Question

What is the integral of ${\tan }^{4}x$?

Answer 1

Find the integral of ${\tan }^{4}x$

$$\begin{gathered} \int {\tan }^{4}xdx=\int {\tan }^{4}x-1+1dx \\ \Rightarrow \int {\tan }^{4}xdx=\int \left[{\left({\tan }^{2}x\right)}^2-{\left(1^2\right)}^2\right]+1dx \\ \Rightarrow \int {\tan }^{4}xdx=\int \left({\tan }^{2}x+1^2\right)\left({\tan }^{2}x-1^2\right)+1dx \\ \Rightarrow \int {\tan }^{4}xdx=\int se{c}^{2}x\left({\tan }^{2}x-1\right)+1dx\left\{\mathbf{\because }\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{x}\mathbf{-}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\mathit{x}\mathbf{=}\mathbf{1}\right\} \\ \Rightarrow \int {\tan }^{4}xdx=\int se{c}^{2}x\left({\tan }^{2}x-1\right)dx+\int 1dx \\ \Rightarrow \int {\tan }^{4}xdx=\int se{c}^{2}x\left({\tan }^{2}x-1\right)dx+x+C\left\{\mathrm{where}C\mathrm{is}\mathrm{a}\mathrm{constant}\right\}...\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Take}\tan \mathit{}x=t \\ \Rightarrow se{c}^{2}xdx=dt \end{gathered}$$.

Put the above values in the equation$\left(1\right)$:

$\begin{array}{rcl}\int {\tan }^{4}xdx& =& \int \left(t^2-1\right)dt+x+C\\ & \Rightarrow & \int {\tan }^{4}xdx=\frac{t^3}{3}-t+x+C\\ & \Rightarrow & \int {\tan }^{4}xdx=\frac{{\tan }^{3}x}{3}-\tan x+x+C\left\{\mathrm{Put}t=\tan x\right\}\end{array}$

Hence, the required integral is $\begin{array}{rcl}& & \int \end{array}\begin{array}{rcl}& & \tan \end{array}\begin{array}{rcl}& & \end{array}$ where $C$ is a constant of integration .