calculus
posted by rox .
find the integral:
∫tan^6 (x) dx

Put tan(x) = u >
x = arctan(u) >
dx = du/(1+u^2)
Integral of u^6/(1+u^2) du =
Integral of
(u^6 + u^4  u^4  u^2 + u^2 + 1 1)/
(u^2 + 1) du =
Integral of
(u^4  u^2 + 1) du  Integral of
1/(1+u^2) du =
1/5 u^5  1/3 u^3 + u  arctan(u) + c=
1/5 tan^5(x)  1/3 tan^3(x) + tan(x)  x + c
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