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- DanJS

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- DanJS

@katie00

- DanJS

For
\[\int\limits_{3}^{4} \frac{ 2 }{ \sqrt{3x-7} }dx\]
i would just put u = 3x - 7 , du = 3
then you wont have to change the interval of the definite integral

- DanJS

u = 3 x - 7
dy = 3 dx, or dx = du/3

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## More answers

- DanJS

\[\int\limits \frac{ \cos~x }{ \sin^4 x }dx\]
u misstyped a power when substituting...
u = sin x
du = cos x dx
\[\large \int\limits \frac{ 1 }{ u^4 }du = \frac{ -1 }{ (3*u^3) } = \frac{ -1 }{ 3*\sin^3(x) }\]

- DanJS

\[\int\limits (x+1)\sqrt{x-2}~~dx\]
i would just put
u = x -2 ---> x = u + 2
du = dx
\[\large \int\limits_{3}^{6}[u + 3]*u^{1/2}du = \large \int\limits_{3}^{6}[ u^{3/2} + 3u^{1/2} ]du\]
no change in integration interval

- DanJS

All the probs look good though,
doesnt matter how you get to the answer if it is right i guess, just was a couple things i noticed..

- anonymous

no, the method does matter. The prof grades the method the method is more imp than the answer.

- anonymous

so what would d be after i put u=3x-7 and du=3? @DanJS

- DanJS

That is how i remember to do it, just less writing maybe

- DanJS

did they show you probs replacing stuff under a root with a squared term like that t^2

- anonymous

no, not that I rember. But can you complete d through out if you think that's the answer than plz complete your problem so I can see what I did worng

- DanJS

\[\sqrt{t^2 }=\left| x \right|\] ,
yeah was just looking at wha tyou did again

- DanJS

\[\int\limits\limits_{3}^{4} \frac{ 2 }{ \sqrt{3x-7} }dx = \int\limits\limits_{3}^{4}\frac{ 2 }{ \sqrt{u} }*\frac{ du }{ 3 }\]

- DanJS

when u = 3x-7 and du = 3 dx

- DanJS

\[\frac{ 2 }{ 3 } *\int\limits_{3}^{4}u^{-1/2}*du\]

- DanJS

you get
\[(2/3)*2u^{1/2} \] from 3 to 4

- anonymous

I though there wont be a u in the answer

- DanJS

right , put back in the sub, u = 3x - 7
\[\frac{ 4 }{ 3 }\sqrt{3x - 7}\]

- DanJS

evaluated for the interval 3 to 4, looks the same final answer you have

- anonymous

hm okay for the next one that u mentioned. u said I missed a power. can you do that one too plz.

- anonymous

i never get the cos/sin ones anyway so

- DanJS

TO decide where a u-sub may be good for a prob, if you notice things in the function that are derivatives of other things , that is a good hint
remember d/dx (sin x) = cos (x)

- DanJS

so if you let the u= sin, the derivative will be cos, and take out the numerator term with the du,
u = sin(x)
du = cos(x) dx
\[\int\limits \frac{ 1 }{ u^4 }du = \int\limits u^{-4}du \]

- DanJS

= (-1/3) * u^-3
= -1/( 3 * sin^3(x))

- anonymous

okay so in this case what do i plug in at the end? or do i even d that?

- DanJS

is it a def or indef integral?

- anonymous

indef

- DanJS

yeah just simplify the function as much as you can , that is the answer ,

- anonymous

so -1/3(sin^3 x) is my answer

- DanJS

yeah,
\[\huge \frac{ -1 }{ 3\sin^3(x) }\]
you had the process right, you just put a u^3 in the bottom to integrate, shoul dbe a 4

- anonymous

okay last one that u mentioned (F) def 3-6

- anonymous

tysm for pointign out my erros :) I would have never realized it ahah but tysm for everything

- anonymous

lets finish the last one so I can work on my hw for other classes after we are doen with clac

- DanJS

ok, they arent huge errors, you have the right idea pretty much

- DanJS

yeah, you just did that let t^2 sub thing again for the quantity under the root,

- DanJS

let
u = x -2 --->x= u + 2
du = dx

- DanJS

\[\int\limits_{3}^{6}(u + 2 + 1)\sqrt{u}*du\]

- DanJS

multiplies out to the integral of
u ^ (3/2) + 3u^(1/2)

- anonymous

what answer did u got I thin k i did the whole prob wrong

- DanJS

no, yours actually looks good, probably less writing too in this case

- anonymous

so do you want me to keep it the way it is? I am changing my work to what you are showing me here

- DanJS

\[\large \frac{ 2 }{ 5 }u^{5/2}+2u^{3/2}\]
replace u=x-2
evaluated from 3 to 6
=132/5

- DanJS

either way looks fine

- DanJS

show them to your teacher, and see what they have to say

- anonymous

oh okay tysm yay done! alright thanks man. I will send you few next time. thnaks bye tc

- DanJS

you are changing the prob from the domain in x, to domain in u
you can change it to whatever you want, change from x to t^2 , just have to keep eveyuthing consistent and defined

- anonymous

alright sure I will try to do that!

- DanJS

subbing variables by how you change from x-space, to another space (u)
so you can have a simpler function to integrate in another domain

- anonymous

okay will try to keep that in mind :)

- DanJS

you just have to keep your Map from one to the othe rconsistent

- DanJS

k goodluck

- anonymous

okay I will but the next assignmnet is going to be hard I am worried about that one

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