# How do you integrate int 4arccosx by parts?

Jun 14, 2017

$\int 4 \arccos x \mathrm{dx} = 4 x \arccos x - 4 {\left(1 - {x}^{2}\right)}^{\frac{1}{2}} + C$

#### Explanation:

IBP formula

$I = \int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$

we have

$I = \int 4 \arccos x \mathrm{dx}$

$I = \int 4 {\cos}^{- 1} x \mathrm{dx} = 4 \int \left(1 \times {\cos}^{- 1} x\right) \mathrm{dx}$

the success of using IBP is the correct identification of the $u \text{ }$&$\text{ "(dv)/(dx)" }$

in this case

u=cos^(-1)x=>(du)/(dx)=-1/sqrt(1-x^2

$\frac{\mathrm{dv}}{\mathrm{dx}} = 1 \implies v = x$

$I = \int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$

becomes

I=4[xcos^(-1)x-int(-x/sqrt(1-x^2)dx]#

now the second integral can be done by inspection

$\int \left(- \frac{x}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx} = \int \left(- x {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}}\right) \mathrm{dx}$

the outside is a multiple of the bracket differentiated

so we can guess the integral as the bracket to the power +1

ie$\text{ } {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$

try this out

$\frac{d}{\mathrm{dx}} \left({\left(1 - {x}^{2}\right)}^{\frac{1}{2}}\right)$

$= \frac{1}{2} \times \left(- 2 x\right) {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}}$

$= - x {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}}$

which is the integral we want

so the final integral is;

$\int 4 \arccos x \mathrm{dx} = 4 x \arccos x - 4 {\left(1 - {x}^{2}\right)}^{\frac{1}{2}} + C$