## karisos 2 years ago Find the derivative: d/dt (0to Sin t)∫[(1/(4-u^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.

1. karisos

$\frac{ d }{dt }\int\limits_{0}^{\sin t } \frac{ 1 }{ 4-u^2} du$

2. ZeHanz

First, try the integral:$\int\limits_{}^{}\frac{ 1 }{ 4-u^2 }du = \int\limits_{}^{}\frac{ 1 }{ 4 }\frac{ 1 }{ 1-\frac{ u^2 }{ 4 } }du=\frac{ 1 }{ 4 }\int\limits_{}^{}\frac{ 1 }{ 1-\left( \frac{ u }{ 2 } \right)^2 }du$Substitute v = u/2 and du = 2dv:$\frac{ 1 }{ 4 }\int\limits_{}^{}\frac{ 2 }{ 1-v^2 }dv=\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ 1-v^2 }dv$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:$\frac{ 1 }{ 2 }\int\limits_{0}^{\frac{ 1 }{ 2 }\sin t}\frac{ 1 }{ 1-v^2 }dv$The integrand is the derivative of arctanh(v). This can also be written as $arctanh(v) =\frac{ 1 }{ 2 }\ln \frac{ 1+v }{ 1-v }$So now we have$\frac{ 1 }{ 2 }\int\limits\limits_{0}^{\frac{ 1 }{ 2 }\sin t}\frac{ 1 }{ 1-v^2 }dv =\frac{ 1 }{ 2 }\frac{ 1 }{ 2 }\left[ \ln \frac{ 1+v }{ 1-v } \right]_{0}^{\frac{ 1 }{ 2 }\sin t}$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

3. karisos

LOLOL thank you!! Looks good =) much appreciated!

4. ZeHanz

Thanks also! I'm learning myself from solving these crazy problems!

5. karisos

So its a win-win situation...AWESOME!