Note 1 - 13/02/2023
TABLE OF CONTENTS
Review integrations technics of one-variable functions
I. Subtitution Rule
\[\int f(g(x))g'(x)~dx = \int f(u)~du ~~\text{with}~ u=g(x)\]Example:
Evaluate the integral: $\mathrm{I} = \displaystyle \int \sin x \sin(\cos x)~dx$
First approach:
$\displaystyle \text{Let}~ u=\cos x \Rightarrow du=-\sin x~dx$
$\displaystyle \Rightarrow \text{I} = - \int \sin u ~du = \cos u + C = \cos(\cos x) + C$
Second approach:
$\displaystyle \frac{d}{dx}(\cos x) = -\sin x \Rightarrow d(\cos x) = -\sin x ~dx$
$\displaystyle \Rightarrow \mathrm{I} = - \int \sin(\cos x) ~d(\cos x) = \cos(\cos x) + C$
II. Integration by Parts
Let $u=f(x)$ and $v=g(x)$ then
\[\int u~dv = uv - \int v~du\]Example:
Evaluate the given integral: $\displaystyle \mathrm{I} = \int \frac{xe^{2x}}{(1+2x)^2} ~dx$
Solution:
$\displaystyle \begin{cases} u = xe^{2x}
dv = \dfrac{dx}{(1+2x)^2} \end{cases} \Rightarrow \begin{cases} du = (1+2x)e^{2x}~dx
\text{choose}~ v=\dfrac{-1}{2(1+2x)} \end{cases}$
$\displaystyle \Rightarrow \mathrm{I} = \frac{-xe^{2x}}{2(1+2x)} + \frac{1}{2} \int e^{2x} ~dx = \frac{-xe^{2x}}{2(1+2x)} + \frac{e^{2x}}{4} + C$
III. Additional methods
Evaluate: $\displaystyle \mathrm{I} = \int \dfrac{ax+b}{Ax^2+Bx+C} dx \ \text{with} \ A \neq 0$
Let $\Delta = B^2-4AC$, we consider three cases:
Case 1: $\Delta = 0$
$\Rightarrow Ax^2+Bx+C=0$ has one real root $\alpha.$
$\displaystyle \therefore \mathrm{I} = \int \frac{ax+b}{A(x-\alpha)^2} ~dx = \int \left[ \frac{E}{(x-\alpha)^2} + \frac{F}{x-\alpha} \right] ~dx$
$\displaystyle \Rightarrow \mathrm{I} = -\frac{E}{x-\alpha} + F\ln \lvert x-\alpha \rvert + C$
($E$ and $F$ are constant values)
Example:
Evaluate the given integral: $\displaystyle \mathrm{I} = \int \frac{x+3}{x^2+2x+1} ~dx$
Solution:
In scratch: $\dfrac{x+3}{x^2+2x+1} = \dfrac{x+3}{(x+1)^2} = \dfrac{A}{(x+1)^2} + \dfrac{B}{x+1} = \dfrac{Bx+A+B}{(x+1)^2}$
Comparing the coefficients: $\begin{cases} B=1
A+B=3 \end{cases} \Rightarrow \begin{cases} A=2
B=1 \end{cases}$So: $\dfrac{x+3}{x^2+2x+1}=\dfrac{2}{(x+1)^2} + \dfrac{1}{x+1}$
$\therefore \displaystyle \mathrm{I} = \int \left[\frac{2}{(x+1)^2} + \frac{1}{x+1}\right] ~dx = -\frac{2}{x+1}+\ln \lvert x+1 \rvert +C$
Case 2: $\Delta > 0$
$\Rightarrow Ax^2+Bx+C=0$ has two real roots $x_1$ and $x_2.$
$\displaystyle \therefore \mathrm{I} = \int \frac{ax+b}{A(x-x_1)(x-x_2)} ~dx = \int \left(\frac{E}{x-x_1} + \frac{F}{x-x_2}\right) ~dx$
$\Rightarrow \mathrm{I} = E\ln \lvert x-x_1 \rvert + F\ln \lvert x-x_2 \rvert + C$
Example:
Evaluate the given integral: $\displaystyle \mathrm{I} = \int_0^1 \frac{2}{2x^2+3x+1} ~dx$
Solution:
In scratch: $\dfrac{2}{2x^2+3x+1} = \dfrac{2}{(x+1)(2x+1)} = \dfrac{A}{x+1} + \dfrac{B}{2x+1}$
$A(2x+1)+B(x+1)=(2A+B)x+A+B$
Comparing the coefficents: $\begin{cases} 2A+B=0
A+B=2 \end{cases} \Rightarrow \begin{cases} A=-2
B=4 \end{cases}$So: $\dfrac{2}{2x^2+3x+1} = \dfrac{4}{2x+1} - \dfrac{2}{x+1}$
$\therefore \displaystyle \mathrm{I} = \int_0^1 \left(\frac{4}{2x+1} - \frac{2}{x+1}\right)~dx = \left[2\ln \lvert 2x+1\rvert - 2\ln \lvert x+1 \rvert \right] \bigg \vert _0^1 = 2\ln\frac{3}{2}$
Case 3: $\Delta < 0$
$\Rightarrow Ax^2+Bx+C=0$ has no real root.
$\displaystyle \therefore \mathrm{I} = \int \left(\frac{E(\text{denominator})’}{Ax^2+Bx+C} + \frac{F}{(x+\alpha)^2+\beta^2} \right) ~dx$
$\Rightarrow \mathrm{I} = E\ln \lvert \text{denominator} \rvert + \dfrac{F}{\beta} \arctan\left(\dfrac{x+\alpha}{\beta}\right) + C$
Example:
Evaluate the given integral: $\displaystyle \mathrm{I} = \int_0^1 \frac{x}{x^2+4x+13} ~dx$
Solution:
\[\begin{align} \mathrm{I} &= \int_0^1 \frac{x}{(x+2)^2+3^2}~dx \\ &= \int_0^1 \frac{x+2}{(x+2)^2+3^2}~dx - \int_0^1 \frac{2}{(x+2)^2+3^2}~dx \\ &= \left.\frac{1}{2} \ln \lvert x^2+4x+13 \rvert \right \vert _0^1 - \left.\frac{2}{3}\arctan\left(\frac{x+2}{3}\right) \right \vert _0^1 \\ &= \frac{1}{2} \ln\frac{18}{13} - \frac{2}{3}\left(\frac{\pi}{4}-\arctan\frac{2}{3}\right) \end{align}\]