Question

What is the integral of `int tan^5(x)`?

Answer 1

`int tan^(5)(x) dx = 1/4sec^(4)(x)-sec^(2)(x)+ln|sec(x)|+C`

Explanation:

`int tan^(5)(x) dx`

Knowing the fact that `tan^(2)(x) = sec^2(x)-1`, we can rewrite it as

`int (sec^2(x)-1)^(2) tan(x) dx`, which yields

`int sec^3(x) sec(x) tan(x) dx-2int sec^2(x) tan(x) dx + int tan(x) dx`

First integral:
Let `u=sec(x) -> du = sec(x)tan(x) dx`

Second integral:
Let `u =sec(x) -> du = sec(x)tan(x) dx`

Therefore

`int u^3 du - 2int u du + int tan(x) dx`

Also note that `int tan(x) dx = ln|sec(x)| + C`, thus giving us

`1/4 u^4 - 1/2 u^2 + ln|sec(x)| + C`

Substituting `u` 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`