Integration by parts and substitution - two methods, step by step

The idea: run the derivative rules in reverse

The product rule and the chain rule give you the derivative of a product and of a composite function. Integration has no such rules: the integral of a product is not the product of the integrals. Instead, two methods take those derivative rules and read them backwards:

• the product rule gives integration by parts,
• the chain rule gives integration by substitution.

So when you meet an integral of a product, ask whether one factor gets simpler when you differentiate it. When you meet a composite function, ask whether the derivative of its inner function (up to a constant) sits next to it.

Integration by substitution: the same area in a new variable

Go through the animation step by step with the buttons or the arrow keys. We find the area under $x\,e^{-x^2}$ from $0$ to $2$.

Notice the limits: $0$ and $2$ have become $0$ and $4$, and the graph of the integrand is different too. The value we were after has not changed.

Integration by parts: the product rule in reverse

The product rule, $(uv)' = u'v + uv'$, gives $uv' = (uv)' - u'v$. Integrate both sides and write $v'\,dx$ as $dv$ and $u'\,dx$ as $du$. That is the formula for integration by parts:

The method pays off when the integral $\int v\,du$ on the right is simpler than the one you started with.

Try it yourself: the two terms of the formula

Pick a function and move the upper limit $b$. Below the graph you see three numbers: the first term of the formula, $[uv]$ between the limits; the second term, $\int v\,du$; and the integral itself. The integral is always the first number minus the second.

What to notice:

• in all three cases, $u$ is the factor that gets simpler when differentiated: the derivative of $x$ is $1$, and the derivative of $\ln x$ is $\frac{1}{x}$,
• $dv$ is the rest of the integrand, and it has to be something we can integrate,
• in $\int \ln x\,dx$ there is no visible product at all. So we write $\ln x$ as $\ln x \cdot 1$ and take $dv = dx$.

How to choose $u$ and $dv$

There is no rule that always works, but there is a simple guideline: choose them so that $\int v\,du$ is no harder than the original integral. In practice:

• take as $u$ the factor that gets simpler when differentiated, such as $x$ or $\ln x$,
• take as $dv$ a factor you can integrate.

For $\int x \sin x\,dx$ the choice $u = x$, $dv = \sin x\,dx$ gives $du = dx$ and $v = -\cos x$, and the new integral is just $\int (-\cos x)\,dx$. The opposite choice, $u = \sin x$, $dv = x\,dx$, gives $v = \frac{x^2}{2}$ and the integral $\int \frac{x^2}{2}\cos x\,dx$, which is harder than the original.

For $\int \ln x\,dx$ take $u = \ln x$ and $dv = dx$. Then $du = \frac{dx}{x}$, $v = x$, and the second integral is $\int 1\,dx = x$. The result is $x \ln x - x + C$.

The rules

Integration by parts. For a definite integral the first term is evaluated between the limits:

Integration by substitution. If $u = g(x)$, then $du = g'(x)\,dx$ and

Why does it work? Let $F$ be an antiderivative of $f$. By the chain rule, $\big(F(g(x))\big)' = f(g(x))\,g'(x)$, so the integral on the left equals $F(g(b)) - F(g(a))$. That is exactly the integral of $f(u)$ from $u = g(a)$ to $u = g(b)$.

There are two ways to handle the limits of a definite integral:

• convert them to $g(a)$ and $g(b)$ and finish in $u$ - as the animation did,
• or find the indefinite integral in $u$ first, go back to $x$, and use the original limits $a$ and $b$.

Do not mix the two: the old limits belong to $x$, and the new ones to $u$.

Both methods reverse rules for taking derivatives, and they help you compute a definite integral when the antiderivative is not obvious.

Example 1: integration by parts - the integral of $x\,e^x$

We want $\int_0^1 x\,e^x\,dx$. The integrand is $x$ times an exponential function, so we try integration by parts.

Example 2: integration by substitution - the integral of $x\,e^{-x^2}$

We want $\int_0^2 x\,e^{-x^2}\,dx$, the area from the animation.

Your turn

Work out $\int_0^{\pi} x \sin x\,dx$. Which factor should be $u$?

Common mistakes

• The old limits after a substitution. If you write $\frac{1}{2}\int_0^2 e^{-u}\,du$, you get $\frac{1 - e^{-2}}{2} \approx 0.432$ instead of $0.491$. The limits $0$ and $2$ belonged to $x$, not to $u$.
• A leftover $x$ in the integral. After a substitution, nothing in $x$ may remain, and that includes the $dx$. If a factor does not cancel, the substitution is probably not the right one. For example, $\int e^{x^2}\,dx$ cannot be done this way, because the factor $x$ is missing.
• A lost constant factor. From $du = 2x\,dx$ you get $x\,dx = \frac{1}{2}\,du$, not $2\,du$.
• A lost minus sign in integration by parts. The formula is $uv - \int v\,du$. It is easy to drop this minus, or the sign inside $v$ itself, as in $v = -\cos x$.
• A bad choice of $u$ and $dv$. If the new integral is harder than the old one, swap the roles. If no split helps, the method may simply not fit this integral.

At school and at university

In England, both methods are part of A-level Mathematics, though not of the AS: the national subject content asks for simple cases of each and describes them as the chain and product rules run in reverse. In the United States, integration by parts belongs to AP Calculus BC, not AB. At university, both methods are a standard part of a first calculus course, and textbooks such as Active Calculus introduce them soon after the fundamental theorem of calculus.

Sometimes one integral needs integration by parts twice, as with $\int x^2 e^x\,dx$. Some functions, such as $e^{x^2}$, have no antiderivative that can be written with elementary functions, so neither method helps there.

Check yourself

Flashcards for review

Flip a card and say whether you knew the answer. The ones you did not know come back at the end of the deck. Why it pays to come back to them in a few days is explained in the guide Spaced repetition.