Practical:Integration by parts

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ORIGINAL FULL PAGE: Integration by parts
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Statement

Statement in multiple versions

Version type Statement
Indefinite integral 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 and f = F'. Then we have:
\int F(x)g(x) \, dx = F(x)G(x) - \int f(x)G(x) \, dx
Definite integral in function notation Suppose F and g are continuous functions on a closed interval [a,b] such that F is a differentiable function at all points of [a,b] and we want to integrate F(x)g(x). Suppose G is an antiderivative for g and f is the derivative F'. 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
The part [F(x)G(x)]_a^b is shorthand for F(b)G(b) - F(a)G(a), in keeping with the standard evaluate between limits notation used for definite integrals.
Indefinite integral in dependent-independent variable 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), du/dx=f(x), dv/dx = g(x), and v = G(x).
Verbal version 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 integral of the second function).

Associated jargon

The function F in the function notation (or the variable u in in the variable notation) is termed the part to differentiate. The function g in the function notation (which is dv/dx in the variable notation) is termed the part to integrate.

As a process

For further information, refer: Practical:Integration by parts

We can think of integration by parts overall as a five- or six-step process. The really hard discretionary parts (i.e., the parts that are not purely procedural but require decision-making) are Steps (1) and (2):

  1. Identify the function being integrated as a product of two functions
  2. Figure out which is the part to differentiate and which is the part to integrate
  3. Determine the antiderivative of the part to integrate and the derivative of the part to differentiate
  4. Apply integration by parts
  5. Now, do the second integral
  6. (Optional) Verify your answer by differentiating the antiderivative obtained

Basic procedural observations

Observation More information
Integration by parts essentially reverses the product rule for differentiation applied to F(x)G(x) (or uv). proof section
Solving a problem through a single application of integration by parts usually involves two integrations -- one to find the antiderivative G for g (which in the u-v notation is equivalent to finding v given dv/dx) and then doing the right side integration of f(x)G(x) (or v\frac{du}{dx}). Note that:
(1) For the first integration (the one involving finding G or v), we just need to find one antiderivative and we do not put a +C or try to evaluate between limits.
(2) On the other hand, at the end of the second integration (the integration of f(x)G(x) or v\frac{du}{dx}), we do put a +C (for indefinite integration) or evaluate between limits (for definite integration).
See the section choice of antiderivative does not matter. For more details, see choice of antiderivative does not matter for integration by parts
The key way that the original integral \int F(x)g(x) \, dx (or \int u \frac{dv}{dx} \, dx) and the new integral \int f(x)G(x) \, dx (or \int v \frac{du}{dx} \, dx) differ is that in the new integral, one part has been differentiated and the other part has been integrated. The section equivalence of integration problems. For more detailed information, see equivalence of integration problems arising from integration by parts


Explicit procedure

The procedure for applying integration by parts, explicitly, is as follows:

  1. Identify the function being integrated as a product of two functions: In some cases,we may take one of the factors to be the function 1, and the other one to be the whole integrand.
  2. Figure out which is the part to differentiate and which is the part to integrate: In the explicit function notation, the part to differentiate is F and the part to integrate is g. In the u-v notation, the part to differentiate is u and the part to integrate is what we'll eventually call dv/dx. This is the step where we need to use explicit heuristics. The general heuristic used is the ILATE or LIATE rule. It says that inverse trigonometric and logarithmic functions are most preferable for differentiation, algebraic functions (polynomials, rational functions, and radicals) are in between, and exponential and trigonometric functions are most preferable for integration.
  3. Determine the antiderivative of the part to integrate and the derivative of the part to differentiate: In the function notation, these are F' and G. In the u-v notation, these are du/dx and v respectively.
  4. Apply integration by parts
  5. Now, do the second integral: This may require integration by parts again, or we may do it some other way, or we may use the recursive version of integration by parts.
  6. (Optional) Verify your answer by differentiating the antiderivative obtained: You should find yourself using the product rule for differentiation, and further, you should find some of the terms canceling and the left-over term should be the integrand.

Example of x \cos x

Consider the integration problem:

Compute the antiderivative \int x \cos x \, dx
  1. Identify the function being integrated as the product of two functions: Here, the two factor functions are x and \cos x respectively.
  2. Figure out which is the part to differentiate and which is the part to integrate: Using the ILATE rule or through direct strategic thinking, we see that x (the polynomial, algebraic part) should be the part to differentiate and \cos x should be the part to integrate. Explicitly:
    • In the function notation, we have F(x) = x and g(x) = \cos x.
    • In the u-v notation, we have u = x and dv/dx = \cos x.
  3. Determine the antiderivative of the part to integrate and the derivative of the part to differentiate: The antiderivative of \cos x is \sin x and the derivative of x is 1. Explicitly:
    • In the function notation, we have F'(x) = 1 and G(x) = \sin x.
    • In the u-v notation, we have du/dx = 1 and v = \sin x.
  4. Apply integration by parts: We get
    \int x\cos x \, dx = x\sin x - \int (1)(\sin x) \, dx = x \sin x - \int \sin x \, dx
  5. Now, do the second integral: We have \int \sin x = -\cos x, so plugging it back in, we get x \sin x - 
(- \cos x) + C = x \sin x + \cos x + C.

Products and typical strategies

Product type Preliminary thoughts Strategy for this product type Examples
polynomial function times basic trigonometric or exponential function (such as sin, cos, exp, cosh, sinh, or a polynomial in such expressions) 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. Trig: x \cos x, x \sin x, x^2 \cos x (see here for worked out examples)
Exp: xe^x, x^2e^x, x \cosh x, x\sinh x (see here for worked out examples)
inverse trigonometric function or logarithmic function (or composite of such a function with a polynomial 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. Simple products:\ln x, x \ln x, \arctan x, x \arctan x, \arcsin x (see here for worked out examples)
Products involving composites: \ln(x^2 + 1), x\ln(x^3 + 1), x \arctan(x^3 - x)
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^2x, e^x \cos x
The function \sec^3x can also be done using a similar method, though it falls outside the basic trigonometric function products.


The examples here are done quickly, without the explicit function/letter naming procedure.

Polynomial times basic trigonometric

In all these, we choose to differentiate the polynomial and integrate the trigonometric function. We use the following formulas:

Function Antiderivative
\sin -\cos
\cos \sin
Integration problem What we choose to differentiate and integrate Full solution (quick version) In video (embedded below)?
\int x\cos x \, dx Differentiate x to get 1, integrate \cos x to get \sin x \int x\cos x \, dx = x \sin x - \int (1) (\sin x) \, dx = x\sin x + \cos x + C Yes
\int x \sin x \, dx Differentiate x to get 1, integrate \sin x to get -\cos x \int x \sin x \, dx = x(-\cos x) - \int (1)(-\cos x) \, dx = -x\cos x + \sin x + C Yes
\int x^2 \cos x \, dx Differentiate x^2 to get 2x, integrate \cos x to get \sin x, then again differentiate the polynomial and integrate the trigonometric function \int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx = x^2\sin x - 2[x(-\cos x) - \int (1)(-\cos x) \, dx]
This simplifies to x^2 \sin x + 2x \cos x - 2 \sin x + C
Yes

Polynomial times exponential

In all these, we choose to differentiate the polynomial and integrate the exponential function. We use that the exponential function is its own antiderivative.

Integration problem What we choose to differentiate and integrate Full solution (quick version) In video (embedded above)?
\int xe^x \, dx Differentiate x to get 1, integrate e^x to get e^x \int xe^x \, dx = xe^x - \int (1)(e^x) \, dx = xe^x - e^x + C Yes
\int x^2e^x \, dx Differentiate x^2 to get 2x, integrate e^x to get e^x, then again differentiate the polynomial and integrate the exponential \int x^2e^x \, dx = x^2e^x - \int 2x e^x \, dx = x^2e^x - 2[xe^x - \int (1)(e^x) \, dx] = x^2e^x - 2xe^x + 2e^x + C Yes

Polynomial times logarithmic

In all these, we choose to integrate the polynomial and differentiate the logarithmic function. We use that d(\ln x)/dx = 1/x.

Integration problem What we choose to differentiate and integrate Full solution (quick version) In video (embedded below)?
\int \ln x \, dx Think of integrand as (\ln x)(1), differentiate \ln x to get 1/x, integrate 1 to get x \int \ln x \, dx = \int (\ln x)(1) \, dx = (\ln x)(x) - \int (1/x)(x) \, dx = x\ln x - x + C Yes
\int x \ln x \, dx Differentiate \ln x to get 1/x, integrate x to get x^2/2 \int x \ln x \, dx = (\ln x)(x^2/2) - \int \frac{1}{x} \frac{x^2}{2} \, dx = \frac{x^2 \ln x}{2} - \int \frac{x}{2} \, dx = \frac{x^2 \ln x}{2} - \frac{x^2}{4} + C Yes

Polynomial times inverse trigonometric

In all these, we choose to integrate the polynomial and differentiate the inverse trigonometric function. We use that \frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}.

For many of these examples, we need to use some tricks/techniques related to the integration of rational functions to simplify the new rational function or algebraic expressions we obtain.

Integration problem What we choose to differentiate and integrate Full solution (quick version) In video (embedded above)
\int \arctan x \, dx Think of integrand as (\arctan x)(1), integrate 1 to get x, differentiate \arctan x to get 1/(1 + x^2) \int \arctan x \, dx = (\arctan x)(x) - \int \frac{1}{1 + x^2}x \, dx = x \arctan x - \frac{1}{2} \int \frac{2x}{1 + x^2} \, dx
This simplifies to x \arctan x - \frac{1}{2}\ln(1 + x^2) + C
Note that for the latter integration, we use the \int g'/g = \ln|g| rule.
Yes
\int x \arctan x \, dx integrate x to get x^2/2, differentiate \arctan x to get 1/(1 + x^2). \int x \arctan x \, dx = (\arctan x)(x^2/2) - \int \frac{1}{1 + x^2} \frac{x^2}{2} \, dx = \frac{x^2\arctan x}{2} - \frac{1}{2} \int \frac{x^2}{1 + x^2} \, dx
We now use that the fraction is an improper fraction to convert it to the mixed fraction 1 - \frac{1}{1 + x^2}, and get: \frac{x^2\arctan x}{2} - \frac{1}{2} \int [1 - \frac{1}{1 + x^2}] \, dx.
Simplfying gives \frac{x^2 \arctan x}{2} - \frac{1}{2}(x-\arctan x) + C = \frac{(x^2 + 1)\arctan x}{2} - \frac{x}{2} + C
Yes