Question

Integrate ${\sec }^{3}x$ w.r.t. $x$.

Answer 1

We can write, $\int {\sec }^{3}xdx=\int \sec x{\sec }^{2}xdx$

Solving by integration by parts,

$=\sec x\tan x-\int \tan x\sec x\tan xd$ ....$[\because \int {\sec }^{2}x=\tan x\text{and}\frac{d\left(\sec x\right)}{dx}=\sec x\tan x]$

$=\sec x\tan x-\int \sec x({\sec }^{2}x-1)dx$ ....$[{\sec }^{2}x={\tan }^{2}x+1]$

$=\sec x\tan x+\int \sec xdx-\int {\sec }^{3}xdx$

$2\int {\sec }^{3}xdx=\sec x\tan x+\log |\sec x+\tan x|+c$

$\int {\sec }^{3}xdx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\log |\sec x+\tan x|+c$