Integrate intsec^3(x) by using integral by parts??

May 1, 2018

$I = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln | \sec x + \tan x | + C$

Explanation:

Here,

color(red)(I=intsec^3xdx...to(1)

$= \int \sec x {\sec}^{2} x \mathrm{dx}$

$\text{Using "color(blue)"Integration by Parts}$

$\int \left(u \cdot v\right) \mathrm{dx} = u \int v \mathrm{dx} - \int \left(u ' \int v \mathrm{dx}\right) \mathrm{dx}$

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

$\implies u ' = \sec x \tan x \mathmr{and} \int v \mathrm{dx} = \tan x$

$I = \sec x \times \tan x - \int \sec x \tan x \times \tan x \mathrm{dx}$

$= \sec x \tan x - \int \sec x {\tan}^{2} x \mathrm{dx} + c$

$= \sec x \tan x - \int \sec x \left({\sec}^{2} x - 1\right) \mathrm{dx} + c$

$I = \sec x \tan x - \textcolor{red}{\int {\sec}^{3} x \mathrm{dx}} + \int \sec x \mathrm{dx} + c$

$I = \sec x \tan x - \textcolor{red}{I} + \int \sec x \mathrm{dx} + c \ldots \to$[ from $\left(1\right)$]

$I + I = \sec x \tan x + \int \sec x \mathrm{dx} + c$

$2 I = \sec x \tan x + \ln | \sec x + \tan x | + C$

$I = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln | \sec x + \tan x | + C$