anonymous
  • anonymous
in Substitution Rule
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
1 Attachment
myininaya
  • myininaya
\[\int\limits\limits_{0}^{\frac{1}{2}}\frac{\sin^{-1}(x)}{\sqrt{1-x}} dx\]
dumbcow
  • 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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anonymous
  • anonymous
but I don't know how to do it
myininaya
  • myininaya
Do you know integration by parts?
anonymous
  • anonymous
yeah i do but that question is on substitution not in integration by parts
myininaya
  • myininaya
Why can't we do both?
anonymous
  • anonymous
you can do :)
anonymous
  • anonymous
it
myininaya
  • 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
  • anonymous
ok Thank you a lot guys :)
TuringTest
  • TuringTest
one thing I am confused on here, where did we get u^2 ? are you saying arcsin(sin^2u)=u^2 ?
dumbcow
  • dumbcow
yes arcsin(sin u) = u by definition of the inverse function
TuringTest
  • TuringTest
somehow the idea that that extends to arcsin(sin^n(u))=u^n has never been brought to my attention. thanks
anonymous
  • anonymous
could we have just used the substitution:\[u = \sin^{-1}(x)\]
anonymous
  • anonymous
ah disregard, i didnt look at the denominator properly.
myininaya
  • myininaya
yeah i don't know if that is true turing
TuringTest
  • TuringTest
yeah it can't be
TuringTest
  • TuringTest
arcsin(sin(u^n))=u^n arcsin(sin^n(u))=?
myininaya
  • myininaya
i agree with the first equation you have turing
dumbcow
  • 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
  • anonymous
I got
TuringTest
  • TuringTest
Oh man, I thought I was loosing it :/
anonymous
  • anonymous
i got |dw:1327181038255:dw|
anonymous
  • anonymous
but i'm not sure about my answer
TuringTest
  • 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
  • anonymous
I don't know man that question is hard to me :(
myininaya
  • myininaya
is it written correctly on the file you posted?
TuringTest
  • TuringTest
yeah, if there was an x^2 in the denom our lives would be about a million times easier
anonymous
  • anonymous
let me check it again, I may write it wrong
anonymous
  • anonymous
yeah you were right TuringTest I'm really sorry guys
anonymous
  • anonymous
i'm sorry guys there is X^2
TuringTest
  • 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
  • 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
  • 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
  • anonymous
Thank You so much Turing that was very helpful
TuringTest
  • TuringTest
anytime :)

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