in Substitution Rule

- anonymous

in Substitution Rule

- schrodinger

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- anonymous

##### 1 Attachment

- myininaya

\[\int\limits\limits_{0}^{\frac{1}{2}}\frac{\sin^{-1}(x)}{\sqrt{1-x}} dx\]

- dumbcow

i would make the substitution:
x = sin^2(u)
dx = 2sin*cos du
\[\rightarrow 2\int\limits_{0}^{.5}u^{2} \sin(u) du\]
Then use integration by parts

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## More answers

- anonymous

but I don't know how to do it

- myininaya

Do you know integration by parts?

- anonymous

yeah i do but that question is on substitution not in integration by parts

- myininaya

Why can't we do both?

- anonymous

you can do
:)

- anonymous

it

- myininaya

\[\int\limits_{}^{}u^2 \sin(u) du=u^2(- \cos(u))-\int\limits_{}^{}2u (-\cos(u)) du\]
you will need to do integration by parts one more time

- anonymous

ok
Thank you a lot guys :)

- TuringTest

one thing I am confused on here, where did we get u^2 ?
are you saying arcsin(sin^2u)=u^2 ?

- dumbcow

yes
arcsin(sin u) = u by definition of the inverse function

- TuringTest

somehow the idea that that extends to arcsin(sin^n(u))=u^n has never been brought to my attention.
thanks

- anonymous

could we have just used the substitution:\[u = \sin^{-1}(x)\]

- anonymous

ah disregard, i didnt look at the denominator properly.

- myininaya

yeah i don't know if that is true turing

- TuringTest

yeah it can't be

- TuringTest

arcsin(sin(u^n))=u^n
arcsin(sin^n(u))=?

- myininaya

i agree with the first equation you have turing

- dumbcow

wait now that i look at it again, i think i have it backwards.
sorry everyone
sin^2(arcsin u) = u^2 , thats what i thought it was

- anonymous

I got

- TuringTest

Oh man, I thought I was loosing it :/

- anonymous

i got |dw:1327181038255:dw|

- anonymous

but i'm not sure about my answer

- TuringTest

wolfram disagrees
http://www.wolframalpha.com/input/?i=integral+from+0+to+0.5+%28arcsin%28x%29%2Fsqrt%281-x%29
but I don't trust wolfram anymore, lol

- anonymous

I don't know man that question is hard to me :(

- myininaya

is it written correctly on the file you posted?

- TuringTest

yeah, if there was an x^2 in the denom our lives would be about a million times easier

- anonymous

let me check it again, I may write it wrong

- anonymous

yeah you were right TuringTest
I'm really sorry guys

- anonymous

i'm sorry guys
there is X^2

- TuringTest

\[\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}\]so \[u=\sin^{-1}x\to du=\frac{1}{\sqrt{1-x^2}}dx\]giving\[\int udu\]

- anonymous

yeah that what i did so far , but i wanna know if i'm doing right TuringTest. Can you do until the end of it ?

- TuringTest

\[\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}\]I'm going to be lazy and not use the proper bound notation, just plug numbers in at the end. \[u=\sin^{-1}x\to du=\frac{1}{\sqrt{1-x^2}}\]giving\[\int udu=\frac{u^2}{2}=\frac{1}{2}(\sin^{-1}x)^2|_{0}^{1/2}=\frac{1}{2}[(\sin^{-1}(1/2))^2-(\sin^{-1}(0))^2]\]\[=\frac{1}{2}[(\frac{\pi}{6})^2-0]=\frac{\pi^2}{72}\]

- anonymous

Thank You so much Turing that was very helpful

- TuringTest

anytime :)

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