Why is this true? $\int \frac{du}{\tan u \cos^2 u} = \int \frac{d(\tan u)}{\tan u}$

Question

I can't understand why $$\int \frac{du}{\tan u \cos^2 u} = \int \frac{d(\tan u)}{\tan u}$$

Can anybody tell me what I'm missing? Is it a trigonometric identity?

Answer 1

$$\frac{d(\tan(u))}{du}=\sec^2(u)=\frac{1}{\cos^2(u)}$$

That's all.

Answer 2

let $y = \tan(u)$, then $dy = \sec^2(u).du $ first expression is nothing but $\int \frac{\sec^2 (u).du}{\tan(u)} = \int \frac{dy}{y}$ and so is second .

okay it was my first time in mathjax please edit if something goes wrong

Answer 3

Because the first integral $\int \frac{du}{\tan u \cos^2 u}$ is equal to $\log(tan(x))+C$ and in the second integral $\int \frac{d(\tan u)}{\tan u}$ we have that $\frac{d(\tan u)}{\tan u}$ is exactly the closed differential form $d(log(tan(u))$ (as say $\int dF=F+Constante)$