# How do you integrate int xarcsecx by parts from [2,4]?

Dec 9, 2016

The answer is $= 7.39$

#### Explanation:

If $y = a r c \sec x$

$\sec y = x$

$\frac{1}{\cos} y = x$

$\cos y = \frac{1}{x} = {x}^{- 1}$

$\sin y = \frac{\sqrt{{x}^{2} - 1}}{x}$

$\left(- \sin y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{x}^{2} \sin y}$

$= \frac{1}{{x}^{2} \cdot \frac{\sqrt{{x}^{2} - 1}}{x}} = \frac{1}{x \sqrt{{x}^{2} - 1}}$

Let's do the integration by parts

$\int u v ' = u v - \int u ' v$

$u = a r c \sec x$, $u ' = \frac{1}{x \sqrt{{x}^{2} - 1}}$

$v ' = x$, $v = {x}^{2} / 2$

$\int x a r c \sec x \mathrm{dx} = {x}^{2} / 2 a r c \sec x - \int \frac{{x}^{2} \mathrm{dx}}{2 x \sqrt{{x}^{2} - 1}}$

$\int \frac{{x}^{2} \mathrm{dx}}{2 x \sqrt{{x}^{2} - 1}} = \frac{1}{2} \int \frac{x \mathrm{dx}}{\sqrt{{x}^{2} - 1}}$

Let $u = {x}^{2} - 1$, $\mathrm{du} = 2 x \mathrm{dx}$, $x \mathrm{dx} = \frac{\mathrm{du}}{2}$

$\frac{1}{2} \int \frac{x \mathrm{dx}}{\sqrt{{x}^{2} - 1}} = \frac{1}{4} \int \frac{\mathrm{du}}{\sqrt{u}}$

$= \frac{1}{4} \cdot \frac{\sqrt{u}}{\frac{1}{2}}$

$= \frac{1}{2} \sqrt{u} = \frac{1}{2} \cdot \sqrt{{x}^{2} - 1}$

So,

$\int x a r c \sec x \mathrm{dx} = {x}^{2} / 2 a r c \sec x - \frac{1}{2} \cdot \sqrt{{x}^{2} - 1}$

${\int}_{2}^{4} x a r c \sec x \mathrm{dx} = {\left[{x}^{2} / 2 a r c \sec x\right]}_{2}^{4} - {\left[\frac{1}{2} \cdot \sqrt{{x}^{2} - 1}\right]}_{2}^{4}$

$= \frac{16}{2} a r c \sec 4 - \frac{4}{2} a r c \sec 2 - \left(\frac{1}{2} \cdot \sqrt{15} - \frac{1}{2} \sqrt{3}\right)$

$= 8 \cdot 1.32 - 2 \cdot 1.05 - \frac{1}{2} \left(\sqrt{15} - \sqrt{3}\right)$

$= 10.56 - 2.1 - 1.07$

$= 7.39$