U Substitution¶

On this page, we assume that $f$ is a continuous function and $F$ is one of its antiderivatives. (According to part 1 of the fundamental theorem of calculus, a continuous function has an antiderivative.) We also assume that $g$ is a function whose derivative $g'$ exists and is continuous.

Substitution for antiderivatives¶

With the chain rule, we get $$ \frac{d}{dx} F(g(x)) = F'(g(x))g'(x) = f(g(x))g'(x), $$ so we have $$ \int f(g(x))g'(x)\ dx = F(g(x))+C. $$ This is useful, but let's turn this into something that is easier to use. The right side is what we get if we calculate $\int f(u)\ du$ and then replace $u$ with $g(x)$, so we can write $$ \int f(g(x))g'(x)\ dx = \int f(u)\ du. $$ For the left side, we can also plug in $g(x)=u$ and $g'(x)=\frac{du}{dx}$ (Leibniz's notation), and we get $$ \int f(u)\frac{du}{dx}\ dx = \int f(u)\ du. $$ This looks like canceling away the $dx$, but that's not what's happening; one $dx$ is a part of notation for derivatives, and the other is a part of notation for integrals. We have shown that in this case, you can cancel $dx$ as if it was a number, and it will work.

This technique is known as u substitution, because $u$ is the variable name that is most commonly used for this. Of course, you can use any other variable name you want.

Note that for using this technique, the derivative of the inner function $g(x)$ must be multiplied in the integral. If it isn't, then you need to use some other technique instead.

Example: $\int 2x\cos(x^2)\ dx$¶

Because the derivative of $x^2$ is $2x$ and we have that in the integral, we can use the substitution $u=x^2$. Then we get $\frac{du}{dx}=2x$ and $$ \begin{align} \int 2x\cos(x^2)\ dx &= \int \cos(x^2)2x\ dx \\ &= \int\cos(u)\frac{du}{dx}\ dx \\ &= \int\cos(u)\ du \\ &= \sin(u)+C \\ &= \sin(x^2)+C. \end{align} $$ We can check this by calculating $$ \frac{d}{dx} \left(\sin(x^2)+C\right) = \cos(x^2)2x + 0 = 2x\cos(x^2). $$

Example: $\int x\cos(x^2)\ dx$¶

Because the derivative of $x^2$ is $2x$ and we instead have $x$ in the integral, we can't substitute $u=x^2$. However, we can cleverly write $$ \int x\cos(x^2)\ dx = \int \frac{1}{2} 2x\cos(x^2)\ dx = \int \frac{1}{2}\cos(x^2)2x\ dx, $$ and because we now have $2x$ in the integral, we can substitute $u=x^2$ and $\frac{du}{dx}=2x$ to get $$ \int \frac{1}{2}\cos(u)\ du = \frac{1}{2}\int \cos(u)\ du=\frac{1}{2}\sin(u)+C=\frac{1}{2}\sin(x^2)+C. $$

Example: $\int \cos(x^2)\ dx$¶

Now the derivative of $x^2$ is not in the integral, but just like in the above example, we can try to be clever and we get $$ \int \cos(x^2)\ dx = \int \frac{1}{2x} \cos(x^2) 2x\ dx. $$ For substituting $u=x^2$, we now need to write $\frac{1}{2x}\cos(x^2)$ using only $u$, and assuming that $x$ is positive, that's $\frac{1}{2\sqrt{u}}\cos(u)$. Now our integral becomes $$ \int \frac{1}{2\sqrt{u}}\cos(u)\frac{du}{dx}\ dx = \frac{1}{2}\int\frac{\cos(u)}{\sqrt{u}}\ du. $$ This integral is not easy, because even though we moved the constant $\frac{1}{2}$ to front, we can't move all of $\frac{1}{2\sqrt{u}}$ to front because that isn't a constant. In this case, substituting turned one difficult integral into another difficult integral.

Example: $\int \tan(x)\ dx$¶

First we notice that $$ \tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{1}{\cos(x)}\sin(x). $$ The derivative of $\cos(x)$ is $-\sin(x)$, and this time, we can get it into the integral nicely: $$ \int \frac{\sin(x)}{\cos(x)}\ dx = \int \frac{1}{\cos(x)}\sin(x)\ dx = \int \frac{1}{\cos(x)}(-1)(-\sin(x))\ dx $$ If we now substitute $u=\cos(x)$, we get $\frac{du}{dx}=-\sin(x)$ and $$ \begin{align} \int \frac{\sin(x)}{\cos(x)}\ dx &= \int \frac{1}{u}(-1)\frac{du}{dx}\ dx \\ &= (-1)\int \frac{1}{u}\ du \\ &= -\ln\abs{u}+C \\ &= -\ln\abs{\cos(x)}+C. \end{align} $$

Treating $dx$ as a number¶

It's quite common to "multiply" both sides of $\frac{du}{dx}=g'(x)$ by $dx$ to get $du=g'(x)\ dx$, as if $dx$ was a number and not a part of the derivative notation, and then replace $g'(x)\ dx$ in the integral with $du$. That works too, because it gives $$ \int f(g(x))g'(x)\ dx = \int f(u)\ du, $$ which is same as what we get by canceling.

Some people don't like treating $dx$ as a number, because it's difficult to tell when and why that actually works. However, it's very common in the context of u substitutions.

Substitution for integrals (as in not antiderivatives)¶

From the above calculations, or by using the chain rule directly, we get $$ \int f(g(x))g'(x)\ dx = F(g(x))+C, $$ where $F$ is an antiderivative of $f$. By using the fundamental theorem of calculus twice, we get $$ \int_a^b f(g(x))g'(x)\ dx = F(g(b))-F(g(a)) = \int_{g(a)}^{g(b)} f(u)\ du. $$

For example, by substituting $u=x^2$ and $\frac{du}{dx}=2x$, we get $$ \begin{align} \int_0^{\sqrt{\pi}} 2x\cos(x^2)\ dx &= \int_0^{\sqrt{\pi}} \cos(u)\frac{du}{dx}\ dx \\ &= \int_{0^2}^{(\sqrt{\pi})^2} \cos(u)\ du \\ &= \sin(\pi)-\sin(0) = 0. \end{align} $$