How do I evaluate the indefinite integral int(tan^2(x)+tan^4(x))^2dx ?

1 Answer
Jul 26, 2014

\int(\tan^2(x)+\tan^4(x))^2 dx=\int((tan^2 x(1+tan^2 x))^2 dx
=\int tan^4 x(1+tan^2 x) (1+tan^2x) dx
=\int(tan^4 x+tan^6 x) (1+tan^2 x) dx
Now, note that (tan x)'=sec^2 x=1+tan^2 x (see derivatives of trig functions ).

Thus, the substitution u=tan x yields du=(1+tan^2 x) dx and
\int(\tan^2(x)+\tan^4(x))^2 dx=\int u^4+u^6 du=u^5/5+u^7/7+C
=(\tan^5 x)/5+(\tan^7 x)/7+C
using the power rule for integration (see, e.g, antiderivatives )