Question

integrate sec^2(1-4x) please :) i'm doing horrible with integration

Answer 1

do you know what the integral of sec^2(x) w.r.t x is?

Answer 2

isn't it tan x?

Answer 3

\[\int\limits_{}^{}\sec^2(x) dx=? \]

Answer 4

yes tan(x)+c

Answer 5

so just do a sub here
let u=1-4x

Answer 6

should this be (tanx/1-4x)+c?

Answer 7

use the substitution u=1-4x

Answer 8

ok 1 sec
lol sec

Answer 9

-tanx/4 + c?

Answer 10

wait

Answer 11

-tan(1-4x)/4 + c

Answer 12

my bad
is that right?

Answer 13

looks good.

Answer 14

yes ,right

Answer 15

ah man integrating trig functions is tough :( :(

Answer 16

Its not too bad. You just have to practice a bit.

Answer 17

how about \[\int\limits_{}^{?}\frac{ x }{ (4x^2+3 }\]

Answer 18

what is derivative of 4x^2+3 w.r.t x?

Answer 19

cause i have nothing at all for that one. been working on it for like three days. and im too broke to pay for wolfram lol. so thank you very much guys i am super grateful

Answer 20

no,it is easy ,you can use integration by parts,substitution,_________________

Answer 21

4/3(x^3)+c

Answer 22

wait

Answer 23

look if you let u=4x^2+3
isn't du/dx=8x
don't you have an x on top?

Answer 24

4/3(x^3) + 3x + c

Answer 25

yes there is an x on top

Answer 26

but then i end up with x=sqrt(u+3)/4

Answer 27

is it really that sloppy?

Answer 28

\[\frac{1}{8} \int\limits_{}^{}\frac{8x}{4x^2+3} dx\]

Answer 29

you should the derivative of the bottom on top
so this has form \[\frac{1}{8} \int\limits_{}^{}\frac{du}{u}\]

Answer 30

sorry!!! oh my. the (4x+3) is (4x+3)^6

Answer 31

wait (4x^2+3)^6

Answer 32

that x squared, or that 6th power, is defeating me. i could solve it if one of them were missing but alas, there they are

Answer 33

then you have this form which still do -able
\[\frac{1}{8} \int\limits_{}^{}\frac{du}{u^6}\]

Answer 34

ok i will try that right now. one sec

Answer 35

doesn't du=8xdx?

Answer 36

yes

Answer 37

so dx=1/8x du?

Answer 38

what am i supposed to do with that extra x?

Answer 39

there was an x on top already

Answer 40

\[\int\limits_{}^{}\frac{x}{(4x^2+3)^6} dx \]
u=4x^2+3
du=8x dx or du/8=xdx

Answer 41

replace the x dx with du/8

Answer 42

AHHHHH

Answer 43

you just blew my mind!!!

Answer 44

I just divided both sides of du=8x dx by 8

Answer 45

to solve for x dx should be

Answer 46

in terms of u

Answer 47

i know it was so obvious in front of my face thank you so much

Answer 48

this is for a test review, the test is next monday. the questions the professor gave us in the review are much tougher than in the book. there was no example like that in the book, with a fraction with exponent on x and the whole term in the denominator

Answer 49

one sec, trying to solve it now

Answer 50

so it should be 1/40(4x^2+3)^5?

Answer 51

-1/40(4x^2+3)^5

Answer 52

ah thank you guys so much

Answer 53

how about for \[\int\limits_{}^{}\frac{ x^2 }{ x^3+5 }\], should i set u=x^3+5?

Answer 54

yes

Answer 55

k, that seemed wrong to me for some reason, but i will try it now

Answer 56

ah im beginning to see it

Answer 57

i end up with 1/3 int u^-1 du

Answer 58

ok and that equals?

Answer 59

1/3u^0du

Answer 60

so would that be 1/3 (x^3+5)

Answer 61

\[\frac{1}{3} \int\limits_{}^{}u^{-1} du \text{ or } \frac{1}{3}\int\limits_{}^{}\frac{1}{u} du \\ =\frac{1}{3}\ln|u|+C\]

Answer 62

omg wow i'm so ashamed i didn't see that

Answer 63

deriv and int of ln u are my faves :(

Answer 64

yeah you can't use the power rule for when the n is -1

Answer 65

yeah it ends up being u^0=1 right

Answer 66

as in
\[\int\limits_{}^{}u^{n} du=\frac{u^{n+1}}{n+1} +C , n \neq -1 \]

Answer 67

well mainly because this fraction I just mentioned isn't defined at n=-1

Answer 68

right

Answer 69

lol

Answer 70

so when n=-1 we have
\[\int\limits_{}^{}u^{-1} du =\ln|u|+c \\ \text{ and not } \frac{u^{0}}{0}+c\]
because fractions aren't defined when their bottom is 0

Answer 71

right, i forgot the n+1 in the denom

Answer 72

what about for \[\int\limits_{}^{}\frac{ 1 }{ 1+(e^-2x) }\], can this be done with u=1+e^-2x?

Answer 73

is that your ques\[\int\limits_{}^{}\frac{ 1 }{ 1+e^{-2x} }\]

Answer 74

yup

Answer 75

sorry, new to this site's format

Answer 76

so can this be done with u sub?

Answer 77

honestly this might require an algebraic manipulation for the algebraic substitution

Answer 78

this might help (try getting rid of that neg exponent on bottom)
multiply top and bottom by e^(2x)

Answer 79

so i end up with e^2x/2?

Answer 80

no

Answer 81

\[\int\limits_{}^{}\frac{1 \cdot (e^{2x})}{1 \cdot (e^{2x})+e^{-2x} \cdot (e^{2x})} dx\]

Answer 82

you can multiply by e^(2x) :\[\frac{ 1 }{ 1+e^{-2x}} to get\frac{ e^{2x} }{ e^{2x}+1 }\],then use u=e^x

Answer 83

i would actually use the sub u=e^(2x)+1

Answer 84

but you can always do two subs too

Answer 85

what if i did not algebraically manipulate it and used u=1+e^-2x? would it still work?

Answer 86

if you differentiate that would you get the thing that is on top? (or some constant multiple times what you have on top)?

Answer 87

no :( ok i see now

Answer 88

is the answer ln|e^2x|?

Answer 89

no wait

Answer 90

du=2e^2xdx right?

Answer 91

so e^2xdx=1/2du

Answer 92

so it should be 1/2(ln|e^2x|)
yes?

Answer 93

\[\int\limits\limits_{}^{}\frac{1}{1+e^{-2x}} dx=\int\limits\limits_{}^{}\frac{e^{2x}}{e^{2x}} \cdot \frac{1}{1+e^{-2x}} dx =\int\limits\limits_{}^{}\frac{e^{2x}}{e^{2x}+1} dx \\ \text{ Let } u=e^{2x}+1 \\ \text{ so } du=2 e^{2x} dx \\ \text{ We see } e^{2x} \text{ on top so we decide to solve } du=2e^{2x} dx \\ \text{ for } e^{2x} dx \\ \text{ So this means we have } \frac{1}{2} du =e^{2x} dx \\ \int\limits\limits_{}^{}\frac{1}{2} \frac{1}{u} du =\frac{1}{2} \int\limits\limits_{}^{}\frac{1}{u} du =\frac{1}{2} \ln|u| +C \]

Answer 94

And what was u again?

Answer 95

You should be able to see that in my second line.

Answer 96

e^2x+1

Answer 97

oops. i forgot the +1

Answer 98

SO MANY THINGS IN ADVANCED CALC ;_;
MY POOR BRAIN