## MATH 221

Last updated: 18 August 2014

## Lecture 22

### Integrals

$f⟶ \frac{df}{dx} ⟶dfdx dfdx⟶ \int \phantom{\rule{1em}{0ex}}dx ⟶f$ The integral with respect to $x$ undoes the derivative with respect to $x\text{.}$ $∫5x4dx=x5, sincedx5dx =5x4.$ ALSO $∫5x4dx= x5+2,since d(x5+2)dx =5x4.$ This is like $\sqrt{9}=3$ and $\sqrt{9}=-3\text{.}$

We usually write $∫5x4dx=x5+c,$ where $c$ is a constant, so that we can write all possible answers to $\int 5{x}^{4}dx$ at once. $∫x2dx = x33 +c,sinced(x33)dx =133x2=x2. ∫x3dx = x44+c,since d(x44)dx =144x3=x3. ∫x4dx = x55+c,since d(x55)dx =155x4=x4. ∫x5dx = x66+c,since d(x66)dx =166x5=x5. ∫x642dx = x643643+c,since d(x643643)dx =1643643x642=x642. ∫x-2dx = x-1-1+c,since d(x-1-1)dx =1-1(-1)x-2 =x-2. ∫x-3dx = x-2-2+c,since d(x-2-2)dx =1(-2)(-2)x-3 =x-3. ∫x-642dx = x-641-641+c,since d(x-641-641)dx =1-641(-641)x-642 =x-642. ∫dx = ∫1dx= ∫x0dx= x11+c,since dxdx=1. ∫x-1dx = ∫1xdx= ln x+c,since dln xdx= 1x.$ Note: $\int {x}^{-1}dx$ is NOT $\frac{{x}^{0}}{0}+c\text{;}$ $\frac{{x}^{0}}{0}$ DOESN'T MAKE SENSE.

$∫xdx= ∫x12dx= x3232+c= 23x32+c,$ since $d23x32dx= 2332x12=x12.$

$∫x3dx=∫x13 dx=x4343+c =34x43+c,$ since $d34x43dx= 34·43x43-1 =x13.$

$∫x362431dx= x362431+1 362431+1 +c= x793431 793431 +c= 431793x793431+c,$ since $d431793x793431dx= 431793793431x793431-1 =x362431.$

$∫exdx = ex+c,since dexdx=ex. ∫e10xdx = e10x10+c,since d(e10x10)dx =110e10x·10=e10x. ∫e-31xdx = e-31x-31+c, since d(e-31x-31)dx= 1-31e-31x (-31)=e-31x.$

$∫exdx= ∫(eln 2)xdx= ∫exln 2dx= exln 2ln 2+c= 2xln 2+c,$ since $d2xln 2dx= dexln 2ln 2dx= 1ln 2exln 2·ln 2= exln 2=2x.$

$∫38xdx= ∫(eln 38)xdx= ∫exln 38dx= exln 38ln 38+c,$ since $dexln 38ln 38dx= 1ln 38exln 38·ln 38= exln 38=38x.$

$∫sin xdx = -cos x+c,since d(-cos x)dx= -(-sin x)=sin x. ∫cos xdx = sin x+c,since dsin xdx=cos x. ∫sec2xdx = tan x+c,since dtan xdx= sec2x. ∫csc2xdx = -cot x+c,since d(-cot x)dx =-(-csc2x)= csc2x. ∫tan x sec xdx = sec x+c,since dsec xdx= tan x sec x. ∫cot xcsc xdx = -csc x+c,since d(-csc x)dx= -(-cot x csc x)= cot x csc x.$

$∫cos xsin2xdx = ∫cos xsin x 1sin xdx = ∫cot xcsc xdx = -csc x+c.$

$∫11+cos xdx = ∫1(1+cos x) (1-cos x)(1-cos x)dx = ∫1-cos x1-cos2xdx = ∫1-cos xsin2xdx = ∫ ( 1sin2x- cos xsin2x ) dx = ∫ ( csc2x- cot x csc x ) dx = -cot x+csc x+c.$

## Notes and References

These are a typed copy of Lecture 22 from a series of handwritten lecture notes for the class MATH 221 given on October 30, 2000.