- 1}{4} \cdot \tan z + K$
We first undo the change $z = \arccos t$ and then $2 x = t$ : $\frac{- 1}{4} \tan z + K = - \frac{1}{4} \tan (\arccos t) + K = - \frac{1}{4} \tan (\arccos \frac{x}{2}) + K$

So we obtain: $\int \frac{d x}{x^{2} \sqrt{4 - x^{2}}} = - \frac{1}{4} \tan (\arccos \frac{x}{2}) + K$

Example

To calculate the following integral by the method of the change of variable: $\int x \cdot e^{x^{2}} d x$

We will do the change of variable $t = x^{2}$ .
We have then $d t = 2 x \cdot d x$ , and therefore, $x \cdot d x = \frac{d t}{2}$
$\int x e^{x^{2}} d x = \int e^{t} x d x = \int e^{t} d t$
$\int e^{t} d t = e^{t} + C$
$\int e^{t} d t = e^{t} + C$

Then:

$\int x e^{x^{2}} d x = e^{x^{2}} + C$

Defined integrals by change of variable

How to proceed:

To decide the change of variable to be used ( $t$ as a function of $x$ ).
To calculate $d t$ according to $x$ and $d x$ . Calculate also the new limits of the interval of integration in the new variable.
To substitute $t$ and $d t$ in the integral, so that they eliminate $x$ . And change the integration limits.
To calculate the integral with the new variable, without need to undo the change of variables if the integration interval has been done correctly.