Please explain the steps in the solution of this integral

Question

After unsuccessfully solving an integral, I turned to WRA for some guidance. Unfortunately, I don't understand what happen in the solution. More particularly, I don't understand why the numerator has tan^1x and not tan^2x (see red highlights)

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Answer 1

Note that $$ \dfrac{1}{\sec{u}} = \cos{u} $$ so substituting everything gives: $$ \begin{align*} \int\dfrac{\sqrt{4x^2-1}}{x^2} \, dx &= \int \dfrac{(\tan(u))}{(\sec(x)/2)^2} \cdot \dfrac{1}{2} \tan(u) \sec(u) \, du \\ &= \dfrac{1}{2}\int \dfrac{4 \tan^2(u) \sec(u)}{\sec(u)^2} \,du\\ &= 2\int \dfrac{\tan^2(u)}{\sec(u)} \,du\\ &= 2\int \tan(u) \cdot \dfrac{\sin(u)}{\cos(u)} \cdot\cos(u) \,du\\ &= 2\int \tan(u) \sin(u) \,du. \end{align*} $$ The trick is in turning one of the tangents into sine over cosine.