Question

I've been doing pretty good with these Definite Integrals, but this one has me stumped. Can someone help me understand how to work through this? The problem:

g(x) = (integral)[(u^2 - 1)/(u^2 + 1)]du, from 2x to 3x.

Any and all help is greatly appreciated!

Answer 1

The problem:
\[g(x) = \int\limits_{2x}^{3x}(\frac{u^2 - 1}{u^2 + 1})du\]

Answer 2

Does this hint help?

w = u^2 + 1
w-2 = u^2 + 1-2
w-2 = u^2 - 2

Answer 3

hmm now that I think of it, if we derived both sides of the equation with respect to u, then

dw/du = 2u
dw = 2u*du

but there is no u*du term in the integrand

Answer 4

I'm not really sure where to go with your hint @jim_thompson5910 :/

Answer 5

i cheated n checked out the integrand in Wolfram ;)

u can use the substitution u=tan(theta)

Answer 6

that trick will work

Answer 7

u = tan(theta)
du/dtheta = sec^2(theta)
du = sec^2(theta)*dtheta

u^2 + 1 = tan^2(theta) + 1 = sec^2(theta)

so we'll have a pair of "sec^2(theta)" terms cancel

Answer 8

Wouldn't it be inverse tan(x)? As in,
\[\frac{d}{dx}\tan^{-1} x = \frac{1}{1 + x^2}\]

Answer 9

correct @jim_thompson5910 and @amonoconnor

Answer 10

What about those boundaries... Do I need to split it up into two integrals, with 2x as the upper for one and 3x as the upper for another? Can 3x=b and 2x=a and I just use it as is?

Answer 11

well u=tan(theta) so finish the integrand in theta first; then subsitute u back in; last evaluate u=2x and 3x

Answer 12

Can you walk me through this? I'm going to try on my end...

Answer 13

Do I do it like a chain rule? Plugging it in, then multiply the "new" integrand by d/dx[tan(theta)]

Answer 14

promise u wont click till u try to solve the intregrand 1st:

http://www.wolframalpha.com/input/?i=integrate+%28u%5E2++-+1%29%2F%28u%5E2+%2B+1%29du

Answer 15

Deal.

Answer 16

follow the steps in the last post by @jim_thompson5910 and substitute u and du into the integrand

Answer 17

Here's what I'm understanding... tell me if I'm wrong!
\[g(x) = \int\limits_{2x}^{3x}\frac{u^2-1}{u^2+1}*du\]
\[= \int\limits_{2x}^{3x}\frac{u^2+1}{1}*\frac{-1}{u^2+1}*du\]
\[f(x) = \frac{u^2+1}{1}*\frac{-1}{u^2+1}\]
\[F(x) = \frac{1}{\tan(x)}*(-\tan(x))\]

Answer 18

you can't jump from \[\Large g(x) = \int\limits_{2x}^{3x}\frac{u^2-1}{u^2+1}*du\] to \[\Large \int\limits_{2x}^{3x}\frac{u^2+1}{1}*\frac{-1}{u^2+1}*du\]

Answer 19

\[\Large \frac{u^2-1}{u^2+1} \ne \frac{u^2+1}{1}*\frac{-1}{u^2+1}\]

Answer 20

Ugg. Thank you, I'll keep working...

Answer 21

it should go like: u=tan(th), du=sec^2(th)d(th)

so the original integrand becomes: (tan^2(th)-1)/(tan^2(th)+1) sec^2(th)d(th)

sorry im super lazy in typing things out in latex!

Answer 22

Alright... tell me if this is right:)
\[g(x) = \int\limits_{2x}^{3x}\tan^2(x)-1\]
And from there, I evaluate by:
\[F(b) - F(a)\]
???

Answer 23

forget x, g(x) and focus on u and theta first....

Answer 24

plugging in 2x and 3x?

Answer 25

no...forget the x part completely. for now - solve the integrand in u by substitution w/ theta

Answer 26

\[\Large u = \tan(\theta)\]

\[\Large \frac{du}{d\theta} = \sec^2(\theta)\]

\[\Large du = \sec^2(\theta)*d\theta\]


-------------------------------------------------------------------------


\[\Large \int\left(\frac{u^2 - 1}{u^2 + 1}\right)du\]

\[\Large \int\left(\frac{(\tan(\theta))^2 - 1}{(\tan(\theta))^2 + 1}\right)\sec^2(\theta)*d\theta\]

\[\Large \int\left(\frac{\tan^2(\theta) - 1}{\tan^2(\theta) + 1}\right)\sec^2(\theta)*d\theta\]

\[\Large \int\left(\frac{\tan^2(\theta) - 1}{\sec^2(\theta)}\right)\sec^2(\theta)*d\theta\]

\[\Large \int\left(\frac{\tan^2(\theta) - 1}{\cancel{\sec^2(\theta)}}\right)\cancel{\sec^2(\theta)}*d\theta\]

\[\Large \int\left(\tan^2(\theta) - 1\right)d\theta\]

Answer 27

Gotcha... okay, what's after that?

Answer 28

ty @jim_thompson5910

Answer 29

@amonoconnor continue w/ the integration!

Answer 30

use the identity

tan^2(x) + 1 = sec^2(x)
tan^2(x) + 1-1 = sec^2(x)-1
tan^2(x) = sec^2(x) - 1

Answer 31

Why do I need to go back to sec^2(x) stuff, if I have tan^2(theta) - 1 as the only term under the integral?

Answer 32

because tan^2 is a pain to integrate compared to sec^2

Answer 33

y = tan(x)
dy/dx = sec^2(x)

you have to think backwards a bit

Answer 34

Gotcha.... so, from there, I need to find F(x), THEN plug in boundaries?

Answer 35

yes

Answer 36

Alright, I'm doing better... let me work this step!

Answer 37

f(x) = sec^2(theta) - 2
F(x) = tan(theta) - 2(theta)
???

Answer 38

Just checking about that 2.

Answer 39

I agree with the `tan(theta) - 2(theta)` part

Answer 40

now go back to
`u = tan(theta)`
and isolate theta to get
`theta = arctan(u)`

Answer 41

Wait... what?
\[F(x) = \tan(\tan^{-1}(u)) - 2(\tan^{-1}(u))\]
???

Answer 42

I thought I had gotten the antiderivative already? :/

Answer 43

What is tan of arctan(u)?

Answer 44

the limits of integration (2x and 3x) are in terms of u

ie,

u = 2x is the lowest u goes
u = 3x is the highest u goes

which is why I'm resubstituting theta for an expression in terms of u

Answer 45

tan(arctan(u)) = u

Answer 46

they are inverses of each other, so in a sense, they cancel out

Answer 47

I'm so sorry for asking a lot of questions, but I'm just struggling on this one...
So..
2(theta) = 2(tan^-1(u)) = 2(tan^-1(3x)) = ?

Answer 48

I get how tan(tan^-1(u)) = u.

Answer 49

@amonoconnor alright - click on the wolfram link in my earlier post. it gives u the integrand in terms of u.

no more theta to create any more confusion ;)

then substitute u=2x and u=3x to get the ans for g(x). good luck!

Answer 50

you should find that

\[\Large \int\left(\frac{u^2 - 1}{u^2 + 1}\right)du = u-2\arctan(u) + C\]

Answer 51

now plug in the limits of integration

\[\Large \int_{\color{red}{2x}}^{\color{blue}{3x}}\left(\frac{u^2 - 1}{u^2 + 1}\right)du \\
\Large \left. = u-2\arctan(u) + C\right]_{\color{red}{2x}}^{\color{blue}{3x}}\\
\Large = \left[\color{blue}{3x}-2\arctan(\color{blue}{3x}) + C\right]-\left[\color{red}{2x}-2\arctan(\color{red}{2x}) + C\right]\\
\Large =??\]

Answer 52

Or you can think of it as

\[\Large h(u) = u-2\arctan(u) + C\]
and then compute `h(3x) - h(2x)` to finish up

Answer 53

x - 2(arctan(3x) - arctan(2x)) ?

Answer 54

that or \[\Large g(x) = x - 2\arctan(3x) + 2\arctan(2x)\]

Answer 55

Alright, cool.. I'm tired, and will likely look back at this one in the morning and have more enlightenment.. Thank you for all of your patient help:)

Answer 56

special thanx to @jim_thompson5910 - i would have given u 5 medals if i can :)

Answer 57

Ditto! You are both Rockstars! ;)