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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

is -2pi/3 to pi = to pi to -2pi/3 ???

then given the interval [a,b] you work it out: F(b)-F(a)

|dw:1326817635699:dw|

|dw:1326817802065:dw|

\[u=\frac{x}{4}\to du=\frac{dx}{4}\to4du=dx\]\[\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 ;)

lol

|dw:1326818150512:dw| i did this process n got that ans

is the problem\[\int_{\frac{-2\pi}{3}}^{\pi}\tan^3(\frac x4)\sec^2(\frac x4)dx\]?

yes