anonymous
  • anonymous
Calculus integration problem involving arccosx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{0}^{1/\sqrt{2}}arccosx/\sqrt{1-x^2} dx\]
anonymous
  • anonymous
im not very sure how to handle the arccosx on the top
jango_IN_DTOWN
  • jango_IN_DTOWN
put x=cos y

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More answers

jango_IN_DTOWN
  • jango_IN_DTOWN
@YadielG
jango_IN_DTOWN
  • jango_IN_DTOWN
the limit turns to be \[\int\limits_{\pi/2}^{\pi/4} (-ydy)=\int\limits_{\pi/4}^{\pi/2} ydy =(1/2) ((\pi/2)^2-(\pi/4)^2)\]
jango_IN_DTOWN
  • jango_IN_DTOWN
= \[3\pi^2/32\]
jango_IN_DTOWN
  • jango_IN_DTOWN
hello @YadielG are you checking it?
anonymous
  • anonymous
i solved it
anonymous
  • anonymous
i did the u substitution for arccosx and it left me with \[\int\limits_{0}^{1/\sqrt{2}}-u du\]
anonymous
  • anonymous
then it was simple power rule and plugging in the values from there
jango_IN_DTOWN
  • jango_IN_DTOWN
the limits are wrong
anonymous
  • anonymous
i left them like that because i substituted x back in, I usually don't like converting the limits to fit the u form
jango_IN_DTOWN
  • jango_IN_DTOWN
ok... then you do the indefinite integral first and then convert it to x and then take the limits.. dont put the limit of x when you are doing work of u, or else its a mistake
anonymous
  • anonymous
yes i understand i left them off when i was showing my work but i put them here simply to restate the limits from the original problem, but anyway i appreciate the help
jango_IN_DTOWN
  • jango_IN_DTOWN
ok.. :)
jango_IN_DTOWN
  • jango_IN_DTOWN
do check that both our answers are same
alekos
  • alekos
that checks with me

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