int tan^(5)(x) dx∫tan5(x)dx
Knowing the fact that tan^(2)(x) = sec^2(x)-1tan2(x)=sec2(x)−1, we can rewrite it as
int (sec^2(x)-1)^(2) tan(x) dx∫(sec2(x)−1)2tan(x)dx, which yields
int sec^3(x) sec(x) tan(x) dx-2int sec^2(x) tan(x) dx + int tan(x) dx∫sec3(x)sec(x)tan(x)dx−2∫sec2(x)tan(x)dx+∫tan(x)dx
First integral:
Let u=sec(x) -> du = sec(x)tan(x) dxu=sec(x)→du=sec(x)tan(x)dx
Second integral:
Let u =sec(x) -> du = sec(x)tan(x) dxu=sec(x)→du=sec(x)tan(x)dx
Therefore
int u^3 du - 2int u du + int tan(x) dx∫u3du−2∫udu+∫tan(x)dx
Also note that int tan(x) dx = ln|sec(x)| + C∫tan(x)dx=ln|sec(x)|+C, thus giving us
1/4 u^4 - 1/2 u^2 + ln|sec(x)| + C14u4−12u2+ln|sec(x)|+C
Substituting uu back into the expression gives us our final result of
1/4sec^(4)(x)-cancel(2)*(1/cancel(2))sec^(2)(x)+ln|sec(x)|+C
Thus
int tan^(5)(x) dx = 1/4sec^(4)(x)-sec^(2)(x)+ln|sec(x)|+C
