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anonymous
 4 years ago
Need help with another intergral. Never really done one like this before. tan^6(3y)sec^2(3y)dy
anonymous
 4 years ago
Need help with another intergral. Never really done one like this before. tan^6(3y)sec^2(3y)dy

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{?}^{?}\tan ^6\left( 3y \right)\sec ^2(3y)dy\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think I need to do u=tan(3y)? Maybe not even close? ,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or u=tanx? So du=sec^2x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Anyone have any clue?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0turing test ..do u knw c++?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yes, that will work, but it's\[u=\tan(3y)\]\[du=3\sec^2(3y)\]remember to use the chain rule all the way through @shiv no, sorry

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1sorry\[u=\tan(3y)\]\[du=3\sec^2(3y)dy\]can't forget the dy! very important

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so would it be \[1/3\int\limits_{?}^{?}\tan(u)^6dy?\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1no,\[u=\tan(3y)\]so after the sub there should be no tangent left

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh... its \[1/3\int\limits_{?}^{?}u^6du\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you are a life saver. Can I just ask you one more quick here

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1sure but I'm headed for dinner so gotta be quick

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ill try. \[\int\limits_{1}^{1/2}t^2 \sin^2(1+1/2)dt sorry its t \to the 2 wont show up though\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Any direction helps. Do i have to set something u ,, then eventually change the limits of integration? I never know when to do that and feel this could be one..

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int t^{2}\sin^2?dt\]sorry, what is the integrand exactly?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the end is sin^2(1+1/t)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that 2 was wrong. you got the rest right. evaluated from 1 to .5

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int t^{2}\sin^2(1+\frac1t)dt\]yes?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0evaluated from 1 to 1/2 if that changes anything

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1\[u=1+\frac1t\]you don't have to change the bounds if you turn it back to t at the end, but if you want to evaluate it in terms of u you need to do this:\[u_1=1+\frac1{1}=0\]\[u_2=1+\frac1{1/2}=1\]so those are your new limits (if you want to evaluate in terms of u the differential is\[du=t^{2}dt\]so the integral is\[\int_{0}^{1}\sin udu=\int_{1}^{0}\sin u du\]Ok gotta eat, good luck!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks a lot! I appreicate it
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