Question

Integrals,

Can you please help me out with these ?

http://screencast.com/t/q7aRmND5gCP

I know they are simple, I just need to remember them

Answer 1

a and b can be split

c looks splitable to, but with an arctangent result

Answer 2

d is just a fancier way to say x^4 ....

Answer 3

which set are you talking about ? there are 3 ets of a b c d

Answer 4

i figured i would just start at random and they just by happenstance lined up in some alphabetical order :/

Answer 5

;s

Answer 6

the first set of course ... starting at the begining tends to be a good place :)

Answer 7

:D

Answer 8

can you see how to split a and b to make them simpler?

Answer 9

a) is 1/x + lnx + c ?

Answer 10

(x+1)/x = 1 + 1/x integrates to: x + lnx + C

Answer 11

i split the fraction addition

Answer 12

whats the int of x/x ? :/

Answer 13

b is similar in that 2=1+1

Answer 14

ah right x/x = 1

Answer 15

\[\frac{x+2}{x+1}\]
\[\frac{x+1+1}{x+1}\]
\[\frac{x+1}{x+1}+\frac{1}{x+1}\]
\[1+\frac{1}{x+1}\]

Answer 16

the only diff is ln(x + 1 )

Answer 17

yes, and for c, i see a split as well; if you know how to integrate the second term:\[\frac{2x+1}{x^2+1}\]
\[\frac{2x}{x^2+1}+\frac{1}{x^2+1}\]

Answer 18

ln|2x| + ln(x^2 +1) + c ?

Answer 19

not ln(2x); think tan^(-1)

Answer 20

but there i a formula Int of f' / f = ln|f| + c

Answer 21

\[y = tan^{-1}(x)\]
\[tan(y) = x\]
\[sec^2(y)y' = 1\]
\[y' = \frac{1}{sec^2(y)}\]
\[y' = \frac{1}{tan^2(y)+1}\]
\[y' = \frac{1}{x^2+1}\]

Answer 22

therefore:\[\int \frac{1}{x^2+1}dx=tan^{-1}(x)+C\]

Answer 23

I see !

Answer 24

and whats 2x/x^2 + 1 ?

Answer 25

the top is the derivative of the bottom ... so ln it up like normal

Answer 26

soo ln|2x| ?

Answer 27

no .. ln |x^2+1|

Answer 28

if it was (x^2 + 1)/(2x) would I be able to apply this formula ?

Answer 29

of course not:

\[[ln(u)]'\to\frac{u'}{u}~\ne~\frac{u}{u'}\]

Answer 30

I see

Answer 31

I solved d x^5/5 + c

Answer 32

yes

Answer 33

sinx + c

Answer 34

2) a)

Answer 35

csc = 1/sin; so set2 is really just: int sin(x) dx

Answer 36

tanx + c next one

Answer 37

sec^2 to tan i agree

Answer 38

and the previews one you were talking about -cosx + c

Answer 39

yes

Answer 40

c has to be an identity right ?

Answer 41

yes; cot^2 + 1 = csc^2 i believe

Answer 42

so -cot + c ?

Answer 43

good

Answer 44

multiply by cos/cos to unveil teh d

Answer 45

hm

Answer 46

then what I do ?

Answer 47

integrate of course :)

Answer 48

I am stuck on how :/

Answer 49

\[\frac{1}{sec+tan}\]
\[\frac{1}{\frac1{cos}+\frac{sin}{cos}}\]
\[\frac{cos}{1+sin}\]
is there any relationship between the top and bottom?

Answer 50

since sine and cosine are related by derivatives; its always best to see if you can apply some rule for it; or run a u-sub for good measure

Answer 51

ln(cosx) + c

Answer 52

not quite ... use the bottom when you ln up

Answer 53

I used u sub

Answer 54

ln(x) to 1/x, not x/1

Answer 55

you u-subbed incorrectly then :)

Answer 56

:( how would you u sub ?

Answer 57

let u = 1 + sin
u' = cos

replace the parts:

u'/u

Answer 58

du/u

Answer 59

thats fine too

Answer 60

so 1/u = lnu ?

Answer 61

good, and what does u equal?

Answer 62

ln(1+ sinx) :D

Answer 63

thats better :) with practice you simply ln it up using the bottom part

Answer 64

How can you intergrate sqrt functions ?

Answer 65

you know power rules; a sqrt is just a ^(1/2) power

Answer 66

add 1, and drop it under

Answer 67

(2(x - 1)^(3/2))/3

Answer 68

or, use u-subs to clear up the distractions

Answer 69

and yes

Answer 70

+C

Answer 71

nc

Answer 72

the next one is the same thing ?

Answer 73

id say multiply by 2/2 to run the next one ... from alot of practice

Answer 74

or let u = 2x+1
du = 2 dx

dx = du/2

Answer 75

c is a chain rule reversal ...

Answer 76

so after sub we have e^u/2 right ?

Answer 77

I know what it is supposed to be but I dont know how to actually find it by intergrating lol

Answer 78

of course

Answer 79

\[e^x\to e^x\]therefore e^{2x+1} has to come from some sort of ke^{2x+1}, so take a derivative:
\[ke^{2x+1}\to 2k~e^{2x+1}=e^{2x+1}\]
when 2k = 1, k = 1/2

Answer 80

ohhhhhh e^(2x+1)/2

Answer 81

\[\int e^{2x+1}dx\]
\[\frac 22\int e^{2x+1}dx\]
\[\frac 12\int 2e^{2x+1}dx\]
\[\frac 12e^{2x+1}\]

Answer 82

c seems a pretty one lol..

Answer 83

its just reversing the chain rule

Answer 84

\[[f(g)]'=f'(g)~g'\]
\[f(g)=\int f'(g)~g'\]

Answer 85

do I have to multiply both parts with dx

Answer 86

it would be best to split the integration into different parts, yes

Answer 87

and since sin and cos are related by derivatives; you should be considering this as chain rule reversal

Answer 88

\[\int (sin)^3~cos\]
let u = sin
du = cos
\[\int (u)^3~u'\]

Answer 89

Ok how did u replace cos^3 ?

Answer 90

consider it a seperate integration and apply the same technique as the first term

Answer 91

let u = cos
u' = -sin

(u^3) (-u')

Answer 92

\[\int(s^3c+sc^3)dx\]
\[\int~s^3c~dx~+~\int~sc^3~dx\]

Answer 93

OK I solved it but its too long to actually type it down :D

Answer 94

oh wait sin^4/4 +cos^4/4 + c

Answer 95

got simplier

Answer 96

(sin^4 - cos^4)/4 + C

Answer 97

ah yea I had a minus too

Answer 98

i got no idea for that last one, at least not without some added thought