Question

What is the integral of $se{c}^{3}x$?

Answer 1

Find the integral of ${\sec }^{3}x$.

$\int {\sec }^{3}xdx=\int \sec x{\sec }^{2}xdx$

From integration by parts: $\mathbf{\int }\mathit{u}\mathit{v}\mathbf{}\mathit{d}\mathit{x}\mathbf{=}\mathit{u}\mathbf{\int }\mathit{v}\mathbf{}\mathit{d}\mathit{x}\mathbf{-}\mathbf{\int }\left(\int vdx\right)\mathbf{}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{}\mathit{d}\mathit{x}$

Let. $u=\sec x,v={\sec }^{2}x$.

To find integral of $se{c}^{3}x$, Put $u=\sec x,v={\sec }^{2}x$ in $\int uvdx=u\int vdx-\int \left(\int vdx\right)\frac{du}{dx}dx$

$\begin{array}{rcl}\int \sec x.{\sec }^{2}xdx& =& \sec x\left(\tan x\right)-\int \tan x\sec x\tan xdx\left\{\because \int se{c}^{2}xdx=\tan x,\frac{d}{dx}\left(secx\right)=\tan xsecx\right\}\\ & \Rightarrow & \int {\sec }^{3}xdx=\sec x\left(\tan x\right)-\int {\tan }^{2}x\sec xdx\\ & \Rightarrow & \int {\sec }^{3}xdx=\sec x\left(\tan x\right)-\int \left({\sec }^{2}x-1\right)secxdx\left\{\mathbf{\because }\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{x}\mathbf{-}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\mathit{x}\mathbf{=}\mathbf{1}\right\}\\ & \Rightarrow & \int {\sec }^{3}xdx=\sec x\left(\tan x\right)-\int \left({\sec }^{3}x-\sec x\right)dx\\ & \Rightarrow & \int {\sec }^{3}xdx=\sec x\left(\tan x\right)-\int {\sec }^{3}xdx+\int \sec xdx\\ & \Rightarrow & 2\int {\sec }^{3}xdx=\sec x\left(\tan x\right)+\ln \left|\sec x+\tan x\right|+C\left\{\because \int \sec xdx=\ln |\sec x+\tan x|\right\}\end{array}$

$\begin{array}{rcl}& \Rightarrow & \int {\sec }^{3}xdx=\frac{1}{2}[\sec x\left(\tan x\right)+\ln \left|\sec x+\tan x\right|]+C\end{array}$

Hence, the required integral is $\begin{array}{rcl}& & \int {\sec }^{3}xdx=\frac{1}{2}[\sec x\left(\tan x\right)+\ln \left|\sec x+\tan x\right|]+C\end{array}$