Question

∫[(e^-x-e^x)/(e^2x+e^-2x+2)] by substitution

Answer 1

Hah! I see you trying to get one more calculus question in myin. ;p

Answer 2

lol you too

Answer 3

true.

Answer 4

god I'm going to miss learning math

Answer 5

i think we might have to do some algebra for doing a substitution let me see

Answer 6

part of this looks kinda like a sinh(x)

Answer 7

Or rather exactly like it.

Answer 8

right i was thinking of but i forgot that part of calculus (or trig)

Answer 9

\[sinh(x) = \frac{e^x - e^{-x}}{2}\]

Answer 10

Although it's actually easier than that. The numerator is just the derivative of the denominator (missing a factor of 2)

Answer 11

so just use that denominator for the substitution and I think it'll work out. too tired to do it fast enough ;p

Answer 12

wait i think i might see something

Answer 13

\[\int\limits_{}^{}\frac{e^{-2x}-1}{e^{2x}+e^{-2x}+2} e^x dx\]
let u=e^x => du=e^x dx
so u^{2}=e^{2x}
so u^{-2}=e^{-2x}
so we have
\[\int\limits_{}^{}\frac{u^{-2}-1}{u^2-u^{-2}+2} du\]

Answer 14

now let me see if i can work with this

Answer 15

yep

Answer 16

\[\int\limits_{}^{}\frac{u^2}{u^2}\frac{u^{-2}-1}{u^2-u^{-2}+2} du\]

Answer 17

\[\int\limits_{}^{}\frac{1-u^2}{u^4-1+2u^2}du\]

Answer 18

now we may be able to use partial fractions lets see if we can factor the denominator

Answer 19

no we may have to use another subsitution

Answer 20

Yeah, my way didn't work either.

Answer 21

\[-\int\limits_{}^{}\frac{u^2-1}{u^4+2u^2-1}du\]
i'm going to try a trig substituion

Answer 22

\[-\int\limits_{}^{}\frac{u^2-1}{u^4+2u^2+1-2}du=-\int\limits_{}^{}\frac{u^2-1}{(u^2+1)^2-2}du\]

Answer 23

\[\sec(\theta)=\frac{u^2+1}{\sqrt{2}}\]

Answer 24

\[\sqrt{2} \sec(\theta)=u^2+1\]
\[\sqrt{2} \sec(\theta)\tan(\theta) d \theta=2u du\]

Answer 25

\[-\int\limits_{}^{}\frac{\sqrt{2}\sec(\theta)-2}{(\sqrt{2}\sec(\theta))^2-2} du\]

Answer 26

\[-\int\limits\limits_{}^{}\frac{\sqrt{2}\sec(\theta)-2}{(\sqrt{2}\sec(\theta))^2-2} \sqrt{2} \sec(\theta) \tan(\theta) d \theta\]

Answer 27

\[-\int\limits_{}^{}\frac{\sqrt{2} \sec(\theta)-2}{2(\sec^2(\theta)-1)} \sqrt{2} \sec(\theta) \tan(\theta) d \theta\]

Answer 28

\[-\int\limits_{}^{}\frac{\sqrt{2}\sec(\theta)-2}{2\tan^2(\theta)} \sqrt{2}\sec(\theta)\tan(\theta) d \theta\]

Answer 29

\[-\frac{\sqrt{2}}{2}\int\limits_{}^{}\frac{\sqrt{2}\sec(\theta)-2}{\tan(\theta)}\sec(\theta) d \theta\]

Answer 30

\[-\frac{\sqrt{2}}{2}\int\limits_{}^{}\frac{\sqrt{2}\sec^2(\theta)-2\sec(\theta)}{\tan(\theta)} d \theta \]

Answer 31

lol ok goodnight

Answer 32

<3

Answer 33

I just couldn't recall the derivatives off the top of my head.

Answer 34

Had to look em up.

Answer 35

i wonder if i was going a long way or a deadend way

Answer 36

I dunno.

Answer 37

wait i can separate that fraction

Answer 38

\[-\frac{\sqrt{2}}{2}\int\limits\limits_{}^{}\frac{\sqrt{2}\sec^2(\theta)-2\sec(\theta)}{\tan(\theta)} d \theta \]
\[=\frac{-\sqrt{2}}{2}(\sqrt{2}\int\limits_{}^{}\frac{\sec^2(\theta)}{\tan(\theta)} d \theta-2\int\limits_{}^{}\frac{\sec(\theta)}{\tan(\theta)}d \theta)\]

Answer 39

\[y=\tan(\theta)=> dy=\sec^2(\theta) d \theta\]

Answer 40

Hrm.. now I'm sure my answer is right. But I'm too tired to look at it now. I'll come back later.

Answer 41

so looking at \[\int\limits_{}^{}\frac{\sec^2(\theta)}{\tan(\theta)}d \theta=\int\limits_{}^{}\frac{du}{u}=\ln|u|+C=\ln|\tan(\theta)|+C\]

Answer 42

btw guys the answer is 1/(e^x+e^-x)+c

Answer 43

\[\int\limits_{}^{}\frac{\sec(\theta)}{\tan(\theta)} d \theta=\int\limits_{}^{}\frac{\sin(\theta)}{\cos^2(\theta)} d \theta\]
\[h=\cos(\theta)=> dh=-\sin(\theta) d \theta\]
\[-\int\limits_{}^{}\frac{dh}{h^2}=-\frac{h^{-1}}{-1}+K=\frac{1}{h}+K=\frac{1}{\cos(\theta)}+K\]

Answer 44

\[\frac{-\sqrt{2}}{2}(\sqrt{2}(\ln|\tan(\theta)|)-2(\frac{1}{\cos(\theta)}))+constant \]
almost there

Answer 45

remember \[\sec(\theta)=\frac{u^2+1}{\sqrt{2}} \]
\[\cos(\theta)=\frac{\sqrt{2}}{u^2+1}\]

\[\sin(\theta)=\frac{\sqrt{(u^2+1)^2-2}}{u^2+1}\]
\[\tan(\theta)=\frac{\sqrt{(u^2+1)^2-2}}{\sqrt{2}}\]

Answer 46

so now we have
\[\frac{-\sqrt{2}}{2}(\sqrt{2}\ln|\sqrt{\frac{(u^2+1)^2-2}{2}}|-2\frac{u^2+1}{\sqrt{2}})\]

Answer 47

ok but u=e^{x}

Answer 48

\[\frac{-\sqrt{2}}{2}(\sqrt{2}*\frac{1}{2}\ln|\frac{(e^{2x}+1)^2-1}{2}|-\frac{2}{\sqrt{2}}(e^{2x}+1))+c\]

Answer 49

now we need to simplify this

Answer 50

\[-\frac{2}{4}\ln|\frac{(e^{2x}+1)^2-1}{2}|+e^{2x}+1+c\]

Answer 51

\[\frac{-1}{2}(\ln|e^{4x}+2e^2x+1-1|-\ln(2))+e^{2x}+1+c\]

Answer 52

\[\frac{-1}{2}\ln(e^{4x}+2e^{2x})+\frac{\ln(2)}{2}+e^{2x}+1+c\]

Answer 53

\[\frac{-1}{2}\ln(e^{2x}(e^{2x}+2))+\frac{1}{2}\ln(2)+e^{2x}+1+c\]
\[\frac{-1}{2}(lne^{2x}+\ln(e^{2x}+2))+\frac{1}{2}\ln(2)+e^{2x}+1+c\]
\[-\frac{1}{2} \cdot 2 -\frac{1}{2}\ln(e^{2x}+2)+e^{2x}+1+c\]

Answer 54

\[-\frac{1}{2}\ln(e^{2x}+2)+e^{2x}+c\]

Answer 55

Why are you still here! ;p

Answer 56

lol

Answer 57

i seen how to finish mine way lol

Answer 58

wonder if i check this what would i get back

Answer 59

-exp(2*x)/(exp(2*x)+2)+2*exp(2*x)
this is what i get back
i wonder if i can rewrite it as what we started out with

Answer 60

\[\frac{-e^{2x}}{e^{2x}+2}+2e^{2x}=\frac{-e^{2x}+2e^{2x}(e^{2x}+2)}{e^{2x}+2}\]

Answer 61

\[=\frac{2e^{4x}+4e^{2x}-e^{2x}}{e^{2x}+1}=\frac{2e^{4x}+3e^{2x}}{e^{2x}+1}\]
:(

Answer 62

Hrm my answer is not right either.

Answer 63

why would i make that 1 a 2 lol

Answer 64

i probably made a mistake somewhere due to all the substitutions

Answer 65

the more lenthiness=probability of mistakes increases

Answer 66

getting out the whiteboard...

Answer 67

ok i'm gonna do this on paper

Answer 68

Ok, so found 2 mistakes so far in m original

Answer 69

i found some mistakes in mine for sure lol

Answer 70

Ah ha!

Answer 71

i keep making mistakes

Answer 72

i think my brain doesn't work anymore

Answer 73

same.

Answer 74

TBH, the first 5 attempts or so I had the original problem written wrong, that's how bad it is getting.

Answer 75

wait

Answer 76

Oh wait. I have it.

Answer 77

ok polpak i think i got it
we need to use partial fractions

Answer 78

I'll look at yours, but I did mine much easier once I had the problem written right.

Answer 79

It's looking pretty good! Check this out though:
\[sinh(x) = \frac{e^x - e^{-x}}{2}\]\[cosh(x) = \frac{e^x + e^{-x}}{2}\]
\[\implies \int \frac{e^{-x}- e^x}{e^{2x} + e^{-2x} + 2}dx\]\[=\int\frac{-2sinh(x)}{(e^x + e^{-x})^2}dx\]\[=\int\frac{-2sinh(x)}{[2cosh(x)]^2}dx\]\[=-\frac{1}{2}\int\frac{sinh(x)}{cosh^2(x)}dx\]Let \(u = cosh(x) \implies du = sinh(x)dx\)\[\implies -\frac{1}{2}\int\frac{sinh(x)}{cosh^2(x)}dx\]\[=-\frac{1}{2}\int\frac{1}{u^2}du = \frac{1}{2}sech(x) +C\]

Answer 80

Yay we got the same answer!

Answer 81

:)

Answer 82

Nice job!

Answer 83

polpak why did we do this

Answer 84

im so tired

Answer 85

Because after a certain point you're too tired to make good decisions.

Answer 86

lol

Answer 87

And this problem came up at just the critical time where we had made a good decision, then saw this a moment later.

Answer 88

once we were no longer capable.

Answer 89

Night Myin.

Answer 90

night

Answer 91

I mean it this time! Go to sleep ;p

Answer 92

are you sure we got the same answer
i can't check this
how did you they were the same

Answer 93

i mean i can check it but i cant right now

Answer 94

Divide top and bottom of yours by e^x