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Integral of tan^4 (x) * sec^2 (x) dx
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derivative of tan(x) is ?
sec^2 (x)
yes, so your u-substitution is u= ?
- thats the way to go
Ohh, how do we determine the u substitution?
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you know that for any \[\int\limits_{ }^{ } F\left(~g(x)~\right)~~\times ~g'(x)~~dx\] then you set: u=g(x) du=g'(x) dx and you get: \[\int\limits_{ }^{ } F\left(~u~\right) ~du\]
is this last post making sense?
YEsssss
thank you
tan^4(x) is same as (tan x)^4
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so your u= ?
tanx
yes, and your new integral is ?
integral of tan^4(x) dx
Yes I understand now, thank you
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no, not really...
\[\int\limits_{ }^{ } (\color{red}{\tan x})^4\cdot \color{green}{(\sec^2x)~dx}\]\[\int\limits_{ }^{ } (\color{red}{u})^4 \color{green}{du}\]
the red is our u-substitution and green is what "du" is
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