integration (1 Viewer)

haboozin

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2 quick questions:

use substitution u = 1/x and evaluate:

ln(x)/(1 +x^2)


second question

evaluate:
x* squareroot(6-x) dx

for the second one i tried using a subsitution like
x = 6sin^2@ so i could get rid of the square root and i can get an answer but its in cos@ and sin@ and it becomes pretty messy. I put it in an online integral and it comes with a much nicer answer without the sin's and cos's.

how would you do this question? btw.. they didnt give u 6sin^2@ as a substitution.

also,
anyone has msn?
i dont like making new threads all the time for small questions like this.
durry_power@hotmail.com
 
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2.
I= integral of x((6-x)^.5) dx

Let u=6-x
-du=dx
So I= Integral of –(6-u)(u^.5) du
= integral of u^1.5-6u^.5
Etc.
 
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It seems like an odd substitution; I tried integrating by parts but it breaks down after the second integration, you get LHS=0=RHS
 

m_isk

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the answer would still be zero if it was an indefinite integral, right?
and is the second question Int[x times sqrt (6-x)] dx??
 
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m_isk

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So if the integral was indefinite, as it seems to be, what then would the answer be?
 

FinalFantasy

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int. ln(x)/(1 +x^2)
let x=tan@, dx=sec²@ d@
I=int. ln tanxsec²@ d@\sec²@=int. ln tanx d@
let u=ln tanx and dv\dx=1
du\dx=sec²x\tanx=1\cosxsinx=2cosec2x and v=x

.: I=xln(tanx)-2int.xcosec2x dx
den integrate 2xcosec 2x dx by parts

but dats not using the substitution o.o
 

haboozin

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FinalFantasy said:
int. ln(x)/(1 +x^2)
let x=tan@, dx=sec²@ d@
I=int. ln tanxsec²@ d@\sec²@=int. ln tanx d@
let u=ln tanx and dv\dx=1
du\dx=sec²x\tanx=1\cosxsinx=2cosec2x and v=x

.: I=xln(tanx)-2int.xcosec2x dx
den integrate 2xcosec 2x dx by parts

but dats not using the substitution o.o


yea you must use the substitution..
 

haboozin

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FinalFantasy said:
do u have the answer?
yea what slide rule said is right...

basicly the question came with a definite integral..
and it said to prove it equals 0.
but i didnt put the definite integral up and only said to evaluate...
 

haboozin

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FinalFantasy said:
oh ok.............

yea i dont really know what he is going on about though...

i think it might be the formatting ..
 

haboozin

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FinalFantasy said:
hey, so the question had limits root3 and 1\root3?

indeed..


btw feel free to look at my conics thread and solve Q4
 

FinalFantasy

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ooer, the limits!
hehe , those are quite important in some integration q's, just the integral or the integral w\ limits make a lot of difference and save a lot of time doing it:p

anyway, didn't shafqat answer ur conics?
 

haboozin

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FinalFantasy said:
ooer, the limits!
hehe , those are quite important in some integration q's, just the integral or the integral w\ limits make a lot of difference and save a lot of time doing it:p

anyway, didn't shafqat answer ur conics?


no, question 4 is new, i just put it up..
 

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