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Mathematics 13 Online
OpenStudy (anonymous):

Evaluate the Definite Integral:

OpenStudy (anonymous):

question

OpenStudy (anonymous):

I don't think it showed up. If you post the integral I'll certainly help you with it :)

OpenStudy (anonymous):

Thanks, I'll put it up again, my net is acting up! :)

OpenStudy (anonymous):

\[\int\limits_{0}^{1/2} \sin^{-1} x/\sqrt{1-x^2}\]

OpenStudy (anonymous):

\[\int\limits_{0}^{\Pi/2} \sin (2Pit/T) - \alpha dt\]

OpenStudy (anonymous):

Well I'll tackle at least the first one to start with. I'm doing it on paper though before I waste time to type it out xD Just to make sure I've got it first :D Should be about 5-10 minutes sir!

OpenStudy (anonymous):

No p, take your time! :)

OpenStudy (anonymous):

I'm sorry if I can keep disappearing on you. My net is still acting up. Argh!

OpenStudy (anonymous):

if I keep*

OpenStudy (anonymous):

Alright gonna start inputing the first one i did. Might take a few minutes.

OpenStudy (anonymous):

No problem

OpenStudy (anonymous):

Same here, that's why it's delayed.. xD slow servers i guess xD

OpenStudy (anonymous):

\[\int\limits_{0}^{1/2} \frac{ \sin ^{-1}(x) }{ \sqrt{1-x ^{2}} }dx\] \[u = \sin ^{-1}(x)\] \[du = \frac{ 1 }{ \sqrt{1-x ^{2}} }dx\] \[\frac{ 1 }{ du } = \sqrt{1-x ^{2}}dx\] \[\int\limits_{0}^{1/2} u*du\] \[\frac{ u ^{2} }{ 2 } | [0 \le x \le 1/2]\] \[\frac{ (\sin ^{-1}(x))^{2} }{ 2 } |[0 \le x \le 1/2]\] \[\frac{ (\sin ^{-1}(1/2))^{2} }{ 2 } - \frac{ (\sin ^{-1}(0))^{2} }{ 2 }\] \[\frac{ 1 }{ 2 } [(30)^{2}-(0)^{2}]\] \[= \frac{ 1 }{ 2 } (900) = 450\] And there you have it, pretty simple, cuz u-substitution works out!

OpenStudy (anonymous):

Thanks, but I can't seem to view the answer :(.

OpenStudy (anonymous):

Wait, never mind. I can see it now.

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