# Integration by Substitution U Substitution

In this section we learn about the method of substitution for integration. In particular, we learn U Substitution, which is often the first technique we learn about in this topic.
The method of substitution for integration is one of the two methods we'll learn to integrate a product of two functions, the other method being integration by parts.

We start by learning the method, with a four-step method. We then watch a tutorial and work through some exercises with their answers.

By the end of this section, we'll know how to find integrals looking like: $\int 4x.sin\begin{pmatrix} x^2 - 3 \end{pmatrix}dx$

or
$\int \frac{4x+2}{\sqrt[3]{x^2+x}}dx$
and many more

Let's get started.

## Formula

Given an integral, if it can be written as: $\int \frac{du}{dx} . f \begin{bmatrix} u(x) \end{bmatrix} dx$ Then we can use the method of substitution, in particular U-Substitution to integrate.
We define the new variable $$u$$ such that:

$$u = u(x)$$, meaning we treat the function $$u(x)$$ as a new variable (like the $$x$$ variable).
The integral then becomes simpler to integrate, with respect to the new variable $$u$$, as the differentials $$dx$$ "cancel each other out": $\int \frac{du}{dx}. f\begin{bmatrix} u(x) \end{bmatrix} dx = \int \frac{du}{dx}.f(u)dx$ Since $$\frac{du}{dx} \times dx = du$$, it becomes: $\int \frac{du}{dx} \times f \begin{bmatrix} u \end{bmatrix} dx = \int f(u) du$ We now integrate with respect to the new variable $$u$$ just as we would do if it were $$x$$, which leads to: \begin{aligned} \int \frac{du}{dx} \times f \begin{bmatrix} u \end{bmatrix} dx & = \int f(u) du \\ & = F(u)+c \\ & = F\begin{bmatrix} u(x) \end{bmatrix}+c \end{aligned} This is best explained in words, the tutorial below reviews this entire method in detail, watch it now to make sure this is understood.

In the following tutorial we learn how to use the method of substitution for integration, with several important examples. Watch it now and then attempt to work through the exercises below.

## Exercise 1

1. $$\int (2x+1) cos \begin{pmatrix}x^2 + x - 3 \end{pmatrix}dx$$

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

3. $$\int (x^2-2x)\sqrt{x^3 - 3x^2 + 1}dx$$

4. $$\int cos(x) \begin{pmatrix}sin(x)+1 \end{pmatrix}^2 dx$$

5. $$\int 4.\begin{pmatrix} 2x-5 \end{pmatrix}^7dx$$

6. $$\int 6x.e^{x^2+4}dx$$

7. $$\int (3x^2-2x).sec^2\begin{pmatrix}x^3 - x^2 \end{pmatrix}dx$$

8. $$\int \frac{6x^2}{\sqrt{x^3+2}}dx$$

1. $$\int (2x+1).cos \begin{pmatrix}x^2 + x -3 \end{pmatrix} dx = sin\begin{pmatrix} x^2 + x - 3 \end{pmatrix} + c$$

2. $$\int x. \sqrt[3]{x^2+1} dx = \frac{3}{8} \sqrt[3]{\begin{pmatrix} x^2+1 \end{pmatrix}^4} + c$$

3. $$\int (x^2 - 2x). \sqrt{x^3 - 3x + 1} dx = \frac{2}{9}.\sqrt{\begin{pmatrix} x^3 - 3x^2 + 1 \end{pmatrix}^3} + c$$

4. $$\int cos(x).\begin{pmatrix} sin(x)+1 \end{pmatrix}^2 dx = \frac{1}{3}.\begin{pmatrix} sin(x) + 1 \end{pmatrix}^3 + c$$

5. $$\int 4.\begin{pmatrix} 2x-5 \end{pmatrix}^7dx = \frac{1}{4}.\begin{pmatrix} 2x - 5 \end{pmatrix}^8 + c$$

6. $$\int 6x.e^{x^2+4}dx = 3.e^{x^2+4}+c$$

7. $$\int (3x^2 - 2x).sec^2\begin{pmatrix} x^3 - x^2 \end{pmatrix} dx = tan\begin{pmatrix}x^3 - x^2 \end{pmatrix} + c$$

8. $$\int \frac{6x^2}{\sqrt{x^3+2}}dx = 4\sqrt{x^3+2}+c$$

## Solution With Working

1. We find $$\int (2x+1).cos \begin{pmatrix} x^2 + x -3 \end{pmatrix} dx$$, using the method of substitution as follows:

Let $$u = x^2 + x - 3$$.
Then $$\frac{du}{dx} = 2x+1$$.

The integral can then be written in terms of $$u$$: \begin{aligned} \int (2x+1).cos \begin{pmatrix} x^2 + x -3 \end{pmatrix} dx & = \int \frac{du}{dx}.cos(u)dx \\ & = \int cos(u) du \end{aligned} We now integrate with respect to $$u$$: $\int cos(u)du = sin(u)+c$ Finally, since $$u=x^2+x-3$$: $$sin(u) = sin \begin{pmatrix} x^2+x-3 \end{pmatrix}$$ and we write the answer in terms of $$x$$: $\int (2x+1).cos \begin{pmatrix} x^2 + x -3 \end{pmatrix} dx = sin\begin{pmatrix}x^2 + x - 3 \end{pmatrix} + c$

2. We find $$\int x. \sqrt[3]{x^2+1}dx$$, using the method of substitution as follows:

Let $$u=x^2 + 1$$.
Then $$\frac{du}{dx} = 2x$$ and $$x = \frac{1}{2}.\frac{du}{dx}$$.

The integral can then be written in terms of $$u$$: \begin{aligned} \int x. \sqrt[3]{x^2+1}dx & = \frac{1}{2}.\frac{du}{dx}.\sqrt[3]{u}dx \\ & = \int \frac{1}{2} \sqrt[3]{u} du \\ & = \frac{1}{2} \int \sqrt[3]{u} du \\ & = \frac{1}{2} \int u^{\frac{1}{3}} du \end{aligned} We now integrate with respect to $$u$$: \begin{aligned} \frac{1}{2}.\int u^{\frac{1}{3}} du &= \frac{1}{2}\times \frac{1}{\frac{1}{3}+1}u^{\frac{1}{3}+1} + c \\ &= \frac{1}{2}\times \frac{1}{\frac{4}{3}}.u^{\frac{4}{3}}+c \\ & = \frac{1}{2}\times \frac{3}{4}.u^{\frac{4}{3}}+c \\ & = \frac{3}{8}.u^{\frac{4}{3}}+c \\ & = \frac{3}{8}.\sqrt[3]{u^4}+c \end{aligned} Finally, using the fact that $$u = x^2 + 1$$, we state $$\frac{3}{8}.\sqrt[3]{u^4} = \frac{3}{8}.\sqrt[3]{\begin{pmatrix} x^2 + 1 \end{pmatrix}^4}$$ and we write the final answer in terms of $$x$$: $\int x. \sqrt[3]{x^2+1}dx = \frac{3}{8} \sqrt[3]{\begin{pmatrix} x^2+1 \end{pmatrix}^4}+c$

3. We find $$\int (x^2 - 2x). \sqrt{x^3 - 3x + 1} dx$$, using the method of substitution as follows:

Let $$u = x^3 - 3x^2+1$$.
Then $$\frac{du}{dx} = 3x^2 - 6x$$ and $$x^2 - 2x = \frac{1}{3}.\frac{du}{dx}$$.

The integral can then be written in terms of $$u$$: \begin{aligned} \int (x^2 - 2x)\sqrt{x^3 - 3x^2+1} dx & = \int \frac{1}{3}.\frac{du}{dx}.\sqrt{u}dx\\ & = \int \frac{1}{3}.\sqrt{u} du \\ & = \frac{1}{3} \int \sqrt{u} du \\ & = \frac{1}{3} \int u^{\frac{1}{2}} du \end{aligned} We now integrate with respect to $$u$$: \begin{aligned} \frac{1}{3} \int u^{\frac{1}{2}} du & = \frac{1}{3} \times \frac{1}{1+\frac{1}{2}}.u^{\frac{1}{2}+1} + c \\ & = \frac{1}{3}\times \frac{1}{\frac{3}{2}}.u^{\frac{3}{2}}+c \\ & = \frac{1}{3} \times \frac{2}{3}.u^{\frac{3}{2}}+c \\ & = \frac{2}{9}.u^{\frac{3}{2}}+c \\ & = \frac{2}{9}.\sqrt{u^3}+c \end{aligned} Finally, using the fact that $$u=x^3 - 3x^2 +1$$, we state $$\frac{2}{9}.\sqrt{u^3} = \frac{2}{9}.\sqrt{\begin{pmatrix} x^3 - 3x^2 +1 \end{pmatrix}^3}$$ and we write the answer in terms of $$x$$: $\int (x^2 - 2x). \sqrt{x^3 - 3x + 1} dx = \frac{2}{9}\sqrt{\begin{pmatrix} x^3 - 3x^2 + 1 \end{pmatrix}^3}+c$

4. We find $$\int cos(x) \begin{pmatrix} sin(x) + 1 \end{pmatrix}^2dx$$, using the method of substitution, as follows:

Let $$u=sin(x)+1$$.
Then $$\frac{du}{dx} = cos(x)$$.

We can now write the integral in terms of $$u$$: \begin{aligned} \int cos(x) \begin{pmatrix} sin(x) + 1 \end{pmatrix}^2dx & = \int \frac{du}{dx}.u^2 dx \\ & = \int u^2 du \end{aligned} We now integrate with respect to $$u$$: \begin{aligned} \int u^2 du & = \frac{1}{2+1}u^{2+1}+c \\ & = \frac{1}{3}u^3+c \end{aligned} Finally, using the fact that $$u=sin(x)+1$$, we state $$\frac{1}{3}u^3 = \frac{1}{3}.\begin{pmatrix} sin(x)+1 \end{pmatrix}^3$$ and we write our answer in terms of $$x$$: $\int cos(x) \begin{pmatrix} sin(x) + 1 \end{pmatrix}^2dx = \frac{1}{3}.\begin{pmatrix} sin(x) + 1 \end{pmatrix}^3 + c$

5. We find $$\int 4 \begin{pmatrix} 2x - 5 \end{pmatrix}^7 dx$$, using the method of substitution, as follows:

Let $$u = 2x-5$$.
Then $$\frac{du}{dx} = 2$$ and $$4 = 2. \frac{du}{dx}$$.

We can now write the integral in terms of $$u$$: \begin{aligned} \int 4 \begin{pmatrix} 2x - 5 \end{pmatrix}^7 dx & = \int 2. \frac{du}{dx}.u^7 dx \\ & = \int 2. u^7 du \\ & = 2 \int u^7 du \end{aligned} We now integrate with respect to $$u$$: \begin{aligned} 2\int u^7 dx & = 2\times \frac{1}{7+1} u^{7+1} + c \\ & = \frac{2}{8}u^8+c \\ & = \frac{1}{4}u^8+c \end{aligned} Finally, using the fact that $$u = 2x-5$$, we state $$\frac{1}{4}u^8 = \frac{1}{4}\begin{pmatrix} 2x - 5 \end{pmatrix}^8 and we write the answer in terms of \(x$$: $\int 4 \begin{pmatrix} 2x - 5 \end{pmatrix}^7 dx = \frac{1}{4}.\begin{pmatrix} 2x - 5 \end{pmatrix}^8 + c$

6. We find $$\int 6x.e^{x^2+4}dx$$, using the method of substitution, as follows:

Let $$u = x^2+4$$.
Then $$\frac{du}{dx} = 2x$$ and $$6x = 3.\frac{du}{dx}$$.

We can now write the integral in terms of $$u$$: \begin{aligned} \int 6x.e^{x^2+4}dx & = \int 3.\frac{du}{dx}.e^udx \\ & = \int 3.e^u du \\ & = 3\int e^u du \end{aligned} We now integrate with respect to $$u$$: $3\int e^u du = 3.e^u + c$ Finally, using the fact that $$u=x^2 + 4$$, we state $$e^u = e^{x^2+4}$$ and we write the answer in terms of $$x$$: $\int 6x.e^{x^2+4}dx = 3.e^{x^2+4}+c$

7. We find $$\int (3x^2 - 2x)sec^2\begin{pmatrix}x^3 - x^2 \end{pmatrix}dx$$, using the method of substitution as follows:

Let $$u = x^3 - x^2$$.
Then $$\frac{du}{dx} = 3x^2 - 2x$$.

The integral can now be written in terms of $$u$$: \begin{aligned} \int (3x^2 - 2x)sec^2\begin{pmatrix}x^3 - x^2 \end{pmatrix}dx & = \int \frac{du}{dx}.sec^2(u)dx \\ & = \int sec^2(u)du \end{aligned} We now integrate with respect to $$u$$: $\int sec^2(u)du = tan(u)+c$ Finally, using the fact that $$u=x^3-x^2$$, we state $$tan(u) = tan\begin{pmatrix}x^3-x^2 \end{pmatrix}$$ and write the final answer: $\int (3x^2 - 2x)sec^2\begin{pmatrix}x^3 - x^2 \end{pmatrix}dx = tan\begin{pmatrix}x^3-x^2 \end{pmatrix}+c$

8. We find $$\int \frac{6x^2}{\sqrt{x^3+2}}dx$$, using the method of substitution, as follows:

Let $$u=x^3+2$$.
Then $$\frac{du}{dx} = 3x^2$$ and $$6x^2 = 2.\frac{du}{dx}$$.

We can now write the integral in terms of $$u$$: \begin{aligned} \int \frac{6x^2}{\sqrt{x^3+2}}dx & = \int \frac{1}{\sqrt{x^3+2}}.6x^2dx \\ & = \int \frac{1}{\sqrt{u}}\times 2.\frac{du}{dx}.dx \\ & = \int \frac{2}{\sqrt{u}}du \\ & = \int \frac{2}{u^{\frac{1}{2}}}du \\ & = 2 \int u^{-\frac{1}{2}}du \end{aligned} We now integrate with respect to $$u$$: \begin{aligned} 2 \int u^{-\frac{1}{2}}du & = 2\times \frac{1}{-\frac{1}{2}+1}.u^{-\frac{1}{2}+1}+c \\ & = 2\times \frac{1}{\frac{1}{2}}.u^{\frac{1}{2}}+c \\ & = 2\times 2.u^{\frac{1}{2}}+c \\ & = 4.u^{\frac{1}{2}}+c \\ & = 4\sqrt{u}+c \end{aligned} Finally, using the fact that $$u=x^3+2$$, we state $$4\sqrt{u} = 4 \sqrt{x^3+2}$$ and write the final answer: $\int \frac{6x^2}{\sqrt{x^3+2}}dx = 4 \sqrt{x^3+2} + c$

## Method for Definite Integrals

When using the method of substitution for definite integrals, since we're changing the variable, with respect to which we're integrating, we need to adjust the lower limit and the upper limit of the integral.

For example, to inegrate: $\int_0^2 4x.cos\begin{pmatrix}x^2+1\end{pmatrix}dx$ we would define $$u=x^2+1$$.
Just as we learnt, further-up the page, we'd find the indefinite integral in terms of $$u$$: $\int 4x.cos\begin{pmatrix}x^2+1\end{pmatrix}dx = \int 2.cos(u) du$ But, for definite integrals, we need to adjust the lower and upper limits of the integral.
Indeed:

• when $$x = 0$$, $$u=1$$

• when $$x = 2$$, $$u=5$$
So our definite integral becomes: $\int_0^2 4x.cos\begin{pmatrix}x^2+1\end{pmatrix}dx = \int_1^5 2cos(u)du$ The method is further explained in the tutorial below.

## Exercise 2

Evauate each of the following:

1. $$\int_1^2 \begin{pmatrix}6x^2 - 2\end{pmatrix}\begin{pmatrix} x^3 - x \end{pmatrix}^3dx$$

2. $$\int_{\frac{\pi}{6}}^{\pi}cos(x) \begin{pmatrix} sin(x)+1 \end{pmatrix}^2dx$$

3. $$\int_0^{\sqrt{3}} \frac{6x}{\sqrt{x^2+1}}dx$$

## Special Case 1

As soon as we $\int f \begin{pmatrix}ax+b \end{pmatrix}dx$ the method of substitution leads to the following result: $\int f \begin{pmatrix}ax+b \end{pmatrix}dx = \frac{1}{a}.F \begin{pmatrix} ax + b \end{pmatrix} + c$ Where $$F(x)$$ is the antiderivative (a.k.a the primitive function) of $$f(x)$$.

## Exercise 3

Integrate each of the following:

1. $$\int cos \begin{pmatrix}3x+1 \end{pmatrix} dx$$

2. $$\int 5.e^{2x-3}dx$$

3. $$\int \sqrt[3]{4x-3}dx$$

4. $$\int 10.\begin{pmatrix} 3x - 2 \end{pmatrix}^4 dx$$

## Solution Without Working

1. $$\int cos\begin{pmatrix}3x+1 \end{pmatrix} dx = \frac{1}{3}.sin\begin{pmatrix} 3x+1 \end{pmatrix}+c$$

2. $$\int 5.e^{2x-3}dx = \frac{5}{2}.e^{2x-3}+c$$

3. $$\int \sqrt[3]{4x-3}dx = \frac{3}{16}\sqrt[3]{\begin{pmatrix} 4x - 3 \end{pmatrix}^4}+c$$

4. $$\int 10.\begin{pmatrix} 3x-2\end{pmatrix}^4 dx = \frac{2}{3}.\begin{pmatrix}3x-2 \end{pmatrix}^5 + c$$

## Solution With Working

1. We integrate $$\int cos \begin{pmatrix} 3x + 1 \end{pmatrix} dx$$ as follows:

We see that this integral is of the type $$\int f \begin{pmatrix} ax+b \end{pmatrix}dx$$, where:

• $$ax+b = 3x+1$$

• $$f(x) = cos(x)$$
Using the special case 1 formula we can state: $\int f \begin{pmatrix} ax+b \end{pmatrix}dx = \frac{1}{a}F\begin{pmatrix} ax+b \end{pmatrix}+c$
Where:
\begin{aligned} F(x) & = \int f(x)dx \\ & = \int cos(x)dx\\ F(x) & = sin(x) + c \end{aligned} We can therefore state the answer: $\int cos \begin{pmatrix}3x+1 \end{pmatrix} dx = \frac{1}{3}.sin\begin{pmatrix} 3x + 1 \end{pmatrix} + c$

2. We integrate $$\int 5.e^{\begin{pmatrix} 2x - 3 \end{pmatrix}} dx$$ as follows:

We start by noticing that $$\int 5.e^{2x-3}dx = 5\int e^{2x-3}dx$$.

We see that this integral is of the type $$\int f \begin{pmatrix} ax+b \end{pmatrix}dx$$, where:

• $$ax+b = 2x - 3$$

• $$f(x) = e^x$$
We can therefore use our special case 1 formula: $\int f\begin{pmatrix} ax+b \end{pmatrix}dx = \frac{1}{a}F\begin{pmatrix}ax+b \end{pmatrix}dx$
Where
\begin{aligned} F(x) &= \int f(x) dx\\ & = \int e^x dx \\ F(x) &= e^x +c \end{aligned} We can therefore state: $\int e^{2x - 3} dx = \frac{1}{2}.e^{2x - 3} + c$ And so our final answer: $5\int e^{2x - 3} dx = \frac{5}{2}.e^{2x - 3} + c$

3. We integrate $$\int \sqrt[3]{4x - 3} dx$$ as follows:

We can see that this is of the type $$\int f\begin{pmatrix} ax + b \end{pmatrix}dx$$, where:

• $$ax+b = 4x - 3$$

• $$f(x) = \sqrt[3]{x}$$
We can therefore use the special case 1 formula: $\int f \begin{pmatrix} ax + b \end{pmatrix} dx = \frac{1}{a} F \begin{pmatrix} ax+b \end{pmatrix} + c$
Where:
\begin{aligned} F(x) & = \int f(x) dx \\ & = \int \sqrt[3]{x} dx \\ & = \int x^{\frac{1}{3}}dx \\ & = \frac{1}{1+\frac{1}{3}}.x^{\frac{1}{3}+1}+c \\ & = \frac{1}{\frac{4}{3}}.x^{\frac{4}{3}}+c \\ F(x) & = \frac{3}{4}.x^{\frac{4}{3}}+c \end{aligned} So we can state the answer: \begin{aligned} \int \sqrt[3]{4x - 3} dx & = \frac{1}{4} \times \frac{3}{4}. \begin{pmatrix} 4x - 3 \end{pmatrix}^{\frac{4}{3}}+c \\ & = \frac{1\times 3}{4\times 4}. \begin{pmatrix} 4x - 3 \end{pmatrix}^{\frac{4}{3}}+c \\ \int \sqrt[3]{4x - 3} dx & = \frac{3}{16}. \begin{pmatrix} 4x - 3 \end{pmatrix}^{\frac{4}{3}}+c \end{aligned} We could also write the answer in raidcal form: $\int \sqrt[3]{4x - 3} dx = \frac{3}{16}.\sqrt[3]{\begin{pmatrix} 4x - 3 \end{pmatrix}^4}+c$

4. We find the integral $$\int 10.\begin{pmatrix} 3x - 2 \end{pmatrix}^4dx$$ as follows:

We start by taking the factor of $$10$$ out of the integral: $\int 10.\begin{pmatrix} 3x - 2 \end{pmatrix}^4dx = 10 \int \begin{pmatrix} 3x - 2 \end{pmatrix}^4dx$ We can see that the integral $$\int \begin{pmatrix} 3x - 2 \end{pmatrix}^4dx$$ is of the type $$\int f \begin{pmatrix} ax+b \end{pmatrix}dx$$, where:

• $$ax+b = 3x - 2$$

• $$f(x) = x^4$$

Using the special case 1 formula we can state: $\int f \begin{pmatrix} ax + b \end{pmatrix}dx = \frac{1}{a}F\begin{pmatrix} ax + b \end{pmatrix} + c$ Where: \begin{aligned} F(x) & = \int x^4 dx \\ & = \frac{1}{4+1}x^{4+1} + c \\ F(x) & = \frac{1}{5}x^5 + c \end{aligned} Using this formula we can state: \begin{aligned} \int 10.\begin{pmatrix} 3x - 2 \end{pmatrix}^4dx & = 10 \int \begin{pmatrix} 3x - 2 \end{pmatrix}^4 dx \\ & = 10\times \frac{1}{3} \times \frac{1}{5} \begin{pmatrix} 3x - 2 \end{pmatrix}^5 + c \\ & = \frac{10\times 1 \times 1}{3\times 5} \begin{pmatrix} 3x - 2 \end{pmatrix}^5 + c \\ & = \frac{10}{15} \begin{pmatrix}3x - 2 \end{pmatrix}^5 + c \\ \int 10.\begin{pmatrix} 3x - 2 \end{pmatrix}^4dx & = \frac{2}{3} \begin{pmatrix} 3x - 2 \end{pmatrix}^5+c \end{aligned}

## Special Case 2

As soon as an integral can be written: $\int \frac{u'(x)}{u(x)} dx$ The method of substitution leads to the following result: $\int \frac{u'(x)}{u(x)}dx = ln \begin{vmatrix}u(x) \end{vmatrix} + c$ Note: we can get rid of the absolute value bars around $$u(x)$$ if $$u(x)>0$$ over its domain.

This special case is explained with some examples in the tutorial below.

## Exercise 4

Integrate each of the following:

1. $$\int \frac{3x}{x^2 + 4}dx$$

2. $$\int \frac{x}{x^2 - 1}dx$$

3. $$\int \frac{x^2 - 2x + 1}{x^3 - 3x^2 + 3x -5}dx$$

4. $$\int cot(x)dx$$

## Solution Without Working

1. $$\int \frac{3x}{x^2+4}dx = \frac{3}{2}.ln \begin{pmatrix} x^2 + 4 \end{pmatrix} + c$$

2. $$\int \frac{x}{x^2 - 1}dx = \frac{1}{2}.ln \begin{vmatrix} x^2 - 1 \end{vmatrix} + c$$

3. $$\int \frac{x^2 - 2x + 1}{x^3 - 3x^2 + 3x -5}dx = \frac{1}{3}.ln \begin{vmatrix}x^3 - 3x^2 + 3x - 5 \end{vmatrix} + c$$

4. $$\int tan(x)dx = -ln \begin{vmatrix} cos(x) \end{vmatrix} + c$$