Question

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int {\tan }^{5}xdx \\ =\int {\tan }^{3}x·{\tan }^{2}xdx \\ =\int {\tan }^{3}x\left({\sec }^{2}x-1\right)dx \\ =\int {\tan }^{3}x·{\sec }^{2}x\mathit{}dx-\int {\tan }^{3}x\mathit{}dx \\ =\int {\tan }^{3}x·{\sec }^{2}xdx-\int \tan x·{\tan }^{2}xdx \\ =\int {\tan }^{3}x·{\sec }^{2}xdx-\int \tan x·\left({\sec }^{2}x-1\right)dx \\ =\int {\tan }^{3}x·{\sec }^{2}xdx-\int \tan x·{\sec }^{2}xdx+\int \tan xdx \\ \mathrm{Putting}\tan x=t\mathrm{in}\mathrm{the}\mathrm{Ist}\mathrm{and}\mathrm{IInd}\mathrm{integral}. \\ \Rightarrow {\sec }^{2}xdx=dt \\ \therefore I=\int t^3·dt-\int t·dt+\int \tan \mathrm{x}\mathrm{d}\mathrm{x} \\ =\frac{t^4}{4}-\frac{t^2}{2}+\ln \left|\sec x\right|+C \\ =\frac{{\tan }^{4}x}{4}-\frac{{\tan }^{2}x}{2}+\ln \left|\sec x\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \mathit{}t=\tan x\right] \end{gathered}$$