calculus - The integral of $sec^4(x)tan(x)$

Consider the integral

$$\int \sec^4(x)\tan(x)$$

Now right off the bat I see two ways of solving this.

1. Let u=$\sec(x)$

2.Use integration by parts

Now doing the first way results in the integrand looking like
$$\int u^3du=\frac{1}{4}\sec^4(x)+C$$

Which is correct but it's not the answer I'm looking for, so instead we'll do it the second way.

$$\int \sec^2(x)\cdot\sec^2(x)\tan(x)dx$$
$$\int\left(\tan^2(x)+1\right)\sec^2(x)\tan(x)dx$$
$$\int\sec^2(x)\tan^3(x)+\sec^2(x)\tan(x)dx$$
Now this is where I got stuck, because I don't know whether to continue with Pythagorean identities or to factor a term out and solve for that. Or perhaps even break the two up and create two integrals.

Answer

Write your integrand in the form $$\tan(x)(\tan^2(x)+1)\sec^2(x)$$ and substitute $$u=\tan(x)$$ and you will get $$\int u(u^2+1)\,du$$