A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 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
 4 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.

This Question is Closed

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

ZeHanz
 4 years ago
Best ResponseYou've already chosen the best response.1First, 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
 4 years ago
Best ResponseYou've already chosen the best response.0LOLOL thank you!! Looks good =) much appreciated!

ZeHanz
 4 years ago
Best ResponseYou've already chosen the best response.1Thanks also! I'm learning myself from solving these crazy problems!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So its a winwin situation...AWESOME!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.