Question

Answer 1

$$\begin{gathered} \int {\tan }^{\frac{3}{2}}x·{\sec }^{2}xdx \\ \mathrm{Let}\tan x=t \\ \Rightarrow {\sec }^{2}x=\frac{dt}{dx} \\ \Rightarrow {\sec }^{2}xdx=dt \\ \mathrm{Now},\int {\tan }^{\frac{3}{2}}x·{\sec }^{2}xdx \\ =\int t^{\frac{3}{2}}dt \\ =\left[\frac{t^{{\displaystyle \frac{3}{2}}+1}}{{\displaystyle \frac{3}{2}}+1}\right]+\mathrm{C} \\ =\frac{2}{5}{t}^{\frac{5}{2}}+\mathrm{C} \\ =\frac{2}{5}{\tan }^{\frac{5}{2}}x+\mathrm{C} \end{gathered}$$