Question

Integrate tan³sec² dx​

Answer 1

Answer:

$\frac{1}{4}\mathrm{tan}^4x+c$

Step-by-step explanation:

Assuming you actually mean:

$\int(\mathrm{tan}^3x)(\mathrm{sec}^2x)dx$.

Let $u=\mathrm{tan}x$.

Then, $\frac{du}{dx}=\frac{d}{dx}(\frac{\mathrm{sin}x}{\mathrm{cos}x})=\mathrm{sec}^2x$.

Hence, $dx=\frac{1}{\mathrm{sec}^2x}du$.

Therefore, we get via substitution:

$\int(\mathrm{tan}^3x)du\\=\int u^3du\\=\frac{u^4}{4}+c\\=\frac{1}{4}\mathrm{tan}^4x+c$