integrate sin^2x / sqrt x dx
We'll replace (sin x)^2 = (1-cos 2x)/2 Int (1-cos 2x)dx/2sqrt x = Int dx/2sqrt x - Int cos 2x dx/2sqrt x Int (1-cos 2x)dx/2sqrt x = sqrt x - Int cos 2x dx/2sqrt x We'll integrate by parts Int cos 2x dx/2sqrt x using formula Int udv = u*v - Int vdu
that still won't work http://www.numberempire.com/integralcalculator.php?function=cos%282x%29%2F%282*sqrt%28x%29%29&var=x&answers=&__utma=164170016.165112442.1306893325.1311847678.1311849610.5&__utmz=164170016.1311849611.5.4.utmcsr%3Dgoogle |utmccn%3D%28organic%29|utmcmd%3Dorganic|utmctr%3Dintegral+calculator&screen_height=768&__utmc=164170016&__utmb=164170016.3.10.1311849611
This seems to be one of those u keep getting an integral that is as difficult as the one u started with. Let me look around, see if there is some tricky way around it.
ok :(
i am having a test and this is a hint
Normally u would do integration by parts twice and then use the result of the second integration to subback which I am just having a look at now but seems I have to go at least 3 times....:-(
:(
I can't be doing all the calculations (doing my head in) but if u use georgia's substitution and integrate by parts twice u will go from Cos 2x to Sin 2x and back to Cos 2x so u should be able to substitute from the first to the last for an answer (I think) Else u will have to wait for the calc wiz's to arrive, sorry...
im telling you there is no solution using elementary functions, the anti-derivative involves the error-function
Strange they would give such a thing in a test, though...
i hope not... he said::: could be in the test
Multiple choice? D-no solution...:-)
jajajaja
hmm i dunno...you have a point though is there more to this problem?
what u mean?
I am going to have to try what I said, I hate number crunching. Where's satellite, he loves this sort of stuff...
is it a definite integral?
i read to solve this problem they use rengels,, something like that
yes
from 0 to pi
oh ok, then you can evaluate it using numerical methods. Even graph it in a graphing calculator then estimate area from 0 to pi
Like 1 and a bit...
yeah...i get 1.34
dumbcow looks right, the problem is the powers of x go up so there is no exact substitution that I can see. I don't know why they bother with all this stuff anyway, real world integrals never work like the ones in the books...:-) They are nearly always have to be solved numerically. Must be for pure math people.
:(
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