Integration by Parts

The substitution rule is the chain rule read backward. Integration by parts is the product rule read backward. Both convert a hard integral into an easier one, and between them they handle a large fraction of the integrals that have elementary antiderivatives.

Start from the product rule for differentiable $f$ and $g$:

Integrate both sides. The left side integrates to $f(x)g(x)$, and the right side splits into two integrals:

Solving for the first integral on the right gives the formula.¹

The formula is easier to remember with the substitutions $u=f(x)$ and $v=g(x)$, so that $\mathbf{d}u=f^{\prime }(x)\mathbf{d}x$ and $\mathbf{d}v=g^{\prime }(x)\mathbf{d}x$:

Reading it as a strategy: an integrand is split into a part $u$ that will be differentiated and a part $\mathbf{d}v$ that will be integrated. The exchange is only worthwhile when $\int v\mathbf{d}u$ is simpler than $\int u\mathbf{d}v$. Choosing $u$ and $\mathbf{d}v$ well is the entire skill.

The formula as areas

For a positive increasing function the formula has a direct reading as areas. Let $v=v(u)$ run from $(a,c)$ to $(b,d)$. The corner rectangle of area $bd$ splits along the curve into the region under it and the region to its left, and subtracting the inner rectangle $ac$ leaves the standard statement $\int u\mathbf{d}v+\int v\mathbf{d}u=uv|$.

Choosing $u$ and $dv$

Two requirements constrain the choice: $\mathbf{d}v$ must be something integrable to get $v$, and $u$ should get simpler (or at least no worse) when differentiated. A useful ordering for picking $u$ is the acronym LIATE: whichever factor comes first on this list is the one to differentiate, so it becomes $u$; the rest is $\mathbf{d}v$.

LIATE is a guideline, not a theorem; it fails on constructed examples, but it picks the right split on nearly every integral encountered in practice.

A polynomial against a trigonometric factor

Consider $\int x\sin x\mathbf{d}x$. The integrand is algebraic times trigonometric, so LIATE selects $u=x$ (the A factor) and $\mathbf{d}v=\sin x\mathbf{d}x$. Then

Applying the formula,

The exchange replaced $\int x\sin x\mathbf{d}x$ with the elementary $\int \cos x\mathbf{d}x$. The opposite choice $u=\sin x$, $\mathbf{d}v=x\mathbf{d}x$ produces $\int \frac{1}{2}{x}^2\cos x\mathbf{d}x$, which is worse: differentiating $\sin x$ did not simplify it, and integrating $x$ raised the degree.

A single logarithm

Some integrands have no obvious second factor. For $\int \ln x\mathbf{d}x$, take $\mathbf{d}v=\mathbf{d}x$ and let $u$ be the whole integrand:

Then

Parts works here because $\ln x$ has a simpler derivative than itself. The same choice evaluates ${\tan }^{-1}x$ and ${\sin }^{-1}x$, the other functions whose derivatives are algebraic.

When a power of $x$ multiplies the logarithm, still differentiate the logarithm; one pass turns it algebraic.

Repeated parts

When one application leaves a still-difficult integral, applying parts again may finish the job. For $\int x^2{e}^x\mathbf{d}x$, the factor $x^2$ becomes simpler on differentiation while $e^x$ is unchanged, so $u=x^2$, $\mathbf{d}v=e^x\mathbf{d}x$:

The remaining $\int x{e}^x\mathbf{d}x$ is simpler but not immediate, so apply parts once more with $u=x$, $\mathbf{d}v=e^x\mathbf{d}x$:

Substituting back,

Each pass lowers the polynomial degree by one; a degree-$n$ polynomial against $e^x$ needs $n$ applications.

The tabular method

When $u$ is a polynomial (so its derivatives reach $0$) and $\mathbf{d}v$ integrates easily any number of times, repeated parts can be organized into a table. Differentiate the left column down to zero, integrate the right column at each step, and multiply along the diagonals with alternating signs.

Written as a procedure:

A trigonometric $\mathbf{d}v$ works the same way, since $\cos x$ integrates cyclically without growing.

Cyclic integrals

Sometimes neither factor simplifies, yet repeated parts still succeeds because the original integral reappears. For $\int e^x\sin x\mathbf{d}x$, choose $u=e^x$, $\mathbf{d}v=\sin x\mathbf{d}x$:

The new integral is no simpler, but apply parts again with $u=e^x$, $\mathbf{d}v=\cos x\mathbf{d}x$:

Substituting the second line into the first returns the starting integral on the right:

This is an equation for the unknown integral. Add $\int e^x\sin x\mathbf{d}x$ to both sides and divide by $2$:

Definite integrals by parts

Combining the formula with Part 2 of the Fundamental Theorem gives a version for definite integrals: evaluate the boundary term and integrate the correction term over the same interval.²

For ${\int }_0^1{\tan }^{-1}x\mathbf{d}x$, take $u={\tan }^{-1}x$, $\mathbf{d}v=\mathbf{d}x$, so $\mathbf{d}u=\mathbf{d}x/(1+x^2)$ and $v=x$:

The remaining integral yields to the substitution $t=1+x^2$, $\mathbf{d}t=2x\mathbf{d}x$, giving $\frac{1}{2}\ln 2$. Therefore

Since ${\tan }^{-1}x\ge 0$ on $[0,1]$, the value is the area under the arctangent curve there.

Reduction formulas

Parts can also express an integral with exponent $n$ in terms of the same integral with a smaller exponent, producing a reduction formula that peels the exponent down by recursion. Take $\int {\sin }^{n}x\mathbf{d}x$ with $u={\sin }^{n-1}x$ and $\mathbf{d}v=\sin x\mathbf{d}x$. Then $\mathbf{d}u=(n-1){\sin }^{n-2}x\cos x\mathbf{d}x$ and $v=-\cos x$, so

Replace ${\cos }^{2}x=1-{\sin }^{2}x$ and the integral splits, reproducing $\int {\sin }^{n}x\mathbf{d}x$ on the right. Solving for it gives the reduction formula.³

Applied repeatedly, this drives the exponent down to $\int \sin x\mathbf{d}x$ when $n$ is odd, or $\int \mathbf{d}x$ when $n$ is even.

The same pattern gives reduction formulas for ${\cos }^{n}x$, ${\tan }^{n}x$, ${\sec }^{n}x$, and $(\ln x{)}^n$, all of which appear among the trigonometric integrals.

When parts applies

Integration by parts and substitution are the two general methods. The other techniques are these two specialized to a particular integrand shape.