Integration by parts

Statement

Formal manipulation version for indefinite integration in function notation

Suppose $F$ and $g$ are continuous functions such that $F$ is a differentiable function and we want to integrate $F(x)g(x)$. Suppose $G$ is an antiderivative for $g$. Then, we have:

$\int F(x)g(x)\,dx=F(x)G(x)-\int F'(x)G(x)\,dx$

Verbally, the integral of a product of two functions is the first function times the integral of the second function minus the integral of the derivative of the first function times the second function.

Formal manipulation version in variable transformation notation

Suppose $u,v$ are variables denoting functions of $x$. Then, we have:

$\int u\,dv=uv-\int v\,du$

More explicitly:

$\int u{\frac {dv}{dx}}\,dx=uv-\int v{\frac {du}{dx}}\,dx$

Compared with the notation of the preceding version, $u=F(x)$, $dv/dx=g(x)$, and $v=G(x)$.

Formal manipulation version for definite integration in function notation

Suppose $F$ and $g$ are continuous functions on a closed interval $[a,b]$ such that $F$ is a differentiable function on the open interval $(a,b)$ and we want to integrate $F(x)g(x)$. Suppose $G$ is an antiderivative for $g$. Then, we have:

$\int _{a}^{b}F(x)g(x)\,dx=[F(x)G(x)]_{a}^{b}-\int _{a}^{b}F'(x)G(x)\,dx$

Verbally, the integral of a product of two functions is the first function times the integral of the second function minus the integral of the derivative of the first function times the second function.

Key observations

Equivalence of integration problems

Integration by parts tells us the following:

PUSH DOWN ONE PART, PUSH UP THE OTHER PART OF THE PRODUCT: Integrating a product of two functions is equivalent to integrating a new product where we have differentiated one of the functions and integrating the other.

Repeated use of integration by parts and the circular trap

Integration by parts can be used multiple times, i.e., the new integration that we obtain from an application of integration by parts can again be subjected to integration by parts. However, we need to make sure that we avoid the circular trap:

AVOID THE CIRCULAR TRAP: When using integration by parts a second time, make sure you don't choose as the part to integrate the thing you got by differentiating the part to differentiate from the original product. Otherwise, you get in a circular trap and don't get any new information. The most typical application of integration by parts a second time is if you choose to differentiate again the expression that you already obtained through differentiation the first time

The typical repeated application of integration by parts looks like:

$F(x)g(x)\to F'(x)\int g(x)\,dx\to F''(x)\int \int g(x)\,dx\,dx\to \dots$

Recursive version of integration by parts

For further information, refer: recursive version of integration by parts

Here, we use integration by parts (one or more times) and combine that with some trigonometric identities or algebraic manipulations to see the original integrand re-appear unexpectedly. We then choose an antiderivative $I$ so that the linear equation holds without any additive constants, and solve the linear equation for $I$ to get the antiderivative $I$. For examples, see sec^3#Integration and sin^2#Integration.

Integration by parts is not the general strategy for products

The first thing to check for when trying to integrate a product of functions is integration by u-substitution, particularly if one of the factors in the product looks like a composite of two functions. Integration by parts should be used if integration by u-substitution does not make sense, which usually happens when it is a product of two apparently unrelated functions.

Also, for trigonometric products, check out integration of product of sinusoidal functions.

Choosing the parts to integrate and differentiate

General heuristics

Recall that for integration by parts, we are trying to integrate a product and in order to do that we differentiate one of the factors and integrate the other factor. Given a product, we need to choose which factor is the part to differentiate and which factor is the part to integrate.

Type of function	How good is it as the part to differentiate and why?	How good is it as the part to integrate and why?
logarithmic function, inverse trigonometric function, or the like	excellent to differentiate, because differentiating this kind of function gets us into the domain of algebraic functions (polynomials and rational functions) which are simpler.	terrible as the part to integrate, because there is no general formula for integrating these functions, and the antiderivatives are as complicated, or more.
algebraic functions (specifically polynomials)	polynomials are good to differentiate, because differentiating a polynomial makes it simpler (reduces the degree).	not terrible to integrate, but integrating a polynomial is costly (in terms of degree gain) and should be done only if there is significant reduction in complexity on the other side.
trigonometric and exponential functions	okay to differentiate; there is usually no gain or reduction in complexity	okay to integrate; there is usually no gain or reduction in complexity

The upshot is that, for the choice of the part to differentiate, we have the following hierarchy:

1. Inverse trigonometric functions and logarithmic functions
2. Algebraic functions (specifically polynomials)
3. Trigonometric and exponential functions

Products and typical strategies

Product type	Preliminary thoughts	Strategy for this product type	Examples
inverse trigonometric function or logarithmic function times polynomial function (the polynomial function could just be the function $1$, which is invisible).	The polynomial function can easily be both differentiated and integrated. The inverse trigonometric or logarithmic function can be differentiated, bringing it into the algebraic domain.	Choose the inverse trigonometric function or logarithmic function as the part to differentiate.
polynomial function times sine function or cosine function (or some similar trigonometric function) or exponential function	The polynomial function can be differentiated, and repeated differentiation keeps making it simpler and simpler until it disappears. The sine or cosine function or exponential function, upon integration, does not get more complicated, and so can be integrated repeatedly.	Take the polynomial function as the part to differentiate, and keep using integration by parts repeatedly till the polynomial disappears.
trigonometric function times exponential function OR product of trigonometric functions or power of a trigonometric function that can be treated as a product (and other techniques such as integration by u-substitution don't seem to solve the problem completely)	both functions are easy to differentiate and to integrate, with the complexity remaining the same.	We need to use the recursive version of integration by parts.	sin^2, sec^3

Proof

The proof uses the product rule for differentiation.

Indefinite integration version in terms of the product rule for differentiation

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Definite integration version in terms of the product rule for differentiation

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