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# How to Integrate by Parts for Multivariable Calculus

## An Introduction to Integration by Parts

If you need some help with integration by parts, then you are in the right place. This review article will give you a simple guide, and link integration by parts to multivariable calculus.

Most people find differentiation easier than integration, because differentiation has familiar and well-defined rules. Integrating functions can be more complicated and requires more experience, particularly because there are several different ways to integrate. Although integrating basic functions can be simple, things can get complicated when integrating products of functions. One of the most common ways to integrate products of functions is the “integrating by parts” method. Note that integration by parts is not used when the integral in question is a product of the integrals of the individual functions.

## The Basics to Integration by Parts for Multivariable Calculus

There are a few things that we need to define before we see how to integrate by parts. If u and v are functions of x, then we have:

$\frac { d\left( uv \right) }{ dx } =u\left( x \right) \frac { dv\left( x \right) }{ dx } +v\left( x \right) \frac { du\left( x \right) }{ dx }$

The above equation is known as the product rule. It says that the derivative of a product of two functions is the sum of the first function times the derivative of the second, plus the second function times the derivative of the first function. Rearranging this equation gives us:

$u\left( x \right) \frac { dv\left( x \right) }{ dx } =v\left( x \right) \frac { du\left( x \right) }{ dx } -\frac { d\left( uv \right) }{ dx }$

Integrating the above equation with respect to x gives us the following:

$\int { udv } =uv-\int { vdu }$

The above equation is the integration by parts formula. This allows us to turn complicated integrals into simpler integrals. The left-hand side of the equation is called the integrand. Unlike the product rule, from which the integration by parts formula is derived, we do not identify u and v but u and dv. Don’t be scared that it’s made up of the derivative of the function v. All you need to do is choose the correct functions for u and dv.

So now that we have the formula, what next? The first thing to do is identify the two basic functions of the product and substitute them for u and dv in the integrand. There are many methods to help us choose which function to use for u.

A good rule of thumb for choosing u is to remember the abbreviation LIATE; this stands for Logarithmic, Inverse circular, Algebraic, Trigonometric (circular), and Exponential. Your first choice should be logarithmic, followed by inverse circular, and so on. The function for u should be chosen such that du is of a simpler form than u. If there is a power function, then it is a good idea to use this as u, as integrating this function will reduce the power by one, resulting in a simpler function on the right-hand side of the integration by parts formula. Obviously, choosing u means that the remaining function is assigned to dv.

Once u and dv have been identified, you need to calculate v and du. To do this, you differentiate u to get du and integrate dv to get v. Then, u and dv are substituted into the right-hand side of the integration by parts equation. The right-hand side presents the result of the integral. If u and v have been chosen correctly, the integral on the right-hand side should just end up being a simple function, or simpler than the original function.

## Integration by Parts Examples

Now, let’s try some integration by parts examples together.

Example 1:

$\int { x\sin { x } } dx$

Following the rule that du should be of a simpler form than u, we chose u = x. Hence, dv = sinxdx.

Integrating u, we have du = dx, and differentiating dv, we have:

$v=\int { \sin { x } dx } =-\cos { x }$

Putting these into the integration by parts equation, we have:

$\int { udv } =uv-\int { vdu }$

$\int { x\sin { x } dx } =(x)\left( -\cos { x } \right) -\int { (-\cos { x )} }$

$=-x\cos { x } +\sin { x } +c$

Example 2:

$\int { { x }^{ 2 }{ e }^{ -x } } dx$

Using the LIATE rule, we can let u = x2 and dv = e-x dx. Hence, du = 2x dx and v = -e-x. Putting this into the integration by parts formula:

$\int { udv } =uv-\int { vdu }$

$\int { { x }^{ 2 }{ e }^{ -x } } dx=({ x }^{ 2 })({ -e }^{ -x })-\int { ({ -e }^{ -x })(2x)dx }$

$={ -x }^{ 2 }{ e }^{ -x }+2\int { { xe }^{ -x }dx }$

The integral above is not a simple function; we need to perform integration by parts again. Let u = x and hence, dv = e-x dx. Therefore, we have du = dx and v = -e-x. Again, working through the integration by parts formula:

$\int { udv } =uv-\int { vdu }$

$\int { { xe }^{ -x }dx } =(x)({ -e }^{ -x })-\int { ({ -e }^{ -x })dx }$

$={ -xe }^{ -x }+\int { ({ e }^{ -x })dx }$

$={ -xe }^{ -x }-{ e }^{ -x }$

Putting both parts together, we have:

$\int { { x }^{ 2 }{ e }^{ -x } } dx={ -x }^{ 2 }{ e }^{ -x }+2({ -xe }^{ -x }-{ e }^{ -x })$

$={ -e }^{ -x }({ x }^{ 2 }+2x+2)+c$

Example 3:

$\int { { x }^{ 2 }\ln { x } } dx$

Following LIATE, we shall choose lnx for u and hence, dv = x2 dx. This means that we have:

$du=\frac { dx }{ x }$

And

$v=\frac { 1 }{ 3 } { x }^{ 3 }$

Substituting this into the integration by parts formula, we have:

$\int { udv } =uv-\int { vdu }$

$\int { { x }^{ 2 }\ln { x } dx } =(\ln { x } )(\frac { 1 }{ 3 } { x }^{ 3 })-\int { (\frac { 1 }{ 3 } { x }^{ 3 }) } (\frac { dx }{ x } )$

$=\frac { 1 }{ 3 } { x }^{ 3 }\ln { x } -\frac { 1 }{ 3 } \int { ({ x }^{ 2 })dx }$

$=\frac { 1 }{ 3 } { x }^{ 3 }\ln { x } -(\frac { 1 }{ 3 }) (\frac {{ x }^{ 3 }}{3})+c$

$=\frac { 1 }{ 3 } { x }^{ 3 }\ln { x } - \frac {{ x }^{ 3 }}{9}+c$

## Integration by Parts and Multivariable Functions

Things can get a bit more complicated when we start dealing with multivariable functions (functions that depend on more than one variable).

Let u(x1, …, xn) and v(x1, …, xn) be functions of n variables x1, …, xn. When using the formula for integration by parts for multivariable functions, instead of integrating over an interval as we have done with one variable, we have to integrate over an n-dimensional set. We also have to replace derivatives with partial derivatives.

Let us denote $\Omega$ as an open, bounded subset of Rn, with a boundary $\Gamma$. If we assume that u and v are continuously differentiable over $\Omega$, then the standard integration by parts formula can be adapted to multivariate functions as follows:

$\int _{ \Omega } \frac { \partial u }{ \partial { x }_{ i } } vd\Omega =\int _{ \Gamma } uv{ v }_{ i }d\Gamma -\int _{ \Omega } \frac { \partial v }{ \partial { x }_{ i } } vd\Omega$

Here, $v\prime$ is the outward unit normal to the boundary $\Gamma$, and vi is its ith component (i ranges from 1 to n).

The above equation can be generalized by substituting vi with v and taking the sum over i:

$\int _{ \Omega } \nabla u.vd\Omega =\int _{ \Gamma } (uv).v\prime d\Gamma -\int _{ \Omega }u\nabla .vd\Omega$

Here, v is a vector function with components v1, …, vn. The above equation leads us to an important theorem when u is set to 1: the divergence/Gauss’ theorem, expressed as follows:

$\int _{ \Omega } \nabla .vd\Omega =\int _{ \Gamma }v.v\prime d\Gamma$

Likewise, we also obtain Green’s formula:

$\int _{ \Omega } \nabla u.\nabla vd\Omega =\int _{ \Gamma } u\nabla v.v\prime d\Gamma -\int _{ \Omega } u\nabla vd\Omega$

The divergence theorem is used to relate surface integrals with triple integrals. For example, it relates a volume integral over volume $\Omega$ (the left-hand side of the equation) to a surface integral over the surface enclosing the volume (the right-hand side). Similarly, Green’s formula can be used to relate line integrals to area integrals.

We hope that you have found this guide to integration by parts useful. After reading this review article, you will have seen where the integration by parts formula comes from, why it is used, and, most importantly, you should be able to tackle even the most difficult integration by parts problems with the tips given above. You will also have read how the integration by parts formula is adapted for multivariable functions so that it can be used for multidimensional problems.

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