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integrate tan^3(x/4)*sec^2(x/4) from-2/3pi to pi help with the steps plz.

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since sec^2 is the derivaive of tan, this is a simple enough arrangement: \[\int f(u)du\]
i thinkk you might be off by a 4 tho
x/4 pops out a spurious little 1/4 that missing

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Other answers:

is -2pi/3 to pi = to pi to -2pi/3 ???
then given the interval [a,b] you work it out: F(b)-F(a)
\[[tan^4(\frac{1}{4}x)]'=\frac{1}{4}tan^3(\frac{1}{4}x)sec^2(\frac{1}{4}x)\] so we have to multiply thru by 4 to begin with
\[u=\frac{x}{4}\to du=\frac{dx}{4}\to4du=dx\]\[\int_0}{|dw:1326817852162:dw|
\[\Pi ^{2}\div9{(65\Pi ^{2}/9)-26)}\] i got this ans. s this correct??i
lol, I just derived tan^4(x/4) and got a spurious 1/4 so that cant be right
hmmm, im off by something
wolf agrees with turing
oh yeah, the 4 pops out to get rid of the 1/4 lol
derivative of (x/4) i think u miss it during derivative
yep, one comes out on top and bottom, so they cancel
a was using an abacus ;)
|dw:1326818150512:dw| i did this process n got that ans
is the problem\[\int_{\frac{-2\pi}{3}}^{\pi}\tan^3(\frac x4)\sec^2(\frac x4)dx\]?
\[\int_{\frac{-2\pi}{3}}^{\pi}\tan^3(\frac x4)\sec^2(\frac x4)dx\]\[u=\frac x4\to du=\frac{dx}{4}\to4du=dx\]so...\[\int_{\frac{-2\pi}{3}}^{\pi}\tan^3(\frac x4)\sec^2(\frac x4)dx=4\int_{\frac{-\pi}{6}}^{\frac{\pi}{4}}\tan^3u\sec^2udu=4(\frac14\tan^4(\frac x4))|_{\frac{-2\pi}{3}}^{\pi}\]\[=\tan^4(\frac x4)|_{\frac{-2\pi}{3}}^{\pi}\]evaluate...
\[\text{Tan}\left[\frac{\pi }{4}\right]^4-\text{Tan}\left[\frac{\frac{-2}{3}\pi }{4}\right]^4=\frac{8}{9} \]

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