anonymous
  • anonymous
Need Help! Evaluate the Definite integral with U- substitution.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1439090251436:dw|
ganeshie8
  • ganeshie8
look up the derivative of \(\sin^{-1}x\)
ganeshie8
  • ganeshie8
substitute \(u=\sin^{-1}x\) then \(du=?\)

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anonymous
  • anonymous
I'm confused on how you got U to be sin-1x, how did you know it was that?
anonymous
  • anonymous
I thought U was supposed what's inside roots or paretheses, such as 1-x^2
ganeshie8
  • ganeshie8
do you know what the derivative of \(\sin^{-1}x\) is ?
anonymous
  • anonymous
Yea...it's 1/srt 1-x^2 * dx
ganeshie8
  • ganeshie8
that means substituting \(u = \sin^{-1}x\) simplifies the integrand, because \(du = \dfrac{1}{\sqrt{1-x^2}}dx\) that entire radical mess can be replaced by \(du\)
anonymous
  • anonymous
So I can't always assume that whats inside the parentheses or roots is always going to be substituted as U, right?
ganeshie8
  • ganeshie8
|dw:1439090695448:dw|
anonymous
  • anonymous
I get that!
ganeshie8
  • ganeshie8
No, with integrals there are no "strict" rules, you will have to do some guessing with each and every integral problem
anonymous
  • anonymous
okay thanks!
ganeshie8
  • ganeshie8
Also, don't forget to change the bounds accordingly
ganeshie8
  • ganeshie8
|dw:1439090839359:dw|
anonymous
  • anonymous
Why would I need to change the bounds?
ganeshie8
  • ganeshie8
because with u-substitution you're changing the variable of integration, so the bounds also change accordingly
ganeshie8
  • ganeshie8
|dw:1439090975245:dw|
ganeshie8
  • ganeshie8
those bounds refer to \(x\), they don't belong to \(u\)
anonymous
  • anonymous
|dw:1439091066367:dw|
anonymous
  • anonymous
|dw:1439091093678:dw|
ganeshie8
  • ganeshie8
no wait, whats the integrand right after substituting ?
anonymous
  • anonymous
|dw:1439091171917:dw|
ganeshie8
  • ganeshie8
Yes, whats antiderivative of \(u\) with respective to \(u\) ?
anonymous
  • anonymous
U(x)?
ganeshie8
  • ganeshie8
nope, whats antiderivative of \(x\) with respect to \(x\) ?
anonymous
  • anonymous
ohh....so would it be U^2/2
ganeshie8
  • ganeshie8
Yes
anonymous
  • anonymous
but isnt U supposed to be thought of a constant, therefore it's antiderivative should be U of x?
ganeshie8
  • ganeshie8
\(u\) is not constant, it is the variable that has replaced \(\sin^{-1}x\)
anonymous
  • anonymous
|dw:1439091490507:dw|
ganeshie8
  • ganeshie8
Looks good!
anonymous
  • anonymous
So do I get a decimal answer when I evaluate the integral?
ganeshie8
  • ganeshie8
plugin the bounds and see
anonymous
  • anonymous
Okay so I got (sin-1(1/2))^2/2? So do I box in that as the answer or its decimal...which is .137077
ganeshie8
  • ganeshie8
recall that \(\sin(\pi/6)=1/2\)
Ac3
  • Ac3
look at the directions in the question
anonymous
  • anonymous
The directions in the questions just say evaluate the integral.
Ac3
  • Ac3
if it asks for EXACT answers which most professors want then you give them that
Ac3
  • Ac3
answer should be pi/6
Ac3
  • Ac3
never do decimal unless it specifically tells you to. We know that arcsin of 1/2 is pi/6 because of the unit circle
anonymous
  • anonymous
ganeshie, is the answer (pi/6)^2/2?
Ac3
  • Ac3
my bad pi/12
Ac3
  • Ac3
wow i'm terrible today
anonymous
  • anonymous
hmm i got 5/2pi
ganeshie8
  • ganeshie8
haha thats not it
anonymous
  • anonymous
lol i need help then..
Ac3
  • Ac3
the answer i gave is wrong it should be pi/72 i'll show simplification right now one sec.
ganeshie8
  • ganeshie8
nope thats still wrong
anonymous
  • anonymous
ah i got it pi^2/72
ganeshie8
  • ganeshie8
looks good! :)
anonymous
  • anonymous
Yes thats what I got too saseal!
Ac3
  • Ac3
|dw:1439091910076:dw|
Ac3
  • Ac3
i meant to say pi^2/72 originally
ganeshie8
  • ganeshie8
yes leave it as \(\dfrac{\pi^2}{72}\) do not convert to decimals unless asked to do so
anonymous
  • anonymous
Okay, cool, thanks ganeshie!
Ac3
  • Ac3
Usually with u-substitution the rule of thumb is to try and get something to go away with your Du. That's where knowing you derivatives really well comes in to play. When in doubt just guess and check.
anonymous
  • anonymous
Thanks Ac3!

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