Question

easy , but i dont know.....

Answer 1

The ok...

Answer 2

What is the question?

Answer 3

a)\[\int\limits_{}^{}\sec3xtan3xdx\]
b)\[\int\limits_{}^{}(x+1)\sqrt{3x+2}\]
c)\[\int\limits_{}^{}\frac{ e ^{x} -e ^{-x}}{ e ^{x} +e ^{-x}}dx\]

Answer 4

help me i by 1 please.....

Answer 5

these are the questions ,not answers

Answer 6

The first is possibly the easiest of the lot, just let u = 3x and work from there :)

Answer 7

First problem :)
let u = 3x
dx = (1/3)du
\[\Large \frac13\int\sec(u)\tan(u)du\]

Answer 8

Oh,I would suggest using Wolfram Alpha, sign up for a free trial, and then you can use the extended keyboard to solve, sorry I am not being much help, I am just starting Pre Calculus

Answer 9

And then, don't forget that
\[\huge \frac{d}{dt}\sec(t)=\sec(t)\tan(t)\]

Answer 10

wait a sec , i wanna try

Answer 11

\[\frac{ 1 }{ 3 }\sec3x d3x\]

Answer 12

u mean i will be like this ?

Answer 13

it*

Answer 14

No... I mean this
\[\Large \frac13\int\sec(u)\tan(u)du\]and just integrate :)

Answer 15

\[\frac{ 1 }{ 3 }\sec u \]

Answer 16

thisss ???

Answer 17

+ C by the way :)
Now just replace u with what you started with ^.^

Answer 18

\[\frac{ 1 }{ 3 } \sec 3x + c \] this is the answer ?

Answer 19

That's right :)
And you can try differentiating it, just to be sure ^.^

Answer 20

ok , what about the nxt quest ? could u help me too?

Answer 21

I'll do one of the other two questions for you, but the last one, I think it's gonna be best if you try it yourself :)
So, which one of the two questions do you wish me to demonstrate for you?

Answer 22

hmm

Answer 23

ok..i choose c)

Answer 24

Oh, surely :)
\[\huge \int\frac{e^x+e^{-x}}{e^x-e^{-x}}dx\]

Answer 25

Now, let
\[\Large u=e^x-e^{-x}\]then\[\Large \frac{du}{dx}=e^x+e^{-x}\]\[\Large du=(e^x+e^{-x})dx\]

Answer 26

\[\Large dx=\frac{du}{e^x+e^{-x}}\]

Answer 27

So, substituting, we get
\[\huge \int \frac{e^x+e^{-x}}{u}\left(\frac{du}{e^{x}+e^{-x}}\right)\]

Answer 28

Cancelling...
\[\huge \int \frac{\cancel{e^x+e^{-x}}}{u}\left(\frac{du}{\cancel{e^{x}+e^{-x}}}\right)=\int\frac{du}{u}\]

Answer 29

Finally
\[\huge = \ln(u)+C=\ln(e^x-e^{-x})+C\]And we're done. Any questions?

Answer 30

ermmm why \[\frac{ e^{x}+e ^{-x} }{ u }\]

Answer 31

because we let
\[\Huge u=e^x-e^{-x}\]

Answer 32

hmm....the actual quest states that \[\int\limits_{}^{}\frac{ e ^{x}-e ^{-x} }{ e ^{x} +e ^{-x}}\]

Answer 33

so why u must be below ?

Answer 34

It shouldn't really make any difference. I must have misread it :)
\[\huge \int\frac{e^x-e^{-x}}{e^x+e^{-x}}dx\]

Answer 35

Okay, so instead let
\[\Large v=e^x+e^{-x}\]so that
\[\Large \frac{dv}{dx}=e^x-e^{-x}\]\[\Large \frac{dv}{e^x-e^{-x}}=dx\]

Answer 36

So substituting...
\[\huge \int \frac{e^x-e^{-x}}{v}\left(\frac{dv}{e^{x}-e^{-x}}\right)\]

Answer 37

Cancelling
\[\huge \int \frac{\cancel{e^x-e^{-x}}}{v}\left(\frac{dv}{\cancel{e^{x}-e^{-x}}}\right)=\int\frac{dv}{v}\]

Answer 38

And finally
\[\huge =\ln(v)+C=\ln(e^x-e^{-x})+C\]

Answer 39

ohhhh i see....just the same concept applied

Answer 40

So, no more questions on that? :)

Answer 41

err...for b) we use the idea of let u ????

Answer 42

Let u = 3x+2 and work from there :)

Answer 43

pls check my answer ....

Answer 44

\[\frac{ 1 }{ 3 }\int\limits_{}^{}(x+1)(\sqrt{u} du\]

Answer 45

With you so far... :)

Answer 46

Try to express (x+1) in terms of u.

Answer 47

\[(x+1)=u-(2x+1)\] what about this ?

Answer 48

But that's not in terms of u, yet :D
There is still an x at the right-side of the equation.

Answer 49

err... how ?

Answer 50

\[\Large u = 3x+2\]\[\Large u+1=3x+3\]\[\Large \frac{u+1}{3}=x+1\]

\[\Huge x+1=\frac13u+\frac13\]

Answer 51

ohhh....ok ok

Answer 52

\[\LARGE \frac{ 1 }{ 3 }\int\limits_{}^{}(x+1)(\sqrt{u}) du=\frac13\int\left(\frac13u+\frac13\right)(u^{\frac12})du \]And from here on in, it's just good old-fashioned power rule ^.^

Answer 53

\[\frac{ 1 }{ 3 } \int\limits_{}^{}(\frac{ 1 }{ 3 } u ^{\frac{ 3 }{ 2 }}+\frac{ 1 }{ 3 }u ^{\frac{ 1 }{ 2 }})du\]

Answer 54

then like this ?

Answer 55

^_^
Yes

And you can even bring out another 1/3, to make things simpler...
\[\Large \frac19\int u^{\frac32}+u^{\frac12}du\]

Much easier now, no?

Answer 56

ok i got it ! well , thanks a lot !!!!!

Answer 57

No problem ^.^

Answer 58

Good thing, too. I'm going to go to sleep now.