Chapter 7 · Integrals

Exercise 7.6 · Q21: $\int e^{2x}\sin x\,dx$

Exercise 7.6Integration by parts · Section 7.6

The question

Integrate the functions in Exercises 1 to 22.

21.

e^{2x}\sin x

Exam answer

Let $I = \displaystyle\int e^{2x}\sin x\,dx$ ⁠.

Taking $e^{2x}$ as the first function, by parts:

\begin{aligned} I &= e^{2x}(-\cos x) \\ &\quad - \int 2e^{2x}(-\cos x)\,dx \\ &= -e^{2x}\cos x \\ &\quad + 2\int e^{2x}\cos x\,dx\quad\ldots(\text{i}) \end{aligned}

Again by parts, with $e^{2x}$ as the first function:

\begin{aligned} &\int e^{2x}\cos x\,dx \\ &= e^{2x}\sin x - \int 2e^{2x}\sin x\,dx \\ &= e^{2x}\sin x - 2I\quad\ldots(\text{ii}) \end{aligned}

Using (ii) in (i):

\begin{aligned} I &= -e^{2x}\cos x \\ &\quad + 2e^{2x}\sin x - 4I \\ \Rightarrow\ 5I &= e^{2x}(2\sin x - \cos x) \end{aligned}

$\therefore\ I = \dfrac{e^{2x}}{5}(2\sin x - \cos x)$ $+ C$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

As in Example 21, neither factor disappears when differentiated or integrated: $e^{2x}$ only gains factors of 2, and $\sin x$ cycles to $\cos x$ and back. Integrating by parts twice, with the same kind of first function both times, brings the original integral $I$ back, multiplied by a number. That gives an ordinary equation for $I$ ⁠, which we solve.

Given, to find, and the plan

Given: $e^{2x}\sin x$ ⁠, defined and continuous for every real $x$ ⁠.
To find: $I = \int e^{2x}\sin x\,dx$ ⁠.
Plan: (1) Parts once ( $e^{2x}$ first). (2) Parts on the new integral ( $e^{2x}$ first again). (3) Substitute and solve for $I$ ⁠. (4) Differentiate back.

Basics you need

1Integration by parts

$\int f g\,dx = f\int g\,dx - \int\big[f'\int g\,dx\big]dx$ (Section 7.6); Example 21 used it twice on $e^x\sin x$ ⁠. Lesson: integration by parts

2Derivatives and integrals used

$\frac{d}{dx}e^{2x} = 2e^{2x}$ (chain rule); $\int\sin x\,dx = -\cos x$ ⁠, $\int\cos x\,dx = \sin x$ (standard formulae (iii), (ii)). Basics: trigonometric identities

3Equations between indefinite integrals

They hold up to a constant, so one constant is added after solving (Section 7.2). Basics: integration undoes differentiation

Solution

Integrate by parts once

Take $e^{2x}$ as the first function and $\sin x$ as the second. The derivative of the first is $2e^{2x}$ and the integral of the second is $-\cos x$ ⁠:
$\begin{aligned} I &= e^{2x}(-\cos x) - \int 2e^{2x}(-\cos x)\,dx \\ &= -e^{2x}\cos x + 2\int e^{2x}\cos x\,dx. \end{aligned}$
Write $I_1 = \int e^{2x}\cos x\,dx$ ⁠, so
$I = -e^{2x}\cos x + 2I_1. \qquad (1)$
This follows Example 21, which takes the exponential as the first function. Both factors are equally easy to integrate, so either choice works, provided the same choice is kept in the second round (Step 2). The memory aid ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential; not in the book) would take $\sin x$ first, since T comes before E; the book notes after Example 21 that this choice also works. Here neither factor ever disappears, so the order changes only the bookkeeping, not the answer.

$\frac{d}{dx}e^{2x} = 2e^{2x}$ ⁠; $\int\sin x\,dx = -\cos x$ ⁠. Lesson: integration by parts
Integrate $I_1$ by parts, keeping $e^{2x}$ as the first function

Now $e^{2x}$ is first and $\cos x$ second; the integral of $\cos x$ is $\sin x$ ⁠:
$\begin{aligned} I_1 &= e^{2x}\sin x - \int 2e^{2x}\sin x\,dx \\ &= e^{2x}\sin x - 2I. \qquad (2) \end{aligned}$
The original integral $I$ has come back. If we had switched roles here (taking $\cos x$ first), this round would simply undo Step 1 and give the useless identity $I = I$ ⁠.

$\int\cos x\,dx = \sin x$ (formula (ii)). Lesson: integration by parts
Substitute and solve for $I$

Put (2) into (1):
$\begin{aligned} I &= -e^{2x}\cos x + 2\big(e^{2x}\sin x - 2I\big) \\ &= -e^{2x}\cos x + 2e^{2x}\sin x - 4I. \end{aligned}$
Add $4I$ to both sides:
$5I = e^{2x}(2\sin x - \cos x),$ $\int e^{2x}\sin x\,dx = \frac{e^{2x}}{5}(2\sin x - \cos x) + C.$
This is a linear equation in the unknown $I$ ⁠: collect the $I$ terms on one side and divide by 5. The equation holds up to a constant, which is why one arbitrary constant $C$ is added at the end. The 5 is $1 + 2^2$ ⁠: the 2 from $e^{2x}$ enters once in each round.

Solving $I = a - 4I$ gives $I = \frac a5$ ⁠. Basics: integration undoes differentiation
Check: differentiate the answer
$\begin{aligned} &\frac{d}{dx}\left[\frac{e^{2x}}{5}(2\sin x - \cos x)\right] \\ &= \frac{2e^{2x}}{5}(2\sin x - \cos x) + \frac{e^{2x}}{5}(2\cos x + \sin x) \\ &= \frac{e^{2x}}{5}\big(4\sin x - 2\cos x + 2\cos x + \sin x\big) \\ &= \frac{e^{2x}}{5}\cdot 5\sin x = e^{2x}\sin x. \end{aligned}$
Product rule: $e^{2x}$ differentiates to $2e^{2x}$ ⁠, and $2\sin x - \cos x$ differentiates to $2\cos x + \sin x$ ⁠. The cosine terms cancel and the sine terms add to $5\sin x$ ⁠, giving the integrand for every real $x$ ⁠.

Product rule; chain rule for $e^{2x}$ ⁠. Basics: integration undoes differentiation

Result

\int e^{2x}\sin x\,dx = \frac{e^{2x}}{5}(2\sin x - \cos x) + C,

valid for every real $x$ ⁠. (Textbook answer: $\frac{e^{2x}}{5}(2\sin x - \cos x) + C$ ⁠, the same.)

Check it with numbers

On $[0, \pi]$ ⁠, using $\sin\pi = 0$ ⁠, $\cos\pi = -1$ ⁠:

\begin{aligned} F(\pi) - F(0) &= \frac{e^{2\pi}}{5}(0 + 1) - \frac15(0 - 1) \\ &= \frac{e^{2\pi} + 1}{5} = 107.2983311050\ldots \end{aligned}

Numerical integration of $e^{2x}\sin x$ over $[0, \pi]$ gives $107.2983311050\ldots$ ⁠, the same.

Common mistakes

Switching the first and second functions in the second round: you get $I = I$ ⁠.
Forgetting the factor 2 from $\frac{d}{dx}e^{2x} = 2e^{2x}$ ⁠. Without it you would get $2I = \ldots$ as for $e^x\sin x$ ⁠, and the wrong answer.
Sign slips: $\int\sin x\,dx = -\cos x$ ⁠, so Step 1 gives $+2\int e^{2x}\cos x\,dx$ ⁠.
Dividing by 4 or 3 instead of 5 when collecting $I + 4I$ ⁠.

What to take away

For $e^{ax}\sin bx$ and $e^{ax}\cos bx$ ⁠: parts twice with the same choice, then solve for $I$ ⁠.
The general result is $\begin{aligned} &\int e^{ax}\sin bx\,dx \\ &= \frac{e^{ax}}{a^2+b^2}(a\sin bx - b\cos bx) + C; \end{aligned}$ here $a = 2$ ⁠, $b = 1$ ⁠, $a^2 + b^2 = 5$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts; same choice in both rounds	Steps 1, 2	Section 7.6; Example 21
$\int\sin x\,dx = -\cos x + C$ ⁠; $\int\cos x\,dx = \sin x + C$	Steps 1, 2	Standard formulae (iii), (ii), Section 7.2
Antiderivatives differ by a constant	Step 3	Section 7.2
Product rule, chain rule; derivatives of $e^{2x}$ ⁠, $\sin x$ ⁠, $\cos x$	Steps 1, 4	Chapter 5, Sections 5.3 and 5.3.1

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Exercise 7.6 · Q21: ∫e2xsin⁡x dx\int e^{2x}\sin x\,dx∫e2xsinxdx

Basics you need

Solution

Integrate by parts once

Integrate I1I_1I1​ by parts, keeping e2xe^{2x}e2x as the first function

Substitute and solve for III

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.6 · Q21: $\int e^{2x}\sin x\,dx$

Integrate $I_1$ by parts, keeping $e^{2x}$ as the first function

Substitute and solve for $I$