Chapter 7 · Integrals

Example 21: $\int e^x\sin x\,dx$

ExamplesIntegration by parts · Section 7.6

The question

Find

\int e^x\sin x\,dx

Exam answer

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

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

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

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

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

Using (ii) in (i):

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

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

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

The one idea this problem is built on

Neither factor ever disappears: $e^x$ stays $e^x$ ⁠, and $\sin x$ cycles through $\cos x$ ⁠, $-\sin x$ ⁠, $-\cos x$ ⁠. So integrating by parts once gives $\int e^x\cos x\,dx$ ⁠, and integrating by parts again brings back $\int e^x\sin x\,dx$ ⁠, the integral we started with. Calling it $I$ ⁠, we get an equation for $I$ ⁠, and we simply solve it.

Given, to find, and the plan

Given: $e^x\sin x$ ⁠, defined and continuous for every real $x$ ⁠.
To find: $I = \int e^x\sin x\,dx$ ⁠.
Plan: (1) By parts once ( $e^x$ first, $\sin x$ second), giving $I_1 = \int e^x\cos x\,dx$ ⁠. (2) By parts on $I_1$ with the same kind of choice. (3) Substitute and solve for $I$ ⁠. (4) See why the choice must be kept. (5) 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). Lesson: integration by parts

2Derivatives and integrals that cycle

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

3Antiderivatives differ by constants

Two antiderivatives of the same function differ by a constant (Section 7.2), so an equation between indefinite integrals holds "up to a constant"; the constant is added at the end. Basics: integration undoes differentiation

Solution

Integrate by parts once

Take $e^x$ as the first function and $\sin x$ as the second. Then the derivative of the first is $e^x$ and the integral of the second is $-\cos x$ ⁠:
$\begin{aligned} I = \int e^x\sin x\,dx &= e^x(-\cos x) - \int e^x(-\cos x)\,dx \\ &= -e^x\cos x + \int e^x\cos x\,dx. \end{aligned}$
Write $I_1 = \int e^x\cos x\,dx$ ⁠, so
$I = -e^x\cos x + I_1. \qquad (1)$
Here neither choice makes the integral simpler; it only turns $\sin x$ into $\cos x$ ⁠. That is fine: the point is to do it twice and come back to $I$ ⁠.

$\int\sin x\,dx = -\cos x$ (formula (iii)). Lesson: integration by parts
Integrate $I_1$ by parts, keeping $e^x$ as the first function

Now $e^x$ is first and $\cos x$ second; the integral of $\cos x$ is $\sin x$ ⁠:
$\begin{aligned} I_1 = \int e^x\cos x\,dx &= e^x\sin x - \int e^x\sin x\,dx \\ &= e^x\sin x - I. \end{aligned}$
The last integral is exactly the original $I$ ⁠. That is not a failure; it is what makes the method work, because now $I$ appears on both sides of an equation.

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

Put $I_1 = e^x\sin x - I$ into (1):
$I = -e^x\cos x + e^x\sin x - I.$
Add $I$ to both sides:
$2I = e^x(\sin x - \cos x),$ $I = \int e^x\sin x\,dx = \frac{e^x}{2}(\sin x - \cos x) + C.$
We treat $I$ like an unknown in an ordinary linear equation. Strictly, each $\int$ stands for an antiderivative, determined only up to a constant, so the equation holds up to a constant; that is why one arbitrary constant $C$ is added after solving.

Solving $I = a - I$ gives $I = \frac a2$ ⁠. Basics: integration undoes differentiation
Why the choice must be kept the second time

Suppose in Step 2 we had switched, taking $\cos x$ first and $e^x$ second. Then
$\begin{aligned} I_1 &= \cos x\cdot e^x - \int(-\sin x)\,e^x\,dx \\ &= e^x\cos x + I, \end{aligned}$
and (1) would become $I = -e^x\cos x + e^x\cos x + I$ ⁠, that is $I = I$ ⁠: true, but useless.

Switching the roles in the second round simply undoes the first round. Keep the same type of function (here the exponential) as the first function both times. The book also notes that taking $\sin x$ first and $e^x$ second in both rounds works equally well and gives the same answer.

Integration by parts is the product rule read backwards; applying it forwards and then backwards returns to the start. Lesson: integration by parts
Check: differentiate the answer
$\begin{aligned} &\frac{d}{dx}\left[\frac{e^x}{2}(\sin x - \cos x)\right] \\ &= \frac{e^x}{2}(\sin x - \cos x) + \frac{e^x}{2}(\cos x + \sin x) \\ &= \frac{e^x}{2}\cdot 2\sin x = e^x\sin x. \end{aligned}$
By the product rule: $e^x$ differentiates to itself, and $\sin x - \cos x$ differentiates to $\cos x + \sin x$ ⁠. The cosines cancel and the sines add, giving exactly the integrand for every real $x$ ⁠.

Product rule; $\frac{d}{dx}\sin x = \cos x$ ⁠, $\frac{d}{dx}\cos x = -\sin x$ ⁠. Basics: integration undoes differentiation

Result

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

valid for every real $x$ ⁠. This is the book's answer.

Check it with numbers

At a point. At $x = \frac\pi2$ ⁠: the integrand is $e^{\pi/2}\cdot 1 = 4.8104773809\ldots$ ⁠; the derivative from Step 5 is $\frac{e^{\pi/2}}{2}(1 - 0) + \frac{e^{\pi/2}}{2}(0 + 1)$ ⁠, the same value.

Over an interval. On $[0, \pi]$ ⁠:

\begin{aligned} F(\pi) - F(0) &= \frac{e^\pi}{2}(0 + 1) - \frac12(0 - 1) \\ &= \frac{e^\pi + 1}{2} \approx 12.0703463164 \end{aligned}

Numerical integration of $e^x\sin x$ over $[0, \pi]$ gives $12.0703463164$ to ten decimal places, the same.

Common mistakes

Switching the first and second functions in the second round. You get $I = I$ (Step 4).
Thinking the method has failed when $I$ reappears. Move it to the left side and solve.
Sign slips. $\int\sin x\,dx = -\cos x$ ⁠, so the first round gives $-e^x\cos x + \int e^x\cos x\,dx$ ⁠, with a plus in front of the integral.
Forgetting to divide by 2 after collecting $2I$ ⁠.

What to take away

For $e^{ax}\sin bx$ or $e^{ax}\cos bx$ ⁠: integrate by parts twice with the same choice, then solve the equation for $I$ ⁠.
$\int e^x\sin x\,dx = \frac{e^x}{2}(\sin x - \cos x) + C$ ⁠; similarly $\int e^x\cos x\,dx = \frac{e^x}{2}(\sin x + \cos x) + C$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts	Steps 1, 2, 4	Section 7.6 (Lesson: integration by parts)
$\int\sin x\,dx = -\cos x + C$ ⁠; $\int\cos x\,dx = \sin x + C$	Steps 1, 2	Standard formulae (iii), (ii), Section 7.2
Antiderivatives differ by a constant	Step 3	Section 7.2
Product rule; derivatives of $e^x$ ⁠, $\sin x$ ⁠, $\cos x$	Step 5	Chapter 5

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Example 21: ∫exsin⁡x dx\int e^x\sin x\,dx∫exsinxdx

Basics you need

Solution

Integrate by parts once

Integrate I1I_1I1​ by parts, keeping exe^xex as the first function

Substitute and solve for III

Why the choice must be kept the second time

Check: differentiate the answer

Every fact used, and where it comes from

Example 21: $\int e^x\sin x\,dx$

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

Substitute and solve for $I$