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

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

Mathematics
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fo u know c++
\[\int\limits_{?}^{?}\tan ^6\left( 3y \right)\sec ^2(3y)dy\]
I think I need to do u=tan(3y)? Maybe not even close? ,

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or u=tanx? So du=sec^2x?
Anyone have any clue?
turing test ..do u knw c++?
yes, 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
sorry\[u=\tan(3y)\]\[du=3\sec^2(3y)dy\]can't forget the dy! very important
so would it be \[1/3\int\limits_{?}^{?}\tan(u)^6dy?\]
no,\[u=\tan(3y)\]so after the sub there should be no tangent left
oh... its \[1/3\int\limits_{?}^{?}u^6du\]
yep :D
you are a life saver. Can I just ask you one more quick here
sure but I'm headed for dinner so gotta be quick
ill try. \[\int\limits_{-1}^{-1/2}t^-2 \sin^2(1+1/2)dt sorry its t \to the -2 wont show up though\]
Any 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..
\[\int t^{-2}\sin^2?dt\]sorry, what is the integrand exactly?
the end is sin^2(1+1/t)
that 2 was wrong. you got the rest right. evaluated from -1 to -.5
\[\int t^{-2}\sin^2(1+\frac1t)dt\]yes?
yes
evaluated from -1 to -1/2 if that changes anything
\[u=1+\frac1t\]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!
Thanks a lot! I appreicate it

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