Calc. II, , Integral of (tan(x))^4?

Question

integral of (tanx)^4

any ideas on how i'd do this?

Answer 1

Integral (tan^4(x) dx )

Your first step is to split this up as tan^2(x) and tan^2(x)

Integral (tan^2(x) tan^2(x) dx)

Now, use the fact that tan^2(x) = sec^2(x) - 1.

Integral (tan^2(x) [sec^2(x) - 1] dx )

Split this up into two integrals.

Integral (tan^2(x) sec^2(x) dx ) - Integral (tan^2(x) dx )

Now, use the same identity on the second integral.

Integral (tan^2(x) sec^2(x) dx) - Integral ( [sec^2(x) - 1] dx)

For the first integral, use substitution.

Let u = tan(x).
du = sec^2(x) dx {which is the tail end of our first integral}, so we now have

Integral (u^2 du) - Integral ( [sec^2(x) - 1] dx )

Integrate as normal using the reverse power rule.

(1/3)u^3 - Integral ( [sec^2(x) - 1] dx)

Substitute back u = tan(x).

(1/3) tan^3(x) - Integral ( [sec^2(x) - 1] dx)

Now, solve the other integral as normal. Keep in mind that
sec^2(x) is a known derivative (It's the derivative of tan(x)).

(1/3) tan^3(x) - [tan(x) - x] + C

(1/3) tan^3(x) - tan(x) + x + C

Answer 2

=tan^2 x (sec^2 x -1)

=tan^2 x sec^2 x - tan^2 x

use u substitution- let u = tanx then du = sec^2 x

=integral(u^2)- integral(sec^2 x - 1)

=1/3 u^3 - tanx + x +C

=1/3(tan^3 x) -tanx + x +C