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5.5)  Integration by substitution

If $F(x) = ln|x^2 - x|$, then by chain rule $F'(x) = {1 \over x^2 - x}(2x 
- 1) = {2x - 1 \over x^2 - x} $


Thus $\int  {2x - 1 \over x^2 - x} dx = $

How do you recognize derivative when chain rule involved?  u -substition

$\int  {2x - 1 \over x^2 - x}dx  = \int {du \over u} = \int {1 \over u}du 
= ln|u| + C =ln| x^2 - x| + C  $

Let $u = x^2 - x$, then $du = 2x - 1$

5.5 Examples.

1.)  $\int 2x e^{x^2} dx$
\vfill

2.)  $\int 3x^2 \sqrt{x^3 + 1} dx$
\vfill

3.)  $\int {x dx \over x^2 + 4}$
\vfill

4.)  $\int x \sqrt{1 + x} dx$
\vfill

5.)  $\int \sqrt{x}(x^2 - 1)dx$
\vfill

6.)  $\int cos^3(x) dx$
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\eject
7.)   $\int_0^\pi cos^3(x) dx =$
 $\int_0^\pi cos^2(x)cos(x) dx =$
 $\int_0^\pi (1 -  sin^2(x))cos(x) dx $

\rightline{ $= \int_0^0 (1 -  u^2)du = 0$}

%%\vskip 5pt

Let $u = sin(x)$, $du = cos(x)dx$,\hfil \break
\line{when $x = 0$, $u = sin(0) = 0$,  
when 
$x = 
\pi$, $u = sin(\pi) = 0$}

Shortcut method:  Use symmetry.

For example:

\line{If $f$ is an odd function $(f(-x) = -f(x))$, then $\int_{-a}^a 
f(x)dx  
= 0$} 
\line{If $f$ is an even function, then $\int_{-a}^a 
f(x)dx  
= 2\int_{0}^a
f(x)dx$} 



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HW 13 (due Thursday 11/30)

%%[5.2) 21, 23, 24 (if you didn't turn in with HW12)]

%%3.10) 20, 22, 26, 32

5.4)  23, 45, 47, 50, 54

5.5)  8, 9, 11, 13, 16, 21, 28, 31, 36, 41, 42, 56, 57, 61, 64, 66, 76 

6.1) 9, 19
and...


HW due Monday 12/4:  Review sheet (include questions/sections you would like reviewed at the 
beginning of the review sheet.  Also include a list of formulas which you would me to 
provide for the final exam).

\end

$(a + b)^3 = (a^2 + 2ab + b^2)(a + b) = a^3 + 3a^2b + 3ab^2 + b^3$

$a^3 + 2a^2b + ab^2 
          a^2b + 2ab^2 + b^3$

\end