# Substitution method

Sometimes, it’s just not possible to evaluate an integral by using only the integration properties and rules. Sad, but true.

However, there’s still hope! Some expressions can be substituted with other expressions that are easier to solve. For example, we need to evaluate the integral $$\int \frac{x}{x^2+1}dx$$. To solve that integral, we need to make the substitution $$u=x^2+1$$.

Let’s see how it works!

## What is the substitution method?

The substitution method is used when a complicated integrand (that is, the function that is being integrated) can be simplified by changing the variable.

If $$\varphi(t)$$ is substituted for $$x$$, then the differential $$dx$$ has to be changed into $$\varphi^{\prime}(t)dt$$ and the following formula will hold:

However, when the substitution is done, we still need to use the integral rules to solve it. Remember these?

Constant multiple property of integrals | $$\int{(c\times f(x))}dx=c\times \int{f(x)}dx$$ |

Sum rule for integrals | $$\int{(f(x) + g(x))}dx=\int{f(x)}dx + \int{g(x)}dx$$ |

Difference rule for integrals | $$\int{(f(x) - g(x))}dx=\int{f(x)}dx - \int{g(x)}dx$$ |

Substitution rule | $$\int{f(\varphi(t))}\varphi^{\prime}(t)dt=\int{f(x)}dx$$ |

Integration by parts | $$\int{u}dv=uv-\int{v}du$$ |

### Why is the substitution method so useful?

There’s a limited amount of integrals in the table of rules and properties, so everything else has to be rewritten as integrals that we know how to solve. This can be done through several different methods, one of which is the substitution method!

Some parts of the integrand expression can be difficult to solve, so by substituting them with known (easier) ones, the integral can be rewritten as one of the table integrals. It’s like a fun workaround for when things get too complicated!

## How to use the substitution method

Okay, let’s dive into a few examples of using the substitution method:

### Example 1

**Find the integral:**

Notice that the expression under the integral is not one of the integrals from our table. That means we need to use the substitution method! By substituting $$t=x^2+1$$ and differentiating it, we get $$dt=2x dx$$, which gives us $$dx=\frac{dt}{2x}$$. BTW: Since we changed the variable, we also need to change $$dx$$ with $$dt$$. Substitute $$t$$ and $$dx$$ to see if we get a simpler expression to integrate:

$$\int \frac{x}{t}\times\frac{dt}{2x}$$

Look! That $$x$$ can be canceled out since it’s in the numerator of the first fraction and the denominator of the second fraction. So, let’s cancel out the $$x$$:

$$\int \frac{1}{t}\times\frac{dt}{2}$$

Multiply the fractions:

$$\int \frac{1}{2t}dt$$

Now we have a much simpler integral to evaluate, and we can use the properties of the integral! Use the property $$\int a\times f(x)dx=a\times\int f(x)dx$$:

$$\frac12\int \frac{1}{t}dt$$

Now, use $$\int \frac1x dx=\ln{(|x|)}$$ to evaluate the integral:

$$\frac12\ln{(|t|)}$$

Substitute $$t=x^2+1$$ back in, which we can do because the substituted values don’t represent the actual solution (we just use them to simplify part of the problem):

$$\frac12\ln{(|x^2+1|)}$$

Notice that the expression $$x^2+1$$ will always be positive, so the absolute value of the expression is positive and the bars can be removed:

$$\frac12\ln{(x^2+1)}$$

Since the derivative of a constant is zero, we need to add the constant of integration $$C$$ to include all possible anti-derivative functions in the result:

$$\frac12\ln{(x^2+1)}+C, C\in \mathbb{R}$$

Woohoo! The indefinite integral $$\int \frac{x}{x^2+1}dx$$ is equal to:

### Example 2

**Find the integral:**

Well, the expression under the integral is not one of our nice table integrals. Time to use the substitution method! The trickiest piece is the $$x^2$$ inside the sine, so let’s try substituting $$t=x^2$$. Because we’re changing the variable, we also need to change $$dx$$ with $$dt$$. So, by differentiating $$t=x^2$$, we get $$dt=2x dx$$; then, by rewriting, $$dx$$ is obtained $$dx=\frac{dt}{2x}$$. Use substitution:

$$\int 2x\sin(t) \frac{dt}{2x}$$

Oh, hey: $$2x$$ can be canceled out! Let’s do that:

$$\int \sin(t) dt$$

Now we have a much simpler integral to evaluate (yay!) — and now we can also use the properties of the integral (double yay!). Let’s use $$\int \sin{x}dx=-\cos{x}$$:

$$-\cos{t}$$

Substitute back $$t=x^2$$. The substituted values don’t represent the actual solution (just a way of simplifying our work), so we can do that:

$$-\cos{x^2}$$

Since the derivative of a constant is zero, we need to add the constant of integration $$C$$ to include all possible anti-derivative functions in the result:

$$-\cos{x^2}+C, C\in \mathbb{R}$$

The indefinite integral $$\int 2x\sin(x^2) dx$$ is equal to:

That wasn’t so bad, right? Let’s review the process so you can try it on any problem:

## Study summary

- Use the appropriate substitution.
- Use the properties of the integral to evaluate the integral.
- Substitute back.
- Add the constant of integration.

## Do it yourself!

Want some more practice? You’re in luck! Try these problems and see if they make sense.

**Find the integral:**

- $$\int\sqrt{2x+3}dx$$
- $$\int\frac{6x-1}{3x^2-x+5}dx$$
- $$\int(5x+3)^9dx$$
- $$\int x\cos(x^2+1)dx$$

*Solutions:*

- $$\frac{(2x+3)\sqrt{2x+3}}{3}+C, C \in \mathbb{R}$$
- $$\ln{(|3x^2-x+5|)}+C, C\in \mathbb{R}$$
- $$\frac{(5x+3)^{10}}{50}+C, C\in \mathbb{R}$$
- $$\frac{\sin{(x^2+1)}}{2}+C, C \in \mathbb{R}$$

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