Today is a major milestone. Until now, we could only integrate functions that looked exactly like standard derivatives we had memorized (like $x^n$, $\sin x$, or $e^x$).
But what if we need to integrate a composite function, like $2x\sqrt{1+x^2}$?
To solve this, we introduce the Substitution Rule (often called u-substitution). This is essentially the Chain Rule running in reverse.
Before we learn the formal rule, let's try to solve a problem using logic and our knowledge of derivatives.
Consider the integral:
$$ \int \cos(3x) \,dx $$
We know that the antiderivative of $\cos(x)$ is $\sin(x)$. So, a reasonable first guess for the answer might be $\sin(3x)$.
Is $\sin(3x)$ the correct antiderivative? Take the derivative of $\sin(3x)$ to check.
Recall the Chain Rule for differentiation:
$$ \frac{d}{dx}[F(g(x))] = F'(g(x)) \cdot g'(x) $$
If we integrate the left side we get:
$$ \int \frac{d}{dx}[F(g(x))] \,dx = F(g(x)) + C $$
Because the integral and derivative are "Inverse" operations, they "Undue" each other.
And the right side becomes:
$$ \int F'(g(x)) g'(x) \,dx $$
Putting it together we have:
$$ F(g(x))+C = \int F'(g(x))g'(x) \,dx $$
If we let $u = g(x)$, then the differential of $u$ is $du = g'(x) \,dx$. This allows us to switch our entire integral from the variable $x$ to the variable $u$.
Throughout this course, we have treated $\frac{du}{dx}$ as a single notation for the derivative. However, the notation was designed by Leibniz to suggest a fraction.
In the method of substitution, we are allowed to treat the differential $du$ and the differential $dx$ as separate algebraic objects related by the derivative:
$$ du = g'(x) \,dx $$
This "splitting" of the derivative notation gives us the mechanism to swap out the $dx$ term in our integrals.
If $u = g(x)$ is a differentiable function whose range is an interval $I$, and $f$ is continuous on $I$, then:
$$ \int f(\underbrace{g(x)}_{u}) \underbrace{g'(x) \,dx}_{du} = \int f(u) \,du $$
Evaluate $\int 2x(x^2 + 1)^4 \,dx$.
This integral looks complicated because of the power of 4. We cannot just distribute the $2x$ inside because of that exponent.
Step 1: Choose $u$.
Look for a "function inside a function." We have $(x^2+1)^4$.
What if we let $u = x^2 + 1$?
Step 2: Find the differential $du$.
First, we take the standard derivative of $u$ with respect to $x$:
$$ \frac{du}{dx} = 2x $$
Now, we convert this to differential form. We treat $du$ and $dx$ as separable quantities and solve for $du$:
$$ du = 2x \,dx $$
Step 3: Substitute into the Integral.
We now look at our original integral and group the terms to see if they match what we found.
$$ \int \underbrace{(x^2+1)^4}_{u^4} \cdot \underbrace{2x \,dx}_{du} $$
The substitution works perfectly!
$$ = \int u^4 \,du $$
Step 4: Integrate with respect to $u$.
Now we have a simple Power Rule integral.
$$ \int u^4 \,du = \frac{1}{5}u^5 + C $$
Step 5: Back-substitute.
The question was asked in terms of $x$, so the answer must be in terms of $x$. Replace $u$ with $x^2+1$.
$$ = \frac{1}{5}(x^2+1)^5 + C $$
Evaluate $\int \sqrt{2x + 1} \,dx$.
Step 1: Choose $u$.
You might be tempted to just choose $u = 2x$. If we did that, we would end up with $\int \sqrt{u+1} \dots$, which we still don't know how to integrate using the Power Rule!
Strategy Tip: When choosing $u$, it is usually best to include the constant terms (like the $+1$) because their derivative is zero. They disappear in the differentiation step but make the integration step much easier.
So, let's choose the entire inside expression: $u = 2x + 1$.
Step 2: Find the differential.
Calculate the derivative $\frac{du}{dx} = 2$.
In differential form, this is $du = 2 \,dx$.
Problem: Look at our integral: $\int \sqrt{2x+1} \,dx$. We have a $dx$, but we need a $2 \,dx$ to substitute for $du$.
We can solve for $dx$ algebraically:
$$ dx = \frac{1}{2} du $$
Step 3: Substitute. Replace $\sqrt{2x+1}$ with $\sqrt{u}$ and replace $dx$ with $\frac{1}{2} du$.
$$ \int \sqrt{u} \cdot \left( \frac{1}{2} \,du \right) = \frac{1}{2} \int u^{1/2} \,du $$
Step 4: Integrate.
$$ \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} u^{3/2} + C $$
Step 5: Back-substitute.
$$ = \frac{1}{3}(2x+1)^{3/2} + C $$
Evaluate $\int 2x e^{x^2} \,dx$.
When dealing with definite integrals (integrals with limits $a$ and $b$), we have two methods.
We will focus on Method 2 because it is usually faster and less prone to notation errors. The rule becomes:
$$ \int_a^b f(g(x))g'(x) \,dx = \int_{g(a)}^{g(b)} f(u) \,du $$
Visualizing the Transformation: The graph on the left shows the area under $f(x)$. The graph on the right shows the area under $f(u)$. Notice that while the shapes (width and height) are different, the total area is exactly the same. Instruction: Move the slider n to see the accumulation of area is the same.
Evaluate $\int_0^1 \sqrt[3]{1 + 7x} \,dx$.
Step 1: Choose $u$. Let $u = 1 + 7x$. Then $du = 7 \,dx$, so $dx = \frac{1}{7}du$.
Step 2: Change Limits.
Step 3: Substitute and Integrate. (Notice we do not go back to $x$!)
$$ \int_1^8 \sqrt[3]{u} \cdot \frac{1}{7} \,du = \frac{1}{7} \int_1^8 u^{1/3} \,du $$
$$ = \frac{1}{7} \left[ \frac{u^{4/3}}{4/3} \right]_1^8 = \frac{1}{7} \cdot \frac{3}{4} [u^{4/3}]_1^8 $$
$$ = \frac{3}{28} (8^{4/3} - 1^{4/3}) $$
Note that $8^{4/3} = (\sqrt[3]{8})^4 = 2^4 = 16$.
$$ = \frac{3}{28} (16 - 1) = \frac{3}{28}(15) = \frac{45}{28} $$
Evaluate $\int_0^{\pi/2} \cos(x) \sin(\sin(x)) \,dx$.
This looks intimidating, but look for the function inside the function. We have $\sin(\sin(x))$.
Let $u = \sin(x)$. Then $du = \cos(x) \,dx$.
The term $\cos(x) \,dx$ exists perfectly in the integrand.
Change Limits:
Evaluate:
$$ \int_0^1 \sin(u) \,du = [-\cos(u)]_0^1 $$
$$ = (-\cos(1)) - (-\cos(0)) $$
$$ = -\cos(1) - (-1) = 1 - \cos(1) $$
Evaluate $\int_1^e \frac{\ln x}{x} \,dx$.
We can use the Substitution Rule to prove convenient properties about integrating symmetric functions over symmetric intervals $[-a, a]$.
Let $f$ be continuous on $[-a, a]$.
This graph illustrates area cancellation for odd functions and the area doubling for even functions.
Evaluate $\int_{-2}^2 (x^6 + 1) \,dx$.
Since $f(x) = x^6 + 1$ is an even function (powers are even), we can split the interval:
$$ \int_{-2}^2 (x^6 + 1) \,dx = 2 \int_0^2 (x^6 + 1) \,dx $$
$$ = 2 \left[ \frac{1}{7}x^7 + x \right]_0^2 $$
$$ = 2 \left[ (\frac{1}{7}(128) + 2) - 0 \right] $$
$$ = 2 \left[ \frac{128}{7} + \frac{14}{7} \right] = 2 \left[ \frac{142}{7} \right] = \frac{284}{7} $$
This is often easier than plugging in $-2$ into the antiderivative.
Evaluate $\int_{-1}^1 \frac{\tan x}{1+x^2+x^4} \,dx$.
This integral looks incredibly difficult to find an antiderivative for. However, let's check for symmetry.
Let $f(x) = \frac{\tan x}{1+x^2+x^4}$.
$$ f(-x) = \frac{\tan(-x)}{1+(-x)^2+(-x)^4} = \frac{-\tan x}{1+x^2+x^4} = -f(x) $$
Since $f$ is an odd function and the interval $[-1, 1]$ is symmetric about the origin, the net area cancels out exactly.
$$ \int_{-1}^1 \frac{\tan x}{1+x^2+x^4} \,dx = 0 $$
Evaluate $\int_{-\pi}^{\pi} \sin(x) \,dx$ using symmetry.
Try these mixed problems to ensure you have mastered the Substitution Rule.
Evaluate $\int x^3 \sqrt{x^2 + 1} \,dx$. (Hint: You might need to be creative with $x^3$).
If $\int_0^4 f(x) \,dx = 10$, find $\int_0^2 f(2x) \,dx$.