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anonymous
 3 years ago
Find the derivative:
d/dt (0to Sin t)∫[(1/(4u^2)][du]
NOTE: What I mean with o to Sin t is that the 0 is below the integral and sin t is up or above the integral.
anonymous
 3 years ago
Find the derivative: d/dt (0to Sin t)∫[(1/(4u^2)][du] NOTE: What I mean with o to Sin t is that the 0 is below the integral and sin t is up or above the integral.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ d }{dt }\int\limits_{0}^{\sin t } \frac{ 1 }{ 4u^2} du\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0First, try the integral:\[\int\limits_{}^{}\frac{ 1 }{ 4u^2 }du = \int\limits_{}^{}\frac{ 1 }{ 4 }\frac{ 1 }{ 1\frac{ u^2 }{ 4 } }du=\frac{ 1 }{ 4 }\int\limits_{}^{}\frac{ 1 }{ 1\left( \frac{ u }{ 2 } \right)^2 }du\]Substitute v = u/2 and du = 2dv:\[\frac{ 1 }{ 4 }\int\limits_{}^{}\frac{ 2 }{ 1v^2 }dv=\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ 1v^2 }dv\]In trying to keep it clear, I have used an indefinite integral, but actually, the range has changed, due to the v = u/2 substitution. This gives the following integral:\[\frac{ 1 }{ 2 }\int\limits_{0}^{\frac{ 1 }{ 2 }\sin t}\frac{ 1 }{ 1v^2 }dv\]The integrand is the derivative of arctanh(v). This can also be written as \[arctanh(v) =\frac{ 1 }{ 2 }\ln \frac{ 1+v }{ 1v }\]So now we have\[\frac{ 1 }{ 2 }\int\limits\limits_{0}^{\frac{ 1 }{ 2 }\sin t}\frac{ 1 }{ 1v^2 }dv =\frac{ 1 }{ 2 }\frac{ 1 }{ 2 }\left[ \ln \frac{ 1+v }{ 1v } \right]_{0}^{\frac{ 1 }{ 2 }\sin t}\]For v = 0 this give 0, so the result is\[\frac{ 1 }{ 4 }\ln \frac{ 1+\frac{ 1 }{ 2 } \sin t}{ 1\frac{ 1 }{ 2 }\sin t }=\frac{ 1 }{ 4 }\ln \frac{ 2+\sin t }{ 2\sin t }\]Finally, differentiate this function:\[\frac{ d }{ dt }\left( \frac{ 1 }{ 4 }\ln \frac{ 2+\sin t }{ 2\sin t } \right)=\frac{ 1 }{ 4 }\frac{ 1 }{ \frac{ 2+\sin t }{ 2\sin t } }*\frac{ \cos t(2\sin t)+(2+\sin t)\cos t }{ (2\sin t)^2 }=\]\[\frac{ 1 }{ 4 }\frac{ 1 }{ 2+\sin t }\frac{ 4\cos t }{ 2\sin t }=\frac{ \cos t }{ (2+\sin t)(2\sin t) }\]This looks so nice, it must be the right answer! ;) ZeHanz

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LOLOL thank you!! Looks good =) much appreciated!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks also! I'm learning myself from solving these crazy problems!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So its a winwin situation...AWESOME!
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