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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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

ZeHanzBest ResponseYou've already chosen the best response.1
First, 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
 one year ago

karisosBest ResponseYou've already chosen the best response.1
LOLOL thank you!! Looks good =) much appreciated!
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Thanks also! I'm learning myself from solving these crazy problems!
 one year ago

karisosBest ResponseYou've already chosen the best response.1
So its a winwin situation...AWESOME!
 one year ago
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