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6.9 Integrating Using Substitution

1 min readjune 18, 2024

6.9 Integrating Using Substitution

Welcome back to AP Calculus with Fiveable! In this key topic, we're delving into the technique of integration using substitution. We’ve worked through definite integrals and finding antiderivatives, so let’s keep building on those skills. 🧱

🔋 The Power of Substitution

Integration using substitution, or u-substitution, is a powerful technique that helps simplify complex integrals by introducing a new variable. This variable is chosen strategically to make the integration process more straightforward and recognizable. The goal is to transform the integral into a simpler form that we know how to work out.

We can use substitution for both definite and indefinite integrals, and it is an invaluable tool in your calculus toolbox. 🧰


🔁 How U-Substitution Works

The substitution method involves replacing a part of the integrand with a new variable to simplify the expression. This new variable is often chosen based on its derivative being present in the integral. The substitution allows us to rewrite the integral in terms of the new variable, making it easier to evaluate.

Remember the Chain Rule? We typically use it for a composite function, which is recognizable in this form: ddx(f(g(x)))=f(g(x))g(x)\frac{d}{dx}(f(g(x)))=f'(g(x))⋅g'(x). Since taking the antiderivative is the “reverse” of a derivative, u-substitution can be thought of as the “reverse” of the chain rule. This is because you need to identify an expression whose derivative is in the integral, which is directly a result of the chain rule. You’ll see this in the examples! 🧐

🪜 U-Substitution Steps

Let's go through the step-by-step process of integration using substitution:

  1. 👀 Identify the Inner Function: Look for a part of the integrand that can be simplified with a new variable.
    1. This usually involves expressions inside square roots, trigonometric functions, or complex algebraic terms that we want to temporarily get rid of.
  2. 🧠 Choose the New Variable: Let the new variable be a function of the existing one. We usually use uu to represent the function.
  3. ✏️ Differentiate the New Variable: Find the derivative of the chosen variable. This will help express dx in terms of the new variable. Don’t forget the chain rule!
  4. 📝 Rewrite the Integral: Substitute the chosen variable and its derivative into the original integral.
    1. Express dx from the integral in terms of the new variable.
    2. If you have a definite integral, you can change the limits of integration to match the new variable. You need to relate the bounds in the same way you relate the original variable. Check this out in the examples later in this guide.
  5. 🤔 Simplify the Integral: Simplify the integrand in terms of the new variable. After this step, you should have an integral that is easier to evaluate than before.
  6. 🖊️ Evaluate the Integral: Integrate the simplified expression with respect to the new variable.
  7. 🔙 Back-Substitute: Replace uu with the original variable in the result obtained from integration. Or, if you changed the limits of integration on your definite integral, you can simply evaluate.

Integration by substitution can be a bit challenging, but with practice, it’s super valuable!


🧮 U-Substitution Practice Problems

Let’s work on a few questions and make sure we have the concept down!

1) Basic Substitution with Indefinite Integrals

Evaluate the following integral using substitution:

2xcos(x2)dx∫2xcos⁡(x^2) dx

👀 Step 1: Identify the Inner Function

First, we try to identify functions we know how to integrate: 2x2x and cos(x)cos(x). But cos(x2)cos(x^2) is a bit more difficult because of the inner function. We have found the item we want to temporarily get rid of: x2x^2!

🧠 Step 2: Choose the New Variable

Let’s let…

u=x2\textcolor{green}{u = x^2}

✏️ Step 3: Differentiate the New Variable

ddxu=ddxx2\frac{d}{dx} u = \frac{d}{dx} x^2
dudx=2x\frac{du}{dx} = 2x

Therefore, expressing dxdx in terms of the new variable, we get du=2xdx\textcolor{blue}{du = 2xdx}! We can recognize that this exists in the original integral.

📝 Step 4: Rewrite the Integral

With some rearrangement of the original integral, we can separate the two parts.

cos(x2)2xdx=cos(u)du∫cos⁡(\textcolor{green}{x^2})\textcolor{blue}{2xdx} = ∫cos⁡(\textcolor{green}{u})\textcolor{blue}{du}

🖊️ Step 5: Evaluate the Integral

cos(u)du=sin(u)+C∫cos⁡(\textcolor{green}{u})\textcolor{blue}{du} = sin(u) + C

🔙 Step 6: Back-Substitute

Our last step is to replace the variable uu with the original function. Therefore, it simplifies to sin(x2)+Csin(x^2) + C.

2xcos(x2)dx=sin(x2)+C∫2xcos⁡(x^2) dx = sin(x^2) + C

Nice work! 🙌 Let’s take it up a level by introducing limits of integration into a problem.

2) Substitution with Definite Integrals

Evaluate the following integral using u-substitution:

122x(x2+1)2dx∫_{1}^{2}\frac{2x}{(x^2+1)^2} dx

This integral may look complex at first, but that is why we’re using u-substitution! Let’s get working.

There are two methods we can use with definite integrals. You can either…

  1. Change the limits of integration as you work through the problem, or
  2. Substitute back into the expression before evaluating.

We can work through both and then you can decide on your own which is more comfortable. ⭐

👀 Step 1: Identify the Inner Function

We start in the same way: identifying an expression whose derivative is present. We can recognize that x2+1x^2 + 1 can be substituted since its derivative, 2x2x, is present in the original integral.

🧠 Step 2: Choose the New Variable

Let’s set uu and take the derivative. Make sure to keep all this information organized!

u=x2+1u = x^2 +1
du=2xdxdu = 2xdx

Now, we can work on the two methods separately.

Method 1: Changing the Limits of Integration

If you choose this method, we have to rewrite the bounds from the original integral. We can use the equation for uu to change the bounds, so that the new integral will be fully in terms of uu.

For the lower bound x=1x = 1, that becomes u=(1)2+1=2u = (1)^2 + 1 = 2.

For the upper bound x=2x = 2, that becomes u=(2)2+1=5u = (2)^2 + 1 = 5

Now, substitute the new bounds and the expression in terms of uu:

251u2du=25u2du \int_{2}^{5} \frac{1}{u^2}du = \int_{2}^{5} u^{-2} \,du
=1u25= \left. \frac{-1}{u} \right|_{2}^{5}

Now, we can evaluate without substituting back into terms of the original variable, since we already changed everything!

1u25=(15)(12)\left. \frac{-1}{u} \right|_{2}^{5} = \left( \frac{-1}{5} \right) - \left( \frac{-1}{2} \right)

Therefore,

122x(x2+1)2dx=310∫_{1}^{2}\frac{2x}{(x^2+1)^2} dx = \frac{3}{10}

Amazing work! Let’s check if the second method comes to the same conclusion.

Method 2: Substituting Back

This method requires us to temporarily forget the bounds. Just don’t forget to include them when evaluating!

Since we already set up uu and dudu, we can jump right into rewriting the integral. For the next couple steps, we can leave off the bounds since they are in terms of xx and the rewritten integral will be in terms of uu. We get the following:

1u2du=u2du=1u \int \frac{1}{u^2}\,du = \int u^{-2} \,du = \frac{-1}{u}

Now, we need to replace uu with its original value.

1u=1x2+1\frac{-1}{u} = \frac{-1}{x^2+1}

Next, we can replace the limits of integration since the expression will all be in terms of xx, and then finally evaluate.

1x2+112=15(12)\left.\frac{-1}{x^2+1}\right|_1^{2} = \frac{-1}{5} - (\frac{-1}{2})

Therefore,

122x(x2+1)2dx=310∫_{1}^{2}\frac{2x}{(x^2+1)^2} dx = \frac{3}{10}

You’re on fire! 🔥 Both methods allowed us to make the same conclusion about the value of the integral while using u-substitution.


🌟 Closing

Awesome work! 🙌 Integration using substitution allows you to navigate through complex integrals and make them much more manageable. Keep practicing, and you'll become better at identifying when and how to apply substitution in different integration scenarios.

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