Math 138 - January 4, 2016 - Lecture 1

Section 7
Prof. Park, MC 5022, bdpark@uwaterloo.ca , Ext. 37016
Office Hours: WF 3:30 - 4:30 pm
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1st assignment due on Friday Jan 15, 2pm

Notation

If F’ = $\frac{dF}{dy}$, then we write $\int \frac{dF}{dy} dx = F + C$ for an arbitrary constant C

Cancel dx’s and obtain:
$\int dF = F + C$
i.e. F = $\int dF$ (up to constant)

Product Rule

We know:
$(fg)' = f'g + fg'$

Integrate both sides:
$\int (fg)' dx = \int f'gdx + \int fg' dx$
$\int fg' dx = \int (fg)' dx- \int f'g dx$
Note: $\int (fg)' dx = fg + C$ so the Cs can be cancelled from the other integral

Integration By Parts Formula
$\int fg' dx = fg - \int gf' dx$

(At first, only g’ part gets integrated. See [section 7.1] for answers)

Easier to remember this if we use differential notation. Let u = f(x) , v = g(x)
$$\begin{equation} \frac{du}{dx} = f'(x) \\\frac{dv}{dx}= g'(x) \\ \implies du = f'(x)dx , dv = g'(x)dx \end{equation}$$

(IP) Memorize!
$\int u dv = uv - \int v dv$

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Ex. Compute: I = $\int x cosx \ dx$
Sol’n. Express $I = \int udv$
Let u = x and dv = cosxdx
$$\begin{equation} \begin{aligned} & ( \frac{du}{dx} = \frac{d}{dx}(x) = 1 \implies)\ du = dx \\ &v = \int dv \\ &= \int cosxdx = sinx + C \\ By \ (IP), \\ I& = uv - \int v du \\ & = x (sinx + C) - \int (sinx + C) dx \\ & = xsinx + Cx - (-cosx + Cx + D) \\ & = xsinx + cosx + E \end{aligned} \end{equation}$$
(where E = -D = constant)

Remark: C's will always cancel. RHS of (IP) looks like:

$$\begin{equation} \begin{aligned} & u(v + c) - \int (v + c) du \\ & = uv + c - \int vdu - \int cdu (which \ is \ cu + D) \\ & = uv + cu - \int vdu - cu - D \\ & = uv + \int vdu - D = uv - \int vdu \end{aligned} \end{equation}$$

Theorem: $\int lnx dx = x lnx - x + C$
Proof: We can differentiate RHS and verify that it is lnx. Alternatively we can use (IP)
Express:
$$\begin{equation} \begin{aligned} & \int lnx dx = \int udv \\ & Let \ u =lnx, \ dv = dx \\ \frac{du}{dx} & = \frac{d}{dx} (lnx) \\ & = \frac{1}{x} \implies du = \frac{1}{x} dx \\ v & = \int dv = \int dx = x (+ c) \\ \int fx & = uv - \int vdu \\ & = (lnx)x = \int x \frac{1}{x}dx \\ & = xlnx - \int fx = xlnx - x + C \ \square \end{aligned} \end{equation}$$

Strategy for picking u

Ex. Compute $I = \int x^2e^xdx$
$$\begin{equation} \begin{aligned} Solution: Set & \ I = \int u dv, Let \ u = x^2, dv = e^x dx \\ Then, & \frac{du}{dx}= 2x \implies du = 2xdx \\ v & = \int dv = \int e^x dx = e^x \\ I & = uv - \int vdu = x^2e^x - \int e^x 2xdx \\ & = x^2e^x - 2 \int x e^xdx (<- \ we \ set \ to \ J) \\ Apply \ (IP) &\ one \ more \ time! \\ Set \ J & = \int udv. Let u = x, dv = e^xdx \\ J & = uv - \int v du \\ & = xe^x - \int e^xdx (<- which \ is \ e^x + C) \\ Thus, \ I & = x^2e^x - 2J \\ &= x^2e^x-2(xe^x - e^x - C) \\ & = e^x (x^2 - 2x + x) + D \end{aligned} \end{equation}$$

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Ex. Compute: $I = \int e^x cosx dx$
$$\begin{equation} Solution: I = \int udv \\ Let \ u = cosx, \ dv = e^xdx \\ \frac{du}{dv} = \frac{d}{dx}cosx = -sinx \implies du = -sinx dx \\ v = \int dv = \int e^x dx = e^x \\ I = uv - \int vdu = e^x cosx - ( - \int e^x sinx dx) (<- J) \end{equation}$$
Apply (IP) once more!
$$\begin{equation} J = \int udv \\ Let \ u = sinx, dv = e^xdx \\ du = cosx dx , v = e^x \\ J = e^x sinx - \int e^x cosx dx (<- Which \ was \ I) \end{equation}$$

Math 138 - January 6 2016 - Lecture 2

Continued from last lecture:

$$\begin{equation} I = e^xcosx + J = e^xcosx + e^xsinx - I \\ I = e^x (cosx + sinx) - I \\ 2I = e^x (cosx + sinx) \\ I = 1/2 e^x (cosx + sinx) + C \end{equation}$$
Remark: Add C at the end to get the most general derivative

FTC (Part 2) says: $$\begin{equation} \int_{a}{b} F'(x)dx = F(x) |a->b = F(b - F(a)\end{equation}$$

Definite Integral by parts

Thm: $\int_{a}{b} udv = uv|a->b - \int_{a}{b} vdu$

Ex: Compute I = $\int_{0}{1} sin^{-1}xdx = \int_{0}{1} arcsinxdx$
Let u = arcsinx , dv = dx $\implies$ v = x
$$\begin{equation} \frac{du}{dx} = \frac{d}{dx} arcsinx = \frac{1}{\sqrt{1 - x^2}} \implies du = \frac{1}{\sqrt{1 - x^2}} dx \end{equation}$$
$I = uv|0->1 - \int_{0}{1} vdu = (arcsinx)x|0->1 - \int_{0}{1} \frac{1}{\sqrt{1- x^2}}xdx$
I = A - B
A = arcsin(1) * 1 - arcsin(0) * 0 = arcsin(1) = $\frac{\pi}{2}$
To compute B, use substitution rule
Let $t = 1-x^2 \ , \ \frac{dt}{dx}= \frac{d}{dx}(1-x^2) = -2x \ \implies \ dt = -2xdx \implies xdx = -\frac{1}{2} dt$
$$\begin{matrix} x = 1 \implies t = 1- 1^2 = 0 \\ x = 0 \implies t = 1 - 0^2 = 1 \end{matrix}$$
$B = \int_{1}{0} \frac{1}{sqrt{t}} (-\frac{1}{2}dt) = -\frac{1}{2} \ int{1}{0} t^{\frac{-1}{2}} dt \\ = -\frac{1}{2} \frac{t^{\frac{1}{2}}}{\frac{1}{2}} | 0->1 = - [t^{\frac{1}{2}}]0->1 = -(0 - 1) = 1$
Hence, I = A - B = $\frac{\pi}{2} - 1$

Iterating integration by parts

Notation:
$$\begin{equation} f^{(o)} = f , f ^{(1)} = f^{'} , f ^{(2)} = f^{''} , ... \int^{(0)} f = f, \int^{(1)} f = \int f dx , \int^{(2)}f = \int (\int f dx )dx), ....\end{equation}$$

Theorem: $$\begin{equation} \int f g dx = \Sigma{k = 0}{\inf} (-1)^{k} f^{(k)} \int^{(k+1)} g \\ = f \int^{(1)} g - f^{'} \int^{(2)} g + f ^{''} \int^{(3)} g - f ^{'''} \int^{(4)} g + ...\end{equation}$$
Remark: If f(x) is a polynomial of degree d, then $f^{(d + 1)} = 0$ hence the sum $\Sigma{k=0}{\inf}$ becomes finite!

Ex: $\int x^2 e^x dx = x^2 e^x - 2x e^x + 2 e^x - 0 e^ + c \\ = e^x ( x^2 -2x + 2) + c$
Ex: $\int x^2 sinx dx = x^x (-cosx) - 2x(-sinx) + 2(- -cosx) - 0 + c \\ (2 - x^2) cosx + 2xsinx + c$

Trig Integrals using substitution rule (sec 7.2)

ex. Compute $I = \int sin^{5}x cos^{2}xdx$
Strategy If the power of sinx is odd, then save one factor of sinx and convert the rest into cosx using $sin^2x = 1 - cos^2 x$ . Then let u = cosx
Solution: $$\begin{equation} I = \int sin^4x \ cos^2x \ sinx dx \\ Let u = cosx . \\ Then \ \frac{du}{dx}=\frac{d (cosx)}{x} = -sinx \\ \implies du = -sinxdx \implies sinxdx = -du \\ I = \int (sin^2x)^2 cos^2x(-du)\end{equation}$$ Set u = cosx . $= \int(1 -u^2)^2u^2(-du) \\ = -\int (1-2u^2 + u^4)u^2du = -\int(u^2-2u^4 + u^6)du \\ = - (\frac{u^3}{3} - 2 \frac{u^5}{5} + \frac{u^7}{7}) + c \\ = - \frac{1}{2}cos^3x + \frac{2}{5} cos^5x - \frac{1}{7}cos^7x + c$
Remark: If you have odd powers of cosx, then use $cos^2x = 1- sin^2x$ and let u = sinx. See p. 481 for more details

Ex. Compute $I = \int_{0}^{\pi}sin^4xdx$
strategy If the powers of sinx and cosx are both even, then you use double angle formulae : $$\begin{equation} cos(2x) = 2cos^2x - 1 = 1- 2sin^2x \end{equation}$$
Solution: From (DA) , $cos^2x = \frac{1 + cos(2x)}{2} , sin^2x = \frac{1 - cos(2x)}{2}$