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anonymous
 5 years ago
Someone please help me real quick. Im going to post the question now
anonymous
 5 years ago
Someone please help me real quick. Im going to post the question now

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0given \[\cos(\sqrt{t})/\sqrt{t}\] If l let u = sqt(t) then does that mean it applies to both cos funtion and root funtion, or just the inner most sqrt inside cos????????

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that doesnt help me for taking the indefinite integral though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because u have given new value to sqrt (t) no matter where it is

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u might need to post your question to have opinions on that problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\cos(\sqrt{t})/\sqrt{t}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to consider \[\frac{\mathbb{d}u}{\mathbb{d}x} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was trying to get ride of the root t in the botom with \[dt = 2\sqrt{t} du\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to substitute in u for root t there aswell..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0after substituting z=t^1/2 u give get as below integration sign ( cosz * 2z^2 dt )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then u can use integration by parts keeping z^2 as first term and cosz as second

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me show you what im trying

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let u = \[\sqrt{t}\] then \[du = 1/(2\sqrt{t})dt\] \[dt = 2\sqrt{t}du\] then i get \[\int\limits_{}^{} \cos(u)/u 2\sqrt{t}du\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0change t^1/2 to u because now we cant have t in new eqation, now we are dealing with u only

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u r on right track ..well done

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bang !!! cheers!!! good luck with other problems too !!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0keep up the gud work !!!
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