What if the derivative is missing a constant?

You can fix it! If your $du$ needs a 5, but you don't have one, you can multiply the inside of the integral by 5 and the outside by $1/5$. Because constants can pass through the integral sign, this balances the equation.

What if I choose the wrong u?

Nothing bad happens! You will just end up with an integral that still has both $x$'s and $u$'s in it, or a $du$ that doesn't match anything. Just cross it out, pick a different part of the function, and try again. Calculus involves a lot of trial and error.

Does u-substitution work for definite integrals?

Yes, but you have to remember to change your limits of integration. If your original limits are $x=a$ and $x=b$, plug those into your $u$ equation to find your new $u$-limits. Then you never have to convert back to $x$ at the end!

How do you choose a u-substitution?

You choose a u-substitution by looking for a specific pattern inside the integral: a function and its derivative. The goal is to pick a part of the expression to be ' $u$ ' so that its derivative, ' $du$ ', exactly matches the remaining parts of the integral (give or take a constant number).

Think of it like finding a matched pair in a game of memory. You are scanning the problem for an 'inner' function (the first card) whose exact rate of change is sitting right next to it (the matching card). Once you spot this pair, you can swap them out for $u$ and $du$ , turning a complicated mess into a simple, basic integral.

What is u-substitution?

In calculus, u-substitution is the reverse of the Chain Rule for derivatives. When you take the derivative of a composite function like $f(g(x))$ , you get $f'(g(x)) \cdot g'(x)$ .

Integration runs that process backwards. You are given an expression that looks like $\int f'(g(x)) g'(x) dx$ , and you need to bundle it back up. By substituting $u = g(x)$ and $du = g'(x)dx$ , the integral simplifies down to $\int f'(u) du$ . This transforms an unrecognizable integral into one of the basic formulas you already know.

The "Inner Function" Strategy

Your first instinct should always be to look for "inner" functions. If part of the expression is trapped inside parentheses, stuck under a square root, or sitting up in the exponent of $e$ , that trapped part is your prime candidate for $u$ .

For example, in the integral $\int (3x+1)^5 dx$ , the expression $3x+1$ is trapped inside the fifth power. Letting $u = 3x+1$ is the best first step. It cleans up the messy inside, leaving you with a simple $u^5$ to integrate.

The "Derivative Match" Strategy

Once you have a suspect for $u$ , do a quick mental check: what is its derivative? The derivative of your chosen $u$ needs to appear elsewhere in the integral, multiplying the rest of the expression.

It is perfectly fine if the derivative is off by a constant multiplier. For instance, if you need $2x dx$ but only have $x dx$ , you can easily multiply and divide by 2 to fix it. However, if your derivative needs an $x^2$ and you only have an $x$ , that u-substitution will not work, and you will need a different strategy.

Where students slip up

The most common mistake is forgetting to substitute the $dx$ . You cannot just change the $x$ terms to $u$ terms and leave the $dx$ hanging around. The differential $dx$ must be fully replaced by $du$ using the relationship you found when taking the derivative.

Another frequent trap is choosing a $u$ that is too large. If you pick the entire denominator or an entire function with its power, the derivative often becomes a massive chain rule mess that isn't anywhere else in the problem. If your $du$ doesn't match the leftovers, back up and try a smaller piece for $u$ .

Worked through

Evaluate the indefinite integral $\int 2x \cos(x^2) dx$ .

First, we look for an inner function. The $x^2$ is trapped inside the cosine function, so it is a great candidate for $u$ .

Let $u = x^2$ .

Next, we take the derivative of $u$ with respect to $x$ to find $du$ : $du = 2x dx$ .

We check our integral to see if we have a match. We have $\cos(x^2)$ and we have $2x dx$ . It matches perfectly! Now we substitute $u$ and $du$ into the integral: $\int \cos(u) du$ .

This is a basic integral. The antiderivative of $\cos(u)$ is $\sin(u) + C$ .

Finally, we replace $u$ with our original expression to get the answer in terms of $x$ : $\sin(x^2) + C$ .

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Where this comes from: OpenStax Calculus Volume 1, Chapter 5: Integration · Stewart Calculus: Early Transcendentals, Chapter 5 · Khan Academy: Integration unit, u-substitution