I'm learning about U-Substitution and I'd like to know where is my mistake in this exercise. I recognize that have a easier way, which is u=tag x substitution. But I would like to understand where the error lies in this resolution.
$\int \tan ^3x\cdot \sec ^2x\ dx=\int \frac{sin ^3x}{cos^ 3x}\cdot \frac{1}{cos^2x}\ dx \\ \int \tan ^3x\cdot \sec ^2x\ dx=-\int \frac{1-u^2}{u^5}\ du, u= \cos x \\ \int \tan ^3x\cdot \sec ^2x\ dx=\frac{1}{4\cos ^4x}-\frac{1}{2\cos ^2x}+C$
\begin{align*} \frac{1}{4 \cos^4 x} - \frac{1}{2 \cos^2 x} &= \frac{(\sin^2 x + \cos^2 x)^2}{4 \cos^4 x} - \frac{\sin^2 x + \cos^2 x}{2 \cos^2 x} \\ &= \frac{\sin^4 x + 2 \sin^2 x \cos^2 x + \cos^4 x}{4 \cos^4 x} - \frac{\sin^2 x + \cos^2 x}{2 \cos^2 x} \\ &= \frac{\sin^4 x}{4 \cos^4 x} + \frac{2 \sin^2 x \cos^2 x}{4 \cos^4 x} + \frac{\cos^4 x}{4 \cos^4 x} - \frac{\sin^2 x}{2 \cos^2 x} - \frac{\cos^2 x}{2 \cos^2 x} \\ &= \frac{\sin^4 x}{4 \cos^4 x} + \frac{\sin^2 x}{2 \cos^2 x} + \frac{1}{4} - \frac{\sin^2 x}{2 \cos^2 x} - \frac{1}{2 } \\ &= \frac{\sin^4 x}{4 \cos^4 x} + \frac{-1}{4} \\ &= \frac{1}{4}\tan^4 x + \frac{-1}{4} \text{.} \end{align*} Of course, the $-1/4$ can be absorbed into the constant of integration.