Differential-Equations · Unit 6 · Video 3 · Interactive Practice

One Division Instead of Two Integrations by Parts: Complexifying the Integral of e^(-x) cos x

IKey Formulas

Formula	Name	What it gives
$e^{-x}\cos x = \operatorname{Re}\, e^{(-1+i)x}$	Complexify the integrand	The real factor $e^{-x}$ scales both parts of $e^{ix} = \cos x + i\sin x$
$\displaystyle\int e^{(-1+i)x}\,dx = \frac{e^{(-1+i)x}}{-1+i}$	Integrate the exponential	Differentiating brings down $-1+i$ , which cancels the denominator
$\dfrac{1}{-1+i} = \dfrac{-1-i}{(-1+i)(-1-i)} = \dfrac{-1-i}{2}$	Divide with the conjugate	The coefficient in the form $a + bi$ : $-\tfrac12 - \tfrac12 i$
$\displaystyle\int e^{-x}\cos x\,dx = \tfrac12 e^{-x}(\sin x - \cos x) + C$	Take the real part	Only $(-1)\cos x$ and $(-i)(i\sin x) = \sin x$ are real

Key Insight: For any $e^{ax}\cos bx$ or $e^{ax}\sin bx$ , integrate $e^{(a+ib)x}$ to get $\dfrac{e^{(a+ib)x}}{a+ib}$ , rewrite $\dfrac{1}{a+ib} = \dfrac{a-ib}{a^2+b^2}$ , and keep the real part (for cosine) or the imaginary part (for sine).

IIThe Integrand Is the Shadow of a Spiral

$e^{(-1+i)x}$ is an arrow of length $e^{-x}$ at angle $x$ ; its real part is exactly $e^{-x}\cos x$ .

x 1.20

IIIOne Division: Multiply by the Conjugate

Multiplying $w$ by its conjugate $\bar w$ gives the real number $|w|^2$ , so $\dfrac{1}{w} = \dfrac{\bar w}{|w|^2}$ .

IVChecking the Answer: Slope of F Equals f

At every $x$ , the slope of the antiderivative $F$ equals the height of the integrand $f$ .

Integrand

VQuiz Questions

Problem 1 · Complexify the Integrand

Given: the integrand $e^{-2x}\cos 3x$ — find the complex exponential whose real part it is.

✅ Correct!

e^{(-2+3i)x} = e^{-2x}e^{3ix} = e^{-2x}(\cos 3x + i\sin 3x)

, and the real factor

e^{-2x}

scales both parts, so the real part is

e^{-2x}\cos 3x

❌ The roles are swapped.

e^{(3-2i)x} = e^{3x}(\cos 2x - i\sin 2x)

has real part

e^{3x}\cos 2x

. The real part of the exponent is the decay rate

-2

; the coefficient of

i

is the frequency

3

❌ The factors must multiply, not add. The real part of

e^{-2x} + e^{3ix}

e^{-2x} + \cos 3x

. The integrand is a product, and

e^{-2x}\cdot e^{3ix} = e^{(-2+3i)x}

❌ The $i$ belongs only to the frequency.

e^{(-2+3)ix} = e^{ix}

has real part

\cos x

. Keep

-2

as the real part of the exponent:

e^{(-2+3i)x}

Show solution

By Euler's definition, $\cos 3x = \operatorname{Re}\, e^{3ix}$ . Multiplying by the real number $e^{-2x}$ scales the real and imaginary parts alike:

e^{-2x}\cos 3x = \operatorname{Re}\big(e^{-2x}(\cos 3x + i\sin 3x)\big) = \operatorname{Re}\big(e^{-2x}e^{3ix}\big) = \operatorname{Re}\, e^{(-2+3i)x}.

The exponent is $a + ib$ with $a = -2$ (decay) and $b = 3$ (frequency).

Problem 2 · The Division Step

Given: the coefficient $\dfrac{1}{2+i}$ — write it in the form $a + bi$ .

✅ Correct!

(2+i)(2-i) = 2^2 + 1^2 = 5

, so

\dfrac{1}{2+i} = \dfrac{2-i}{5}

. Check:

(2+i)\cdot\dfrac{2-i}{5} = \dfrac{5}{5} = 1

❌ A quotient cannot be split term by term.

\tfrac12 + \tfrac1i = \tfrac12 - i

treats

\dfrac{1}{2+i}

\dfrac12 + \dfrac1i

, which is false:

(2+i)(\tfrac12 - i) = 2 - \tfrac32 i \neq 1

. Multiply top and bottom by the conjugate

2 - i

❌ Check the denominator.

(2+i)(2-i) = 4 - i^2 = 4 - (-1) = 5

, not

4 - 1 = 3

: a number times its conjugate is always

a^2 + b^2

❌ Check the sign of the imaginary part. The numerator becomes the conjugate

2 - i

, so the imaginary part is

-\tfrac15

. Check:

(2+i)(\tfrac25 + \tfrac15 i) = \tfrac35 + \tfrac45 i \neq 1

Show solution

Multiply top and bottom by the conjugate of the denominator, $2 - i$ :

\frac{1}{2+i} = \frac{1}{2+i}\cdot\frac{2-i}{2-i} = \frac{2-i}{(2+i)(2-i)} = \frac{2-i}{2^2+1^2} = \frac{2-i}{5} = \frac25 - \frac15 i.

Real part $\tfrac25$ , imaginary part $-\tfrac15$ .

Problem 3 · The Full Method

Given: $\displaystyle\int e^{2x}\cos x\,dx$ — find it by complexifying.

✅ Correct!

\operatorname{Re}\big[\tfrac{2-i}{5}\,e^{2x}(\cos x + i\sin x)\big] = \tfrac15 e^{2x}(2\cos x + \sin x)

, and its derivative is

\tfrac15 e^{2x}(4\cos x + 2\sin x - 2\sin x + \cos x) = e^{2x}\cos x

❌ Watch $(-i)(i)$ .

(-i)(i\sin x) = -i^2\sin x = +\sin x

, because

i^2 = -1

. The two real products are

2\cos x

and

+\sin x

❌ Check the denominator.

(2+i)(2-i) = 2^2 + 1^2 = 5

, so the coefficient is

\tfrac{2-i}{5}

, not

\tfrac{2-i}{3}

❌ The coefficients are swapped. Differentiate to check:

\tfrac15 e^{2x}(2\cos x + 4\sin x - \sin x + 2\cos x) = \tfrac15 e^{2x}(4\cos x + 3\sin x) \neq e^{2x}\cos x

. The real part of

(2 - i)(\cos x + i\sin x)

2\cos x + \sin x

Show solution

Complexify: $e^{2x}\cos x = \operatorname{Re}\, e^{(2+i)x}$ , so

\int e^{2x}\cos x\,dx = \operatorname{Re}\int e^{(2+i)x}\,dx = \operatorname{Re}\,\frac{e^{(2+i)x}}{2+i}.

Divide: $\dfrac{1}{2+i} = \dfrac{2-i}{5}$ .

Real part: the real factors $\tfrac15$ and $e^{2x}$ move outside, leaving

\operatorname{Re}\big[(2-i)(\cos x + i\sin x)\big] = 2\cos x + (-i)(i\sin x) = 2\cos x + \sin x.

\int e^{2x}\cos x\,dx = \tfrac15 e^{2x}(2\cos x + \sin x) + C.

Check: $\dfrac{d}{dx}\big[\tfrac15 e^{2x}(2\cos x + \sin x)\big] = \tfrac15 e^{2x}\big(4\cos x + 2\sin x - 2\sin x + \cos x\big) = e^{2x}\cos x$ ✓

Problem 4 · Sine Takes the Imaginary Part

Given: $\displaystyle\int e^{-x}\sin x\,dx$ , using the same coefficient $\dfrac{1}{-1+i} = \dfrac{-1-i}{2}$ from the video — find the integral.

✅ Correct! The imaginary part of

(-1-i)(\cos x + i\sin x)

-\sin x - \cos x

, and the derivative of

-\tfrac12 e^{-x}(\sin x + \cos x)

e^{-x}\sin x

❌ That is the cosine integral.

\tfrac12 e^{-x}(\sin x - \cos x)

is the real part, the answer for

e^{-x}\cos x

. Since

\sin x = \operatorname{Im}\, e^{ix}

, the sine integral is the imaginary part.

❌ Check the overall sign. Its derivative is

\tfrac12 e^{-x}(-\sin x - \cos x + \cos x - \sin x) = -e^{-x}\sin x

. Both imaginary products,

(-1)(i\sin x)

and

(-i)(\cos x)

, carry a minus sign.

❌ Differentiate to check.

\tfrac12 e^{-x}(\cos x - \sin x)

has derivative

-e^{-x}\cos x

. Take the imaginary part of

(-1-i)(\cos x + i\sin x)

-\sin x - \cos x

Show solution

Since $\sin x = \operatorname{Im}\, e^{ix}$ , we have $e^{-x}\sin x = \operatorname{Im}\, e^{(-1+i)x}$ , so

\int e^{-x}\sin x\,dx = \operatorname{Im}\,\frac{e^{(-1+i)x}}{-1+i} = \operatorname{Im}\Big[\frac{-1-i}{2}\,e^{-x}(\cos x + i\sin x)\Big].

Expand the product: $(-1-i)(\cos x + i\sin x) = (-\cos x + \sin x) + i(-\sin x - \cos x)$ . Keep the imaginary part:

\int e^{-x}\sin x\,dx = -\tfrac12 e^{-x}(\sin x + \cos x) + C.

Check: $\dfrac{d}{dx}\big[-\tfrac12 e^{-x}(\sin x + \cos x)\big] = -\tfrac12 e^{-x}\big(\cos x - \sin x - \sin x - \cos x\big) = e^{-x}\sin x$ ✓

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