U-Substitution in Integrals: A Step-by-Step Guide

Why does $\int 2x\cos(x^2)\, dx$ look impossible, when every piece of it, $2x$ and $\cos$, is something you already know how to integrate?

You know $\int \cos(x)\, dx = \sin(x) + C$. You know $\int 2x\, dx = x^2 + C$. But multiply $2x$ and $\cos(x^2)$ together and none of your rules seem to apply anymore. The $x^2$ trapped inside the cosine breaks everything.

Here's the good news. This integral isn't a new kind of problem. It's a rule you already know, the chain rule, hiding in reverse.

This "impossible" integral is just the chain rule wearing a disguise

Remember the chain rule for derivatives. If $F$ and $g$ are functions, then:

Now let $F(u) = \sin(u)$ and $g(x) = x^2$. Then $F'(u) = \cos(u)$, so:

Look at that closely. It's exactly $2x\cos(x^2)$, our original integrand. The function we couldn't integrate is the derivative of $\sin(x^2)$, we just didn't recognize it yet.

Drag the point and watch the tangent line. At every single value of $x$, the slope of $\sin(x^2)$ matches $2x\cos(x^2)$ exactly. That's the whole secret behind u-substitution: it's a way to spot a disguised derivative and undo it.

So what do we actually substitute?

Once we see the pattern $f(g(x)) \cdot g'(x)$, we give the inner function a new name. We write:

Differentiating both sides with respect to $x$ gives $\dfrac{du}{dx} = g'(x)$, which we usually rearrange into:

For our example, $u = x^2$, so $du = 2x\, dx$. Now look at the original integral again:

Every $x$ is gone. What's left is trivial: $\int \cos(u)\, du = \sin(u) + C$. Swap $u$ back for $x^2$ and we get $\sin(x^2) + C$, exactly the antiderivative we already guessed above.

Why are we even allowed to swap dx for du?

It looks like we just deleted $dx$ and wrote $du$ in its place, which feels a little too easy. But $dx$ and $du$ aren't interchangeable by fiat. They're connected by a stretch factor, and that stretch factor is $g'(x)$.

Picture the $x$-axis as a rubber ruler. The function $g$ grabs that ruler and stretches it, unevenly, into a new ruler marked in units of $u$. A tiny step of width $dx$ on the original ruler doesn't land as a step of the same width on the new one. It stretches or compresses by however fast $g$ is changing at that point, which is exactly $g'(x)$.

Slide the window along the x-ruler and watch the matching window on the u-ruler below it. Near $x = 0.2$, where $g$ is barely changing, the two windows are almost the same size. Near $x = 2$, where $g(x) = x^2$ is climbing fast, the u-window balloons. That difference in size is $g'(x)$, and it's the reason $du = g'(x)\, dx$ isn't a trick. It's just unit conversion between two rulers.

How do you know which piece to call u?

Look for a function trapped inside another function, a composition. Then check whether its derivative, or something close to it, is sitting elsewhere in the integrand, ready to cancel.

A few patterns to watch for:

• Trig function wrapped around a polynomial, like $\cos(x^2)$: let $u$ be the polynomial.
• Something raised to a power, like $(x^3 + 1)^5$: let $u$ be the base.
• An exponent that's itself a function, like $e^{x^3}$: let $u$ be the exponent.

The test is always the same. After you substitute, does every remaining $x$ disappear, replaced entirely by $u$? If yes, you picked well. If stray $x$'s are still hanging around, you picked the wrong piece.

Try all three buttons. Only $u = x^3$ leaves nothing behind. $u = x^2$ leaves a stray $x$ with nowhere to go. And $u = e^{x^3}$ technically "works," but it just renames the entire integrand as $u$ without simplifying anything, so you learn nothing from it.

Definite integrals need new boundaries too

Substitution changes more than the function inside the integral. If you're integrating from $x = a$ to $x = b$, those bounds were measured on the $x$-ruler. Once you switch to the $u$-ruler, you need bounds measured on that ruler instead.

The rule is simple: convert the bounds using the same substitution you already picked.

For our running example, $u = x^2$, so an integral from $x = 0$ to $x = b$ becomes an integral from $u = 0$ to $u = b^2$.

Move the slider and watch both shaded regions. The orange area is computed entirely in $x$, the green area entirely in $u$, with completely different bounds and a completely different curve underneath. They still come out equal every time, because they're measuring the exact same accumulated quantity, just on two different rulers.

Two quick examples to lock it in

Example 1: $\int x\sqrt{x^2+1}\, dx$

Let $u = x^2 + 1$, so $du = 2x\, dx$, which means $x\, dx = \frac{1}{2} du$. The integral becomes $\frac{1}{2}\int \sqrt{u}\, du = \frac{1}{3}u^{3/2} + C = \frac{1}{3}(x^2+1)^{3/2} + C$.

Example 2: $\int \frac{\ln x}{x}\, dx$

Let $u = \ln x$, so $du = \frac{1}{x} dx$. The integral becomes $\int u\, du = \frac{1}{2}u^2 + C = \frac{1}{2}(\ln x)^2 + C$.

Notice the shape both times: something composed inside a function, and its derivative sitting right next to it, just waiting to become $du$.

The mistakes that actually trip people up

Forgetting to convert $dx$ all the way is the biggest one. If a stray $x$ survives after substitution, the integral isn't done, it's stuck, and no amount of algebra rescues it. Go back and pick a different $u$.

For definite integrals, forgetting to swap the bounds is the second biggest. If you substitute the function but leave the original $x$-bounds in place, you'll compute a number that has nothing to do with the actual area. Either convert the bounds to $u$, or substitute back to $x$ before plugging in the original bounds. Just don't mix the two.

And don't forget the $+C$. Once you swap back from $u$ to $x$, that constant of integration comes along for the ride just like always.

U-substitution isn't a separate topic from differentiation. It's the chain rule, read from right to left. Whenever you see a function tangled inside another function, with the inner function's derivative loitering nearby, rename the inner function $u$, rewrite $dx$ in terms of $du$, and let the composition collapse into something you already know how to integrate. Convert the bounds if you're working with a definite integral, and you're done.