Question

$\int {\sec }^{3}xdx$ is equal to?

• A. $\frac{\sec x\tan x}{2}+\frac{1}{2}ln|\sec x+\tan x|+C$
• B. $\frac{\sec x\tan x}{2}-\frac{1}{2}ln|\sec x+\tan x|+C$
• C. $\frac{\sec x\tan x}{2}+\frac{1}{2}ln|\sec x-\tan x|+C$
• D. $\sec x\tan x+ln|\sec x+\tan x|+C$

Answer 1

The correct option is A $\frac{\sec x\tan x}{2}+\frac{1}{2}ln|\sec x+\tan x|+C$

$\int se{c}^{3}xdx-I$

$\int secx$ - I $se{c}^{2}x$ - II

Now integrating by parts

$\int uvdx=u\int vdx-\frac{du}{dx}[\int vdx]dx$

$secx\int se{c}^{2}xdx-\int \frac{dsecx}{dx}-\int \frac{dsecx}{dx}[\int se{c}^{2}x]dx$

$secx.tanx-\int secxtanx\left(tanx\right)dx$ $[\int se{c}^{2}x=tanx]$

$secx.tanx-\int secx.ta{n}^{2}xdx$

$(se{c}^{2}x-ta{n}^{2}x=1,\Rightarrow ta{n}^{2}x=se{c}^{2}x-1)$

$secx.tanx-\int secx(se{c}^{2}x-1)dx$

$secx.tanx-\int (se{c}^{3}x-secx)dx$

$secx.tanx-[\int se{c}^{3}xdx-\int secxdx]$

[ Now $se{c}^{3}x=I$]

$secx.tanx-I+\int secxdx=I$

$secxtanx+log|secx+tanx|=2I$

$I=\frac{secx.tanx}{2}+\frac{1}{2}log|sec+tanx|+c$ [where C is a constant]