MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 6 September 2014

Lecture 24

$$\begin{array}{ccc}{\displaystyle \int \frac{\text{sin}\hspace{0.17em}x}{\text{sin}\hspace{0.17em}x-\text{cos}\hspace{0.17em}x}dx}& =& {\displaystyle \int \frac{\text{sin}\hspace{0.17em}x-\text{cos}\hspace{0.17em}x+\text{sin}\hspace{0.17em}x+\text{cos}\hspace{0.17em}x}{\text{sin}\hspace{0.17em}x-\text{cos}\hspace{0.17em}x}·\frac{1}{2}dx}\\ & =& {\displaystyle \frac{1}{2}\int \frac{\text{sin}\hspace{0.17em}x-\text{cos}\hspace{0.17em}x}{\text{sin}\hspace{0.17em}x-\text{cos}\hspace{0.17em}x}+\frac{\text{sin}\hspace{0.17em}x+\text{cos}\hspace{0.17em}x}{\text{sin}\hspace{0.17em}x-\text{cos}\hspace{0.17em}x}dx}\\ & =& {\displaystyle \frac{1}{2}\int (1+\frac{\text{sin}\hspace{0.17em}x+\text{cos}\hspace{0.17em}x}{\text{sin}\hspace{0.17em}x-\text{cos}\hspace{0.17em}x})dx}\\ & =& {\displaystyle \frac{1}{2}(x+\text{ln}(\text{sin}\hspace{0.17em}x-\text{cos}\hspace{0.17em}x))+c}\end{array}$$ since $$\frac{d\frac{1}{2}(x+\text{ln}(\text{sin}\hspace{0.17em}x-\text{cos}\hspace{0.17em}x))}{dx}=\frac{1}{2}(1+\frac{1}{\text{sin}\hspace{0.17em}x-\text{cos}\hspace{0.17em}x}(\text{cos}\hspace{0.17em}x+\text{sin}\hspace{0.17em}x))\text{.}$$

$$\begin{array}{ccc}{\displaystyle \int \frac{x^3}{1+x^8}dx}& =& {\displaystyle \int \frac{x^3}{1+{\left(x^4\right)}^2}dx}\\ & =& {\displaystyle \int \frac{1}{4}·\frac{4x^3}{1+{\left(x^4\right)}^2}dx}\\ & =& {\displaystyle \frac{1}{4}{\text{tan}}^{-1}\left(x^4\right)+c}\end{array}$$ since $$\frac{d\frac{1}{4}{\text{tan}}^{-1}x{x}^4}{dx}=\frac{1}{4}\frac{1}{1+{\left(x^4\right)}^2}\frac{d{x}^4}{dx}=\frac{1}{4}\frac{4x^3}{1+{\left(x^4\right)}^2}\text{.}$$

$$\begin{array}{ccc}{\displaystyle \int {\text{tan}}^{-1}\left(\frac{\text{sin}\hspace{0.17em}2x}{1+\text{cos}\hspace{0.17em}2x}\right)dx}& =& {\displaystyle \int {\text{tan}}^{-1}\left(\frac{2\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}x}{1+{\text{cos}}^{2}x-{\text{sin}}^{2}x}\right)dx}\\ & =& {\displaystyle \int {\text{tan}}^{-1}\left(\frac{2\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}x}{{\text{cos}}^{2}x+{\text{cos}}^{2}x}\right)dx}\\ & =& {\displaystyle \int {\text{tan}}^{-1}\left(\frac{2\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}x}{2{\text{cos}}^{2}x}\right)dx}\\ & =& {\displaystyle \int {\text{tan}}^{-1}\left(\frac{\text{sin}\hspace{0.17em}x}{\text{cos}\hspace{0.17em}x}\right)dx}\\ & =& {\displaystyle \int {\text{tan}}^{-1}\left(\text{tan}\hspace{0.17em}x\right)dx}\\ & =& {\displaystyle \int xdx}\\ & =& {\displaystyle \frac{x^2}{2}+c\text{.}}\end{array}$$

$$\int {\text{cos}}^{-1}\left(\text{sin}\hspace{0.17em}x\right)dx$$ Let $x={\text{sin}}^{-1}y\text{.}$ Then $\frac{dx}{dy}=\frac{1}{\sqrt{1-y^2}}\text{.}$ $$\begin{array}{ccc}{\displaystyle \int {\text{cos}}^{-1}\left(\text{sin}\hspace{0.17em}x\right)dx}& =& {\displaystyle \int {\text{cos}}^{-1}\left(\text{sin}\hspace{0.17em}x\right)\frac{dx}{dy}dy}\\ & =& {\displaystyle \int {\text{cos}}^{-1}\left(\text{sin}\left({\text{sin}}^{-1}y\right)\right)\frac{1}{\sqrt{1-y^2}}dy}\\ & =& {\displaystyle \int {\text{cos}}^{-1}y\frac{1}{\sqrt{1-y^2}}dy}\\ & =& {\displaystyle -\int {\text{cos}}^{-1}y\frac{-1}{\sqrt{1-y^2}}dy}\\ & =& {\displaystyle -\frac{{\left({\text{cos}}^{-1}y\right)}^2}{2}+c}\\ & =& {\displaystyle -\frac{{\left({\text{cos}}^{-1}\left(\text{sin}\hspace{0.17em}x\right)\right)}^2}{2}+c\text{.}}\end{array}$$ Another way: $$\begin{array}{ccc}{\displaystyle {\text{cos}}^{-1}\left(\text{sin}\hspace{0.17em}x\right)}& =& {\displaystyle y}\\ {\displaystyle \text{sin}\hspace{0.17em}x}& =& {\displaystyle \text{cos}\hspace{0.17em}y\text{.}}\end{array}$$ So $$y=\frac{\pi }{2}-x\text{.}$$ So $${\text{cos}}^{-1}\left(\text{sin}\hspace{0.17em}x\right)=\frac{\pi }{2}-x\text{.}$$ So $$\int {\text{cos}}^{-1}\left(\text{sin}\hspace{0.17em}x\right)dx=\int (\frac{\pi }{2}-x)dx=\frac{\pi }{2}x-\frac{x^2}{2}+c\text{.}$$

$$\begin{array}{ccc}{\displaystyle \int \frac{2x^2+x-2}{x-2}dx}& =& {\displaystyle \int \frac{2x(x-2)+5x-2}{x-2}dx}\\ & =& {\displaystyle \int 2x+\frac{5x-2}{x-2}dx}\\ & =& {\displaystyle \int 2x+\frac{5(x-2)+8}{x-2}dx}\\ & =& {\displaystyle \int 2x+5+\frac{8}{x-2}dx}\\ & =& {\displaystyle x^2+5x+8\text{ln}(x-2)+c\text{.}}\end{array}$$

Definite integrals

Warning: The following is not quite correct, though it is correct most of the time. $${\int }_a^b\frac{df}{dx}dx=f\left(b\right)-f\left(a\right)\text{.}$$

$$\begin{array}{ccc}{\displaystyle {\int }_1^2{x}^{-2}dx}& =& {\displaystyle \frac{x^{-1}}{-1}+c{|}_{1=x}^{2=x}}\\ & =& {\displaystyle (-2^{-1}+c)-({-1}^{-1}+c)}\\ & =& {\displaystyle -\frac{1}{2}+c+1-c}\\ & =& {\displaystyle \frac{1}{2}\text{.}}\end{array}$$

$$\begin{array}{ccc}{\displaystyle {\int }_0^4\sqrt{x}dx}& =& {\displaystyle {\int }_0^4{x}^{\frac{1}{2}}dx}\\ & =& {\displaystyle \frac{2}{3}{x}^{\frac{3}{2}}+c{|}_{0=x}^{4=x}}\\ & =& {\displaystyle \frac{2}{3}{4}^{\frac{3}{2}}+c-(\frac{2}{3}{0}^{\frac{3}{2}}+c)}\\ & =& {\displaystyle \frac{2}{3}{2}^3+c-c}\\ & =& {\displaystyle \frac{2}{3}·8}\\ & =& {\displaystyle \frac{16}{3}\text{.}}\end{array}$$

$$\begin{array}{ccc}{\displaystyle {\int }_1^3(\frac{1}{t^2}-\frac{1}{t^4})dt}& =& {\displaystyle {\int }_1^3(t^{-2}-t^{-4})dt}\\ & =& {\displaystyle \frac{t^{-1}}{-1}-\frac{t^{-3}}{-3}+c{|}_{t=1}^{t=3}}\\ & =& {\displaystyle (\frac{3^{-1}}{-1}-\frac{3^{-3}}{-3}+c)-(\frac{1^{-1}}{-1}-\frac{1^{-3}}{-3}+c)}\\ & =& {\displaystyle -\frac{1}{3}+\frac{1}{3^4}+c+1-\frac{1}{3}-c}\\ & =& {\displaystyle 1+\frac{1}{81}-\frac{2}{3}}\\ & =& {\displaystyle \frac{1}{3}+\frac{1}{81}}\\ & =& {\displaystyle \frac{28}{81}\text{.}}\end{array}$$

$$\begin{array}{ccc}{\displaystyle {\int }_{-3}^0(5y^4-6y^2+14)dy}& =& {\displaystyle (y^5-2y^3+14y+c){|}_{y=-3}^{y=0}}\\ & =& {\displaystyle (0-0+0+c)-({(-3)}^5-2{(-3)}^3+14(-3)+c)}\\ & =& {\displaystyle c+3^5-2·3^3+14·3-c}\\ & =& {\displaystyle 243-54+42}\\ & =& {\displaystyle 231\text{.}}\end{array}$$

BUT ${\int }_a^b\frac{df}{dx}dx$ is not always equal to $f\left(b\right)-f\left(a\right)\text{.}$

What does ${\int }_a^{b}f\left(x\right)dx$ really mean??

${\int }_a^{b}f\left(x\right)dx$ really is $$\begin{array}{c}\underset{\mathrm{\Delta }x\to 0}{\text{lim}}(f\left(a\right)\mathrm{\Delta }x+f(a+\mathrm{\Delta }x)\mathrm{\Delta }x+\cdots +f(b-2\mathrm{\Delta }x)\mathrm{\Delta }x+f(b-\mathrm{\Delta }x)\mathrm{\Delta }x)\\ \begin{array}{c} a b x y y=f(x) ⏟ Δx \end{array}\end{array}$$ Think of ${\int }_a^{b}f\left(x\right)dx$ as saying $$\text{Add up the areas}\hspace{0.17em}f\left(x\right)dx\hspace{0.17em}\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}b$$ where $$\begin{array}{c}f\left(x\right)dx\hspace{0.17em}\text{is the "area" of an infinitesimally}\\ \text{small box with height}\hspace{0.17em}f\left(x\right)\hspace{0.17em}\text{and width}\hspace{0.17em}dx\\ \begin{array}{c} } f(x) ⏟dx \end{array}\end{array}$$

Notes and References

These are a typed copy of Lecture 24 from a series of handwritten lecture notes for the class MATH 221 given on November 4, 2000.

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