Evaluate the integral 1/sqrt[1-x^2], x= 1/2 to sqrt(3)/2
\[\int\limits_{1/2}^{\sqrt{3}/2} 1/\sqrt{1-x^2} d\]
arcsin(x). That's a inverse trig. integral.
Then evaluate, of course :-)
let u=arcsin(x), dv=dx, ,, find du..and v..
ty bmp- i woulda go for trig sub... would be a mess lol-
u would or wouldnt use inverse trig.??
i would if bmp didnt mention that.== it lot easier to use arcsin(x)
@sam30317 You should remember some main integrals: arctan, arcsin, arccos (albeit this is rare), to avoid doing extra work. :-) @iHelp Sure thing, :-). These kind of problem are generally to catch people offguard.
This* kind of problems, damn my "grammer" is bad today.
nvm that quest
rofl, i would thought that was correct grammer--- i HATE ENGLISH =(
I thought i did memorize them!!! Teachers never really give u examples or quiz u on this stuff. But on the test, they;re everywhere! would the answer be pi/6?
no offense but 10 year or English, i still cant learn proper grammar.. it a pain
Americans and Brits dont know their english that well either. Dont worry, its idiotic to us too
Yup, that's correct @sam30317 . @iHelp None taken, I am not a native English speaker, haha.
**Gasp** Omg, really??? Muah ha ha ha!
btw sam, i have faith in u =) dont let us down
lmao!! I'd rather you have faith in your deity and pray for me. I need allll the help i can get
ill pray for us*
Haha, lets do that. Ive been praying for weeks now.
That's right. i have 3 in one day. I WILL RAPE THOSE PAPERS DOWN. amen.
Omg, thats hilarious, lol!!!!! I need to get off here, haha
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