Integration by Substitution

Integration by Substitution Video

According to the Fundamental Theorem of Calculus, integration and differentiation can be thought of as inverse operations. Differentiating an integral and integrating a derivative yield almost identical results, for example:

\(\frac{d}{dx}\left [ \int 3x^{2}dx \right ]=3x^{2}\)

\(\int (\frac{d}{dx}\left [ 2x^{4}+7 \right ])dx=2x^{4}+c\)


Function types such as polynomial functions, exponential functions, and trigonometric functions have their own differentiation and integration formulas. When it comes to techniques, however, integration and differentiation can be quite different. In general, when taking a derivative, if the function is a polynomial, use the Power Rule, if it is a product, use the Product Rule, if it is a quotient, use the Quotient Rule, and if it is a composite, use the Chain Rule.

It doesn’t work that way with integration. There’s no rule specifically for integrating a product or a quotient. Part of the reason is that the Product, Quotient, and Chain rules all incorporate parts of the function being differentiated as well as the derivatives of those parts. So, when looking at an integral, it’s difficult to tell which rule needs to be undone. Fortunately, there is a useful technique that involves substitution. Let’s begin with an example:

\(\int (8x-13)(4x^{2}-13x)^{5}dx\)


At first glance, this seems like a quick polynomial integration, right? Look closely. In order to integrate this as a polynomial, we would need to expand \(4x^{2}-13x\) to the 5th power, then distribute the \(8x-13\) to that gigantic result – a huge amount of work!

Notice that \(8x-13\) is the derivative of \(4x^{2}-13x\). This tells us we can use substitution. Traditionally, the variable \(u\) is used in this technique, so it is often called \(u\)-substitution. So here’s the big idea:

1. First, we’re gonna assign a part of the function to be \(u\).
2. Then we’ll rewrite the integral in terms of \(u\), not \(x\), creating an easily-integrable form.
3. Then we’ll integrate.
4. And finally, we’ll convert the result back to being in terms of \(x\).

First, identify the part of the function that is the derivative (or multiple of the derivative) of the other. Once this is identified, the other piece will be \(u\). Typically, this involves thinking a little bit ahead and making a choice. In this case, since \(8x-13\) is the derivative of \(4x^{2}-13x\), we’ll call \(4x^{2}-13x\) our \(u\).

So \(u=4x^{2}-13x\).

Then we’ll take the derivative of this, and get that \(du=(8x-13)dx\)

This gives us everything we need to rewrite the integral in terms of \(u\). So we’re gonna have our \(u\) is right here, this part \((4x^{2}-13x)\), not including to the 5th power. And then our \(du\) is this part \((8x-13)\) and this part \((dx)\).

So if we rewrite this integral, we get:

\(\int u^{5}du\)


Now this becomes something that can be easily integrated. So this is equal to:

\(\int u^{5}du=\frac{1}{6}u^{6}+c\)


All that’s left is to rewrite the result in terms of \(x\), which involves substituting \(4x^{2}-13x\) for \(u\). So we have, this is equal to:



Remember, you can always take the derivative to check your result.

Let’s try another!

\(\int 6(x-4)\sqrt[3]{x^{2}-8x}dx\)


At first glance, there are many possibilities for \(u\): \(x-4\), \((x-4)^{3}\), \(x^{2}-8x\), \((x^{2}-8x)^{\frac{1}{3}}\), but there is only one that has the necessary relationship. Let’s take a look at:

\(\frac{d}{dx}\left [ x^{2}-8x \right ]\)


This gives us:

\(\frac{d}{dx}\left [ x^{2}-8x \right ]=2x-8\)


Which then can factor to be:

\(\frac{d}{dx}\left [ x^{2}-8x \right ]=2x-8=2(x-4)\)


Since \(x-4\) is a multiple of the derivative of \(x^{2}-8x\), \(x^{2}-8x\) must be our \(u\).

So \(u=x^{2}-8x\).

And \(du=(2x-8)dx\), which is \(2(x-4)dx\).

Notice that there’s no 2 in the original integral. No problem, just divide by 2. So:



Make the substitutions and simplify a little bit.

So notice, up here we have our \(u\) is, \(x^{2}-8x\), so that’s this portion. And then our \(du\), or \(\frac{1}{2}du\), is \((x-4)dx\). Now let’s substitute in our \(u\)’s.

The integral of 6, \(u=x^{2}-8x\), which is this part, cube rooted, or we can say to the \(\frac{1}{3}\) power (it means the same thing, times our blue part right here, which we said is \(\frac{1}{2}du\)).

\(\int 6\cdot u^{\frac{1}{3}}\cdot \frac{1}{2}du\)


When we simplify this, we get:

\(\int 6\cdot u^{\frac{1}{3}}\cdot \frac{1}{2}du=\int 3u^{\frac{1}{3}}du\)


Which is equal to:

\(\int 6\cdot u^{\frac{1}{3}}\cdot \frac{1}{2}du=\int 3u^{\frac{1}{3}}du=\frac{3}{4}\cdot 3u^{\frac{4}{3}}+c=\frac{9}{4}u^{\frac{4}{3}}+c\)


Now we’re gonna rewrite this in terms of \(x\).

\(\int 6(x-4)\sqrt[3]{x^{2}-8x}dx=\frac{9}{4}(x^{2}-8x)^{\frac{4}{3}}+C\)


Let’s work one final problem together.

\(\int cos(9x)dx\)


The temptation here is to say that the answer is \(sin(9x)+c\) since we know the derivative of \(sin(x)\) is \(cos(x)\). But we need to account for the 9x in the parentheses since it is a function.

Clearly, we can’t make \(u\) equal to \(cos(9x)\) because there is no negative sine function in the integral. So if our \(u=9x\), then \(du\) must be equal to \(9dx\). But there’s not a 9 in our function, so we’re gonna take \(\frac{1}{9}du=dx\).

Now let’s work on rewriting our integral. So our \(u=9x\), which is this part right here. And our \(du\), or our \(\frac{1}{9}du\) is this part right here.

So plugging that into our integral gives us:

\(\int \frac{1}{9}cos(u)du\)


Now we’re gonna integrate. So this is equal to:



And finally, we’re gonna substitute in our \(u\)-value again. So this gives us:



Here’s one for you to try! Using substitution, find a formula for the antiderivative of the tangent function.

\(\int tan(x)dx=\int \frac{sin(x)}{cos(x)}dx\)


Pause the video and try this one for yourself! Okay, so the first thing we need to do is choose our \(u\) carefully, because \(du\) cannot be in a denominator.

So our \(u\) is gonna be \(cos(x)\). Which then means that our \(du=-sin(x)dx\). Now, there’s not a \(-sin(x)\) in our function, so we’re gonna make this \(-du\), by dividing both sides by negative 1, \(-du=sin(x)dx\). Now let’s rewrite our function.

So our \(u\)-part, \(cos(x)\), is right here in our denominator. And our \(-du\) is this part right here in our numerator and our \(dx\) right here. So when we rewrite this, we’ll have:

\(\int -\frac{1}{u}du\)


When we integrate this, we get:

\(-ln\left | u \right |+c\)


Finally, we’re gonna rewrite this in terms of \(x\). So this is equal to:

\(-ln\left | cos(x) \right |+c\)


Be sure to integrate the method of substitution into your calculus repertoire! It is one of the key ways of finding antiderivatives.

I hope you enjoyed this video! Thanks for watching and happy studying!


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by Mometrix Test Preparation | This Page Last Updated: July 24, 2023