Evaluating the integral ∫tan^2(x)+tan^4(x)dx

For the integral of...

∫tan²(x) + tan⁴(x)dx

Initially, you may look at this as a sum of  2 separate integrals and sure enough the integral of tan²(x) is easy enough but the integral of tan⁴(x) is a quite a bit more tricky, so frankly this is not a very efficient way of doing it but what if we factor out a tan²(x) from this entire expression?

So we'll get:

tan²(x) + tan⁴(x) = tan²(x)[1 + tan²(x)]

But we can also realize that...

1 + tan²(x) = sec²(x)

And therefore...

tan²(x)[1 + tan²(x)] = tan²(x)sec²(x)

And thus...

∫tan²(x) + tan⁴(x)dx = ∫tan²(x)sec²(x)dx

But we can make this integral even simpler by letting u = tan(x), and thus du = sec²(x)dx

Please watch the video to see how this integral is solved.