A useful trick for integrating tan(x)tan(x) is to use the identity tan^2(x) = 1 + sec^2(x)tan2(x)=1+sec2(x).
This is because of a general integration result it is good to be aware of. By using the substitution u=f(x)u=f(x) we can show that
\int "f"'(x)["f"(x)]^n "d"x = ("f"(x)^(n+1))/(n+1)+C. *
This is useful because "d"/("d"x) tan(x) = sec^2(x).
sec^2(x) = 1 + tan^2(x) means that tan^2(x) = sec^2(x) - 1. Then,
So we write,
tan^4(x) = tan^2(x)*tan^2(x),
tan^4(x) = (sec^2(x)-1)*tan^2(x),
tan^4(x) = -tan^2(x) + sec^2(x)tan^2(x).
tan^4(x) = 1 - sec^2(x) + sec^2(x)tan^2(x).
Armed with the knowledge from * and the fact that sec^2(x) is the derivative of tan^2(x), we are ready to integrate.
\int tan^4(x) "d"x = \int 1 - sec^2(x) + sec^2(x)tan^2(x) "d"x
\int tan^4(x) "d"x = x - tan(x) + (tan^3(x))/3 + C
