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

Question

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

Christopher Agresta · Accepted Answer

#\int( an^2(x)+ an^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( an^2(x)+ an^4(x))^2 dx=\int u^4+u^6 du=u^5/5+u^7/7+C#
#=( an^5 x)/5+( an^7 x)/7+C#
using the power rule for integration (see, e.g, antiderivatives )

Elijah Abram · Answer

undefined To evaluate the indefinite integral ∫(tan^2(x)+tan^4(x))^2 dx, we can use trigonometric identities to simplify the expression. By using the identity tan^2(x) + 1 = sec^2(x), we can rewrite tan^4(x) as (tan^2(x))^2 = (sec^2(x) - 1)^2. Expanding the square of the sum, we get (tan^2(x) + tan^4(x))^2 = (sec^2(x))^2 - 2sec^2(x) + 1. The integral then becomes ∫((sec^2(x))^2 - 2sec^2(x) + 1) dx. Integrating each term separately, we get ∫(sec^4(x) - 2sec^2(x) + 1) dx. The integral of sec^4(x) can be found using integration by parts or by using the power-reduction formula for integrals of secant functions. After integrating each term, the result will be the evaluated indefinite integral.

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