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 23

The chain rule for derivatives: $$\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}$$ Now $$f=\int \frac{df}{dx}dx=\int \frac{df}{du}\frac{du}{dx}dx\text{.}$$ On the other hand $$f=\int \frac{df}{du}du\text{.}$$ So $$\int \frac{df}{du}\frac{du}{dx}dx=\int \frac{df}{du}{\partial }_u\text{.}$$ So $$\int \left(\text{JUNK}\right)\frac{du}{dx}dx=\int \left(\text{JUNK}\right)du\text{.}$$ This is THE CHAIN RULE FOR INTEGRALS.

$\int \frac{4x-5}{2x^2-5x+1}dx$ $$u=2x^2-5x+1,\frac{du}{dx}=4x-5\text{.}$$ So $$\begin{array}{ccc}{\displaystyle \int \frac{4x-5}{2x^2-5x+1}dx}& =& {\displaystyle \int \frac{1}{u}\frac{du}{dx}dx=\int \frac{1}{u}du}\\ & =& {\displaystyle \text{ln}\hspace{0.17em}u+c=\text{ln}(2x^2-5x+1)+c\text{.}}\end{array}$$

$\int \frac{\text{tan}\sqrt{x}{\text{sec}}^2\sqrt{x}}{\sqrt{x}}dx$ $$\begin{array}{ccc}{\displaystyle u}& =& {\displaystyle \text{tan}\sqrt{x},}\\ {\displaystyle \frac{du}{dx}}& =& {\displaystyle {\text{sec}}^2\sqrt{x}\frac{1}{2}{x}^{-\frac{1}{2}}}\\ & =& {\displaystyle \frac{1}{2}\frac{{\text{sec}}^2\sqrt{x}}{\sqrt{x}}\text{.}}\end{array}$$ $$\int 2\text{tan}\sqrt{x}\frac{{\text{sec}}^2\sqrt{x}}{2\sqrt{x}}dx=\int 2u\frac{du}{dx}dx=\int 2udu=u^2+c={\text{tan}}^2\sqrt{x}+c\text{.}$$

$$\begin{array}{ccc}{\displaystyle \int x\sqrt{3x-2}dx}& =& {\displaystyle \int \frac{1}{3}3x\sqrt{3x-2}dx}\\ & =& {\displaystyle \int \frac{1}{3}(3x-2+2)\sqrt{3x-2}dx}\\ & =& {\displaystyle \int \frac{1}{3}({(3x-2)}^{\frac{3}{2}}+2{(3x-2)}^{\frac{1}{2}})dx}\\ & =& {\displaystyle \frac{1}{3}(\frac{2}{5}\frac{{(3x-2)}^{\frac{5}{2}}}{3}+2\frac{{(3x-2)}^{\frac{3}{2}}}{3}\frac{2}{3})+c}\\ & =& {\displaystyle \frac{2}{45}{(3x-2)}^{\frac{5}{2}}+\frac{4}{27}{(3x-2)}^{\frac{3}{2}}+c\text{.}}\end{array}$$

$\int x\sqrt{x^2-1}dx$ $$u=x^2-1,\frac{du}{dx}=2x\text{.}$$ $$\begin{array}{ccc}{\displaystyle \int x\sqrt{x^2-1}dx}& =& {\displaystyle \int \frac{1}{2}2x\sqrt{x^2-1}dx}\\ & =& {\displaystyle \int \frac{1}{2}\frac{du}{dx}\sqrt{u}dx}\\ & =& {\displaystyle \int \frac{1}{2}\sqrt{u}du}\\ & =& {\displaystyle \int \frac{1}{2}{u}^{\frac{1}{2}}du}\\ & =& {\displaystyle \frac{1}{2}\frac{2}{3}{u}^{\frac{3}{2}}+c}\\ & =& {\displaystyle \frac{1}{3}{u}^{\frac{3}{2}}+c}\\ & =& {\displaystyle \frac{1}{3}{(x^2-1)}^{\frac{3}{2}}+c\text{.}}\end{array}$$

$$\begin{array}{ccc}{\displaystyle \int {\text{cos}}^{3}xdx}& =& {\displaystyle \int \text{cos}\hspace{0.17em}x\hspace{0.17em}{\text{cos}}^{2}xdx}\\ & =& {\displaystyle \int \text{cos}\hspace{0.17em}x(1-{\text{sin}}^{2}x)dx}\\ & =& {\displaystyle \int (\text{cos}\hspace{0.17em}x-{\text{sin}}^{2}x\hspace{0.17em}\text{cos}\hspace{0.17em}x)dx}\\ & =& {\displaystyle \text{sin}\hspace{0.17em}x-\frac{{\text{sin}}^{3}x}{3}+c\text{.}}\end{array}$$

$\int \frac{\text{ln}\hspace{0.17em}{x}^2}{x}dx$ $$u=\text{ln}\hspace{0.17em}{x}^2,\frac{du}{dx}=\frac{1}{x^2}2x=\frac{2}{x}\text{.}$$ $$\begin{array}{ccc}{\displaystyle \int \frac{1}{2}2\frac{\text{ln}\hspace{0.17em}{x}^2}{x}dx}& =& {\displaystyle \int \frac{1}{2}\frac{2}{x}\text{ln}\hspace{0.17em}{x}^{2}dx}\\ & =& {\displaystyle \int \frac{1}{2}\frac{du}{dx}udx}\\ & =& {\displaystyle \int \frac{1}{2}udu}\\ & =& {\displaystyle \frac{1}{2}\frac{u^2}{2}+c}\\ & =& {\displaystyle \frac{u^2}{4}+c}\\ & =& {\displaystyle \frac{{\left(\text{ln}\hspace{0.17em}{x}^2\right)}^2}{4}+c\text{.}}\end{array}$$

$$\begin{array}{ccc}{\displaystyle \int \frac{x}{\sqrt{1+x}}dx}& =& {\displaystyle \int \frac{x+1-1}{\sqrt{1+x}}dx}\\ & =& {\displaystyle \int (\frac{x+1}{{(1+x)}^{\frac{1}{2}}}-\frac{1}{{(1+x)}^{\frac{1}{2}}})dx}\\ & =& {\displaystyle \int ({(x+1)}^{\frac{1}{2}}-{(x+1)}^{-\frac{1}{2}})dx}\\ & =& {\displaystyle \frac{2}{3}{(x+1)}^{\frac{3}{2}}-2{(x+1)}^{\frac{1}{2}}+c\text{.}}\end{array}$$

$$\begin{array}{ccc}{\displaystyle \int x\sqrt{x-1}dx}& =& {\displaystyle \int (x-1+1)\sqrt{x-1}dx}\\ & =& {\displaystyle \int ((x-1)\sqrt{x-1}+\sqrt{x-1})dx}\\ & =& {\displaystyle \int {(x-1)}^{\frac{3}{2}}+{(x-1)}^{\frac{1}{2}}dx}\\ & =& {\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+\frac{2}{3}{(x-1)}^{\frac{3}{2}}+c\text{.}}\end{array}$$

$$\begin{array}{ccc}{\displaystyle \int (1-x)\sqrt{1+x}dx}& =& {\displaystyle \int (-(1+x)+2)\sqrt{1+x}dx}\\ & =& {\displaystyle \int -(1+x)\sqrt{1+x}2\sqrt{1+x}dx}\\ & =& {\displaystyle \int (-{(1+x)}^{\frac{3}{2}}+2{(1+x)}^{\frac{1}{2}})dx}\\ & =& {\displaystyle -\frac{2}{5}{(1+x)}^{\frac{5}{2}}+2·\frac{2}{3}{(1+x)}^{\frac{3}{2}}+c}\\ & =& {\displaystyle -\frac{2}{5}{(1+x)}^{\frac{5}{2}}+\frac{4}{3}{(1+x)}^{\frac{3}{2}}+c\text{.}}\end{array}$$

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

Notes and References

These are a typed copy of Lecture 23 from a series of handwritten lecture notes for the class MATH 221 given on November 1, 2000.

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