Question

Compute the integral of 1/(x*sqrt(x^3-25)) Hint: let u=x^(3/2)

Answer 1

I don't know that your hint gets you there. Who gave the hint?

Answer 2

it is on the review sheet

Answer 3

i know the answer too if that helps. its 2/15 arcsec(x^(3/2)/5)+c

Answer 4

In doing it, they manipulated in a way to use trig method.

Answer 5

Yeah, you have to brush up on your trig method (some variation of tan^2=sec2-1). Using hint x under the square root turns to sq root of u^2-25 and you can do the trig method.

Answer 6

put (x^(3/2)-25)=t^2

Answer 7

use this substitution

Answer 8

I think he wants us to use the trig method. I just keep getting stuck.

I have u=x^(3/2)
du=3/2sqrt(x)dx
2/3du=sqrt(x)dx

I would be able to solve from there but i dont have a sqrt(x)

Answer 9

use my method and multiply denominator and numerator by x^0.5
and put x^(3/2) as t^2+25

Answer 10

He's going to test us on the trig method...

Answer 11

who?

Answer 12

the teacher

Answer 13

oh...but even ur hint uses substitution

Answer 14

right, but isnt there a difference between trig substitution and regular substitution?

Answer 15

if u want to use trig then use sec^(2/3)t=x

Answer 16

got it?

Answer 17

not really. Can you go through the process step by step?

Answer 18

use (25*sec (t))^(4/3)=x
so x^(3/2)=25*sec^2(t)
so dx=25^(4/3)*(4/3)*sec^(1/3)(t)*sec(t)*tan(t)

Answer 19

dx=c*sec^(4/3)(t)*tan(t) dt
c=constant

Answer 20

now solve further i m very slow in typing

Answer 21

Jas, give me a minute. I have the method his instructor is expecting.

Answer 22

OK, using hint
\[u=x ^{3/2}\]Convert this all the way through you get \[u ^{2/3}=x\]

Answer 23

if u will solve my expression then it will become simplified

Answer 24

Yes, but he is in school, he is learning a really helpful method, technique.

Answer 25

Just hold on one minute, let me finish, so the work don't get all muddled.

Answer 26

i have already told similar method before but he wants trig method

Answer 27

k......go ahead

Answer 28

\[2/3 du =\sqrt{x}\]

Answer 29

if u will use x^(3/2)-25=t^2
then it will be more easier

Answer 30

New\[\int\limits_{?}^{?}[(2/3) du]/[(u ^{2/3}\sqrt{u ^{2}-25}\]

Answer 31

what happened to 2/3du=sqrt(x)?

Answer 32

it is 2/3du=sqrt(x)dx

Answer 33

\[\int\limits_{?}^{?}[(2/3) du]/25\sec \theta \tan \theta\]

Answer 34

I guess that should be derivative theta

Answer 35

i mean, what happens to the sqrt(x)?

Answer 36

multiply deno and nume by x
and cha... made mistake in calculation

Answer 37

In the substitution there is an extra \[\sqrt{5 \sec \theta}\] and that is = to sq rt of x. That is swallowed up in dtheta.

Answer 38

OK, I just noticed the square root of x is in bottom. How do I get it on top to be swallowed by dtheta, Jas?

Answer 39

you solve complete expression in u first
it will be 1/u*(......)

Answer 40

(.....)=u^2-25

Answer 41

now put u=sec(t)

Answer 42

One second, there is a fix for this, don't change it too much or uswag would get lost in translation.

Answer 43

otherwise change directly which i already proposed earlier

Answer 44

OK, that is already wiped out uswag. when I bring sq rt x and dx on top, I have an extra sq rt x in bottom, they cancel each other out.

Answer 45

there is no sqrt but there is x

Answer 46

Yes, but that x is converted to u^(2/3). In my conversion I keep u^2=5 sec theta and I wipe out sq root 5 theta or sq rt x.

Answer 47

Uswag, you seem to be following along. Does this make any sense?

Answer 48

k...i was doing in other way

Answer 49

i need to go

Answer 50

Can we start the explanation from 2/3 du/ u^(2/3)(sqrt(u^2-25)..... i got lost in the explanation after that.

so its 2/3 du= sqrt(x)dx
so we have sqrt(x)dx on top.

Answer 51

Thanks Jas, you been a great help.

Answer 52

Thanks Jas

Answer 53

thanks 4 being my fan

Answer 54

\[{[2/(3\sqrt{x})]du}/[\sqrt{x}u ^{2}\sqrt{u ^{2}-25}]\]

Answer 55

\[2/75\int\limits_{?}^{?}d \theta/\sec \theta \tan \theta\]

Answer 56

Now you have a function you can integrate. Once you integrate you have to go back and change to x. These trig techniques are explained nicely in MIT lectures http://www.youtube.com/watch?v=CXKoCMVqM9s&feature=BFa&list=PL590CCC2BC5AF3BC1&index=25

Answer 57

I'll write down some of the substitutions I made\[u^{2}-25=5^{2}\sec ^{2}\theta-5^{2}\]

Answer 58

And that is equal to \[5^{2}\tan ^{2}\theta\]

Answer 59

\[u =5\sec \theta \]

Answer 60

\[\sqrt{u ^{2}-25}=\sqrt{5^{2}\sec ^{2}\theta-5^{2}}=\sqrt{5^{2}\tan ^{2}\theta}=5\tan \theta\]

Answer 61

A lot of info, makes sense?

Answer 62

yeah, after seeing the info plus that it does. Thanks!