Module 20: Integration by parts and structural choices

8 exercises, each with a complete solution. 1 figure. 1 note from the engine.

On this page

A. Prerequisites and observable goals
B. The central idea: reverse the product rule
C. The rule and the roles of the pieces
D. Why the transfer can simplify a product
E. Choose a method by the resulting integral
F. Worked teaching examples
G. Independent practice
H. Complete solutions: Module 20
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Product rule, substitution, logarithms, and definite integral evaluation.

After

Recognize when a product suggests integration by parts, choose useful components, repeat the method when necessary, and solve a cyclic integral.

B. The central idea: reverse the product rule#

Substitution reverses a chain rule. Integration by parts reverses a product rule. Neither is triggered by appearance alone: a product such as $x e^{x^2}$ is better handled by substitution, while $x e^x$ is well suited to parts. Your task is to find a transformation that makes the remaining integral simpler or relates it to the original in a solvable way.

C. The rule and the roles of the pieces#

Integrating $(uv)'=u'v+uv'$ gives

\int u\dd v=uv-\int v\dd u.

For differentiable $u,v$ with sufficient continuity on $[a,b]$ ,

\int_a^b u(x)v'(x)\dd x=[u(x)v(x)]_a^b-\int_a^b v(x)u'(x)\dd x.

Here $\dd v$ includes the integration differential; it is not the same object as $v$ . Choosing $u$ requires differentiating it, while choosing $\dd v$ requires finding an antiderivative. A memorized ranking of logarithms, inverse trigonometric functions, polynomials, and exponentials can suggest choices, but it is a heuristic, not a theorem guaranteeing success. [O2Parts]

For indefinite calculations, one arbitrary constant at the end is enough. When using the symbol $I$ to represent an integral in a cyclic calculation, work with a consistently chosen antiderivative up to constants, then restore the arbitrary constant after solving the relation.

Integration by parts as two areas. Plot

v

against

u

. The area under the curve is

\int v\dd u

(blue), the area to its left is

\int u\dd v

(yellow), and together they fill the large rectangle less the small one:

u_2v_2-u_1v_1

. That is the definite formula above, read as a picture.

D. Why the transfer can simplify a product#

In $\int x e^x\dd x$ , set $u=x$ , $\dd v=e^x\dd x$ . Differentiating lowers the polynomial degree and integrating the exponential does not make it harder. The formula transfers one derivative onto $x$ :

\int x e^x\dd x=xe^x-\int e^x\dd x=e^x(x-1)+C.

The product rule checks the answer: $e^x(x-1)+e^x=xe^x$ . In contrast, choosing $u=e^x$ , $\dd v=x\dd x$ gives a new integral containing $x^2e^x$ , which is harder. A legal transformation need not be a useful one.

E. Choose a method by the resulting integral#

Simplify first. Look for a direct derivative pattern or substitution. If neither appears, ask whether one factor becomes simpler under differentiation and the other has a manageable antiderivative. Write a two-column record of $u,\dd u$ and $\dd v,v$ . Apply the formula with its minus sign, then inspect the new integral. Stop when it is elementary, repeat if complexity is decreasing, or solve algebraically if the original integral returns. For definite integrals, evaluate the boundary product as well as the remaining integral.

F. Worked teaching examples#

A hidden product. To integrate $\ln x$ for $x>0$ , write it as $(\ln x)\cdot1$ . Choose $u=\ln x$ , $\dd v=\dd x$ , so $\dd u=\dd x/x$ , $v=x$ . Then $\int\ln x\dd x=x\ln x-\int1\dd x=x\ln x-x+C$ . A product need not have two visibly complicated factors.

Repeated parts. With $I=\int x^2\cos x\dd x$ , choose $u=x^2$ , $\dd v=\cos x\dd x$ . Then $I=x^2\sin x-2\int x\sin x\dd x$ . The second integral, using $u=x$ , $\dd v=\sin x\dd x$ , equals $-x\cos x+\sin x$ . Therefore

I=x^2\sin x+2x\cos x-2\sin x+C.

Notice that $-2$ multiplies both terms of the second result.

A cyclic comparison. Integrals of $e^{ax}\sin bx$ or $e^{ax}\cos bx$ often return to themselves after two applications. That is not failure: it supplies an equation for the unknown antiderivative. Keep the repeated integral named consistently and collect it on one side.

From the engine: AlightThe rocket equation, integrated twice

Alight, the author’s powered-descent guidance lab, bounds what a rocket can do in a vertical burn. With exhaust velocity $v_e$ , starting mass $m_0$ , and propellant flow $\beta$ , the fraction burned by time $t$ is $x = \beta t/m_0$ , and thrust adds the velocity $-v_e\ln(1 - x)$ : the rocket equation. Height needs that velocity integrated once more, and $\int \ln u\,du = u\ln u - u$ is the integration by parts this module opens with.

Alight: closed forms for a vertical burnC++

VerticalState profile(const Channel& ch, double alpha, double gain, double beta, double t) {
    const double x = beta * t / ch.m0;
    const double l = std::log1p(-x);
    VerticalState s;
    s.upMps = ch.w0 - ch.g * t - (gain / alpha) * l;
    s.altitudeM = ch.h0 + ch.w0 * t - 0.5 * ch.g * t * t + (gain * ch.m0 / (alpha * beta)) * ((1.0 - x) * l + x);
    return s;
}

In the code 1/alpha is $v_e$ and gain is the share of thrust that points up. The bracket $(1 - x)\ln(1 - x) + x$ is the by-parts result after the substitution $u = 1 - \beta s/m_0$ . Try $v_e = 3000$ m/s, $m_0 = 1000$ kg, $\beta = 5$ kg/s, and $t = 60$ s, from rest at height zero, with all thrust upward and $g = 9.81\ \mathrm{m/s^2}$ . Then $x = 0.3$ ; thrust adds 1070.0 m/s, gravity removes 588.6 m/s, leaving 481.4 m/s; and the height gained is 12,539 m. The model assumes constant gravity and no drag. log1p(-x) computes $\ln(1 - x)$ without losing digits when $x$ is small.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M20.01FSolution

Evaluate $\int x\cos(2x)\dd x$ .

M20.02CSolution

Evaluate $\int x^2e^x\dd x$ .

M20.03CSolution

Evaluate $\int_1^e\ln x\dd x$ .

M20.04TSolution

Evaluate $\int\arctan x\dd x$ .

M20.05TSolution

Evaluate $\int e^{2x}\sin x\dd x$ by a cyclic integration-by-parts calculation.

M20.06CSolution

Evaluate $\int_0^1x\ln(1+x)\dd x$ .

M20.07TSolution

Choose and execute an efficient method for each integral: $\int x e^{x^2}\dd x$ and $\int x e^x\dd x$ . Explain the structural difference.

M20.08SSolution

For integers $n\ge1$ , derive a reduction relation for $I_n=\int_0^{\pi/2}\sin^n x\dd x$ . Use it to find $I_5$ , treating the base case separately.

Optional small hints

For M20.04, use $\dd v=\dd x$ . For M20.05, name the sine integral $I$ and the cosine integral $J$ . For M20.06, divide $x^2$ by $1+x$ after parts. For M20.08, the standard recurrence is intended for $n\ge2$ , not for the base case $n=1$ .

H. Complete solutions: Module 20#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M20.01Show solution

Remember the antiderivative’s scale factor

Choose $u=x$ , $\dd v=\cos(2x)\dd x$ , so $\dd u=\dd x$ and $v=\sin(2x)/2$ . Then

\int x\cos(2x)\dd x=\frac{x\sin(2x)}2-\frac12\int\sin(2x)\dd x =\boxed{\frac{x\sin(2x)}2+\frac{\cos(2x)}4+C}.

Differentiation makes the two extra sine terms cancel.

Back to M20.01

M20.02Show solution

Lower the polynomial degree twice

First, $I=x^2e^x-2\int xe^x\dd x$ . A second application gives $\int xe^x\dd x=xe^x-e^x$ . Thus $\boxed{I=e^x(x^2-2x+2)+C}$ . Checking: the product derivative is $e^x[(x^2-2x+2)+(2x-2)]=x^2e^x$ .

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M20.03Show solution

Evaluate the boundary product too

The primitive is $x\ln x-x$ . At $e$ it is $e-e=0$ ; at $1$ it is $-1$ . Hence the integral is $0-(-1)=\boxed1$ . The positive answer agrees with $\ln x\ge0$ on the interval.

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M20.04Show solution

A single inverse trigonometric factor is a product with one

Use $u=\arctan x$ , $v=x$ . Then

\int\arctan x\dd x=x\arctan x-\int\frac{x}{1+x^2}\dd x =\boxed{x\arctan x-\tfrac12\ln(1+x^2)+C}.

The remaining integral is a substitution with $w=1+x^2$ . The logarithm argument is positive for every real $x$ , so no additional domain split is required.

Back to M20.04

M20.05Show solution

Solve the equation produced by repetition

Let $I=\int e^{2x}\sin x\dd x$ and $J=\int e^{2x}\cos x\dd x$ . Integrate the exponential in each parts step:

I=\tfrac12e^{2x}\sin x-\tfrac12J,\qquad J=\tfrac12e^{2x}\cos x+\tfrac12I.

Substitution gives

I=\tfrac12e^{2x}\sin x-\tfrac14e^{2x}\cos x-\tfrac14I.

Thus $(5/4)I=e^{2x}(2\sin x-\cos x)/4$ , so

\boxed{I=e^{2x}(2\sin x-\cos x)/5+C}.

Product differentiation verifies the coefficients and signs.

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M20.06Show solution

Parts followed by polynomial division

Take $u=\ln(1+x)$ and $\dd v=x\dd x$ . Then

I=\left[\frac{x^2}2\ln(1+x)\right]_0^1-\frac12\int_0^1\frac{x^2}{1+x}\dd x.

Since $x^2/(1+x)=x-1+1/(1+x)$ , the remaining integral is $1/2-1+\ln2=\ln2-1/2$ . Consequently $I=\tfrac12\ln2-\tfrac12(\ln2-1/2)=\boxed{1/4}$ . The logarithmic terms cancel only after both definite evaluations are included.

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M20.07Show solution

Similar-looking products can encode different rules

For $xe^{x^2}$ , the inner exponent’s derivative is $2x$ , so $u=x^2$ gives $\boxed{\tfrac12e^{x^2}+C}$ . For $xe^x$ , the polynomial factor is not the derivative of the exponent; parts reduces it to $\boxed{e^x(x-1)+C}$ . Differentiating both answers checks that the first produces an extra $2x$ through the chain rule while the second uses product-rule cancellation.

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M20.08Show solution

A recurrence from one parts step and an identity

For $n\ge2$ , write $\sin^n x=\sin^{n-1}x\sin x$ and choose $u=\sin^{n-1}x$ , $\dd v=\sin x\dd x$ . The boundary term $-\sin^{n-1}x\cos x$ vanishes at both ends. Thus

I_n=(n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x\dd x =(n-1)(I_{n-2}-I_n).

Solving gives $\boxed{I_n=\frac{n-1}{n}I_{n-2}}$ . With $I_1=\int_0^{\pi/2}\sin x\dd x=1$ , $I_3=2/3$ and $\boxed{I_5=8/15}$ . For $n=1$ the boundary term is not zero, which is why the recurrence’s derivation must not be blindly extended to that base case.

Back to M20.08

I. Retention and readiness#

Revisit M17.04 and M10.05. Explain which earlier derivative rule each integration method reverses. Advance when you can predict the new integral before committing to a parts choice and can check the final primitive by differentiation.