# How do you find the integral of arccos(x)x?

Aug 9, 2018

$I = {x}^{2} / 2 a r c \cos x + \frac{1}{4} a r c \sin x - \frac{x}{4} \sqrt{1 - {x}^{2}} + C$

#### Explanation:

Here ,

$I = \int a r c \cos x \cdot x \mathrm{dx}$

Using Integration by parts:

$I = a r c \cos x \int x \mathrm{dx} - \int \left(\frac{d}{\mathrm{dx}} \left(a r c \cos x\right) \int x \mathrm{dx}\right) \mathrm{dx}$

$I = a r c \cos x \left({x}^{2} / 2\right) - \int \frac{- 1}{\sqrt{1 - {x}^{2}}} \times {x}^{2} / 2 \mathrm{dx}$

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

color(red)(I=x^2/2arc cosx+1/2I_1............to(A)

where, ${I}_{1} = \int {x}^{2} / \sqrt{1 - {x}^{2}} \mathrm{dx}$

Subst. color(blue)(x=sinu=>dx=cosudu

So,

${I}_{1} = \int {\sin}^{2} \frac{u}{\sqrt{1 - {\sin}^{2} u}} \cos u \mathrm{du}$

${I}_{1} = \int {\sin}^{2} \frac{u}{\cos} u \cos u \mathrm{du}$

${I}_{1} = \int {\sin}^{2} u \mathrm{du}$

${I}_{1} = \int \frac{1 - \cos 2 u}{2} \mathrm{du}$

${I}_{1} = \frac{1}{2} \left[u - \frac{\sin 2 u}{2}\right] + c$

${I}_{1} = \frac{1}{2} \left[u - \sin u \cos u\right] + c$

${I}_{1} = \frac{1}{2} \left[u - \sin u \sqrt{1 - {\sin}^{2} u}\right] + c$

Subst. back color(blue)(sinu=x and u=arcsinx

${I}_{1} = \frac{1}{2} \left[a r c \sin x - x \sqrt{1 - {x}^{2}}\right] + c$

Subst. value of ${I}_{1}$ into color(red)((A) we get

$I = {x}^{2} / 2 a r c \cos x + \frac{1}{2} \left\{\frac{1}{2} \left[a r c \sin x - x \sqrt{1 - {x}^{2}}\right]\right\} + C$

$I = {x}^{2} / 2 a r c \cos x + \frac{1}{4} a r c \sin x - \frac{x}{4} \sqrt{1 - {x}^{2}} + C$