Integration of tan pow4x

Answer 1

integral of (tanx)^4

(tanx)^4 = (tanx)^2 (tanx)^2

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

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

= integral of sec^2 x tan^2 x dx - integral of tan^2 x dx

First, integral of sec^2 x tan^2 x dx

Let u = tanx because that would make du = sec^2 x dx

so then we have integral of u^2 du which is (1/3)u^3

substituting back in tanx we get (1/3)tan^3 x

Next, integral of tan^2 x

tan^2 x = sec^2 x -1

integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx

= tanx - x

so putting it all together we have integral of tan^4 x dx =

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