Quiz:Integration by parts

For background, see integration by parts.

Statement

Formal manipulation version for definite integration in function notation

Suppose $F,G$ are continuous functions defined on all of $\mathbb{R}$. Suppose $a,b$ are real numbers with $a<b$. Suppose, further, that $G(x)$ is identically zero everywhere except on the open interval $(a,b)$. Then, what can we say about the relationship between the numbers $P=\underset{a}{\overset{b}{\int }}F(x)G^{\prime }(x)dx$ and $Q=\underset{a}{\overset{b}{\int }}{F}^{\prime }(x)G(x)dx$?

	$P=Q$
	$P=-Q$
	$PQ=0$
	$P=1-Q$
	$PQ=1$

Key observations

Equivalence of integration problems

Which of the following is not true (the ones that are true can be deduced from integration by parts)?

	We can compute an expression for the antiderivative of the pointwise product of functions $fg$ based on knowledge of expressions for $f$, $g$, and their antiderivatives.
	Suppose $F$ and $G$ are everywhere differentiable. Given an expression for the antiderivative for the pointwise product of functions $F^{\prime }G$, we can obtain an expression for the antiderivative for the pointwise product $F{G}^{\prime }$.
	If $F$ is a one-to-one function, we can find an antiderivative for $F^{-1}$ in terms of $F^{-1}$ and an antiderivative for $F$.

For more quiz questions on the theme of equivalence of integration problems, see Quiz:Equivalence of integration problems.

Repeated use of integration by parts and the circular trap

1 Suppose we start with a product of two functions $F$ and $g$. Apply integration by parts $k$ times. Start off by taking $F$ as the part to differentiate and $g$ as the part to integrate. Each time, take the part to differentiate as the function obtained by differentiation, and the part to integrate as the function obtained by integration. Assuming that the process of repeatedly finding antiderivatives works without a hitch, what can we conclude about the final integrand? Ignore the other times in the antiderivatve that don't involve integral signs.

	It is a product of the $k^{th}$ derivative of $F$ and the $k^{th}$ derivative of $g$.
	It is a product of the $k^{th}$ derivative of $F$ and the $k^{th}$ antiderivative of $g$.
	It is a product of the $k^{th}$ antiderivative of $F$ and the $k^{th}$ derivative of $g$.
	It is a product of the $k^{th}$ antiderivative of $F$ and the $k^{th}$ antiderivative of $g$.

2 Consider the integration $\int p(x)q^{″}(x)dx$. Apply integration by parts twice, first taking $p$ as the part to differentiate, and $q$ as the part to integrate, and then again apply integration by parts to avoid a circular trap. What can we conclude?

	$\int p(x)q^{″}(x)dx=\int p^{″}(x)q(x)dx$
	$\int p(x)q^{″}(x)dx=\int p^{\prime }(x)q^{\prime }(x)dx-\int p^{″}(x)q(x)dx$
	$\int p(x)q^{″}(x)dx=p^{\prime }(x)q^{\prime }(x)-\int p^{″}(x)q(x)dx$
	$\int p(x)q^{″}(x)dx=p(x)q^{\prime }(x)-p^{\prime }(x)q(x)+\int p^{″}(x)q(x)dx$
	$\int p(x)q^{″}(x)dx=p(x)q^{\prime }(x)-p^{\prime }(x)q(x)-\int p^{″}(x)q(x)dx$

3 Which of the following is an incorrect way of applying integration by parts twice?

	After applying integration by parts once, we get a new product. Choose as the part to integrate the factor in the product arising from integration, and as the part to differentiate the factor in the product arising from differentiation.
	After applying integration by parts once, we get a new product. Choose as the part to differentiate the factor in the product arising from integration, and as the part to integrate the factor in the product arising from differentiation.
	Neither method is incorrect in general. The first method is used for straightforward integrations and the second method is used for the recursive version of integration by parts.

Recursive version of integration by parts

See the questions in the next section, #Choosing the parts to integrate and differentiate.

Integration by parts is not the exclusive strategy for products

1 Which of the following correctly describes the relationship between integration by parts and integration by u-substitution when deciding to integrate a pointwise product of functions or a composite of two functions?

	Integration by parts, which is obtained from the product rule for differentiation, is the exclusive strategy for integrating products. Integration by u-substitution, which is obtained from the chain rule for differentiation, is the exclusive strategy for integrating composites.
	Integration by parts, which is obtained from the product rule for differentiation, is the exclusive strategy for integrating composites. Integration by u-substitution, which is obtained from the chain rule for differentiation, is the exclusive strategy for integrating products.
	Both methods are useful for both types of integrations. Specifically, integration by parts helps with certain kinds of products and composites, and integration by u-substitution helps with certain kinds of products.

2 Which of the following functions of $x$ can be integrated with respect to $x$ without integration by parts, and purely using integration by u-substitution and the knowledge of the antiderivative of the cosine function?

	$x^2\cos x$
	$x{\cos }^{2}x$
	$x\cos (x^2)$
	$x^2\cos (x^2)$
	$x^2{\cos }^{2}x$

Choosing the parts to integrate and differentiate

Products and typical strategies

Review the strategies for products and determining the parts to integrate and differentiate: [SHOW MORE]

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: $x{e}^x,x^2{e}^x,x\cosh x,x\sinh x$ (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.	${\sin }^{2}x,e^x\cos x$ The function ${\sec }^{3}x$ can also be done using a similar method, though it falls outside the basic trigonometric function products.

{{#widget:YouTube|id=WQQfIbhusak}}

1 Suppose $p$ is a polynomial function. In order to find the indefinite integral for a function of the form $x\mapsto p(x)\sin x$, the general strategy, which always works, is to take $p(x)$ as the part to differentiate and $\sin x$ as the part to integrate, and keep repeating the process. Which of the following is the best explanation for why this strategy works?
RELATED COMPUTATIONAL PRACTICE: $\int x^2\sin xdx$, $\int x\sin xdx$, $\int (x^2+1{)}^2\cos xdx$, $\int x{\cos }^{2}xdx$, $\int 2x\cos (3x)dx$

	$\sin$ can be repeatedly differentiated and polynomials can be repeatedly integrated, giving polynomials all the way.
	$\sin$ can be repeatedly integrated and polynomials can be repeatedly differentiated, eventually becoming zero.
	$\sin$ and polynomials can both be repeatedly differentiated.
	$\sin$ and polynomials can both be repeatedly integrated.

2 Suppose $p$ is a polynomial function. In order to find the indefinite integral for a function of the form $x\mapsto p(x)\exp (x)$, the general strategy, which always works, is to take $p(x)$ as the part to differentiate and $\exp (x)$ as the part to integrate, and keep repeating the process. Which of the following is the best explanation for why this strategy works? RELATED COMPUTATIONAL PRACTICE: $\int x{e}^{x}dx$, $\int x^2{e}^{x}dx$, $\int (x+3)e^{-4x}dx$

	$\exp$ can be repeatedly differentiated and polynomials can be repeatedly integrated, giving polynomials all the way.
	$\exp$ can be repeatedly integrated and polynomials can be repeatedly differentiated, eventually becoming zero.
	$\exp$ and polynomials can both be repeatedly differentiated.
	$\exp$ and polynomials can both be repeatedly integrated.

3 Consider the function $x\mapsto \exp (x)\sin x$. This function can be integrated using integration by parts. What can we say about how integration by parts works?
RELATED COMPUTATIONAL PRACTICE: $\int e^{-x}\cos xdx$, $\int e^x\cos xdx$, $\int e^x{\sin }^{2}xdx$

	We choose $\exp$ as the part to integrate and $\sin$ as the part to differentiate, and apply this process once to get the answer directly.
	We choose $\exp$ as the part to integrate and $\sin$ as the part to differentiate, and apply this process once, then use a recursive method (identify the integrals on the left and right side) to get the answer.
	We choose $\exp$ as the part to integrate and $\sin$ as the part to differentiate, and apply this process twicce to get the answer directly.
	We choose $\exp$ as the part to integrate and $\sin$ as the part to differentiate, and apply this process twice, then use a recursive method (identify the integrals on the left and right side) to get the answer.

4 Suppose $p$ is a polynomial function. Consider the function $x\mapsto p(x)\ln x$ for $x>0$. This function can be integrated using integration by parts and knowledge of how to integrate polynomials. Which of the following is the best integration strategy?
RELATED COMPUTATIONAL PRACTICE: $\int \ln xdx$, $\int x\ln xdx$, $\int (x^2-2)\ln (x+3)dx$.

	Take $p$ as the part to differentiate and $\ln$ as the part to integrate. Integration by parts needs to be applied just once.
	Take $p$ as the part to integrate and $\ln$ as the part to integrate. Integration by parts needs to be applied just once.
	Take $p$ as the part to differentiate and $\ln$ as the part to integrate. Integration by parts needs to be applied as many times as the degree of $p$.
	Take $p$ as the part to integrate and $\ln$ as the part to integrate. Integration by parts needs to be applied as many times as the degree of $p$.

5 Suppose $p$ and $q$ are polynomial functions. Consider the function $x\mapsto p(x)q(\ln x)$ for $x>0$. This function can be integrated using integration by parts and knowledge of how to integrate polynomials. Using the best strategy, how many times do we need to apply integration by parts?
RELATED COMPUTATIONAL PRACTICE: $\int x^3(\ln x{)}^{2}dx$, $\int (x^2+1)((\ln x{)}^2+1))dx$

	The number of times equals the sum of degrees of $p$ and $q$.
	The number of times equals the product of degrees of $p$ and $q$.
	The number of times equals the degree of $p$.
	The number of times equals the degree of $q$.

6 Suppose $p$ and $q$ are polynomial functions. Consider the function $x\mapsto p(x)\ln (q(x))$. Assume we are integrating over a domain where $q(x)$ is positive. How can we integrate this function, and why does the method work?
RELATED COMPUTATIONAL PRACTICE: $\int \ln (x^2+1)dx$, $\int x\ln (x^3+1)dx$

	We can integrate the function by applying integration by parts repeatedly, taking $\ln (q(x))$ as the part to differentiate and $p(x)$ as the part to integrate. The number of times we need to apply integration by parts is the degree of $q$.
	We can integrate the function by applying integration by parts, taking $\ln (q(x))$ as the part to differentiate and $p(x)$ as the part to integrate, and then integrating the rational function so obtained. We use the fact that there is a strategy for integrating any rational function.
	We can integrate the function by applying integration by parts repeatedly, taking $\ln (q(x))$ as the part to integrate and $p(x)$ as the part to differentiate. The number of times we need to apply integration by parts is the degree of $p$.
	We can integrate the function by applying integration by parts, taking $\ln (q(x))$ as the part to integrate and $p(x)$ as the part to differentiate, and then integrating the rational function so obtained. We use the fact that there is a strategy for integrating any rational function.