Answer:
[tex]\frac{1}{4}\mathrm{tan}^4x+c[/tex]
Step-by-step explanation:
Assuming you actually mean:
[tex]\int(\mathrm{tan}^3x)(\mathrm{sec}^2x)dx[/tex].
Let [tex]u=\mathrm{tan}x[/tex].
Then, [tex]\frac{du}{dx}=\frac{d}{dx}(\frac{\mathrm{sin}x}{\mathrm{cos}x})=\mathrm{sec}^2x[/tex].
Hence, [tex]dx=\frac{1}{\mathrm{sec}^2x}du[/tex].
Therefore, we get via substitution:
[tex]\int(\mathrm{tan}^3x)du\\=\int u^3du\\=\frac{u^4}{4}+c\\=\frac{1}{4}\mathrm{tan}^4x+c[/tex]
