# How do you integrate int arccosx by parts from [0,1/2]?

Jan 2, 2018

$I = \frac{\pi - 3 \sqrt{3}}{6} + 1$

#### Explanation:

$I = {\int}_{0}^{\frac{1}{2}} {\cos}^{- 1} x \mathrm{dx}$

we integrate by parts

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

the key is to choose the appropriate values for $u \text{ }$and $\frac{\mathrm{dv}}{\mathrm{dx}}$

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

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

$I = {\left[x {\cos}^{- 1} x - \int \left(- \frac{x}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}\right]}_{0}^{\frac{1}{2}}$

$I = {\left[x {\cos}^{- 1} x + \int \left(\frac{x}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}\right]}_{0}^{\frac{1}{2}}$

the integral can be done by inspection

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

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

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

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

$\therefore I = {\left[x {\cos}^{- 1} x - {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}\right]}_{0}^{\frac{1}{2}}$

$I = \left[\frac{1}{2} {\cos}^{- 1} \left(\frac{1}{2}\right) - {\left(1 - {\left(\frac{1}{2}\right)}^{2}\right)}^{\frac{1}{2}}\right] - \left[0 - {\left(1 - 0\right)}^{\frac{1}{2}}\right]$

$I = \frac{1}{2} \left(\frac{\pi}{3}\right) - \frac{\sqrt{3}}{2} + 1$

$I = \frac{\pi - 3 \sqrt{3}}{6} + 1$

Jan 2, 2018

${\int}_{0}^{\frac{1}{2}} {\cos}^{-} 1 \left(x\right) \setminus \mathrm{dx} = \frac{\pi + 6 - 3 \sqrt{3}}{6}$

#### Explanation:

I will start by working out the antiderivative. When you have a simple function that you can't rewrite, the classic trick is to use integration by parts.

The formula for integration by parts is:
$\int \setminus f \left(x\right) g ' \left(x\right) \setminus \mathrm{dx} = f \left(x\right) g \left(x\right) - \int \setminus f ' \left(x\right) g \left(x\right) \setminus \mathrm{dx}$

In this case, I will let $f \left(x\right) = {\cos}^{-} 1 \left(x\right)$ and $g ' \left(x\right) = 1$.
$f ' \left(x\right) = - \frac{1}{\sqrt{1 - {x}^{2}}}$
$g \left(x\right) = x$

Plugging this into the formula gives:
$x {\cos}^{-} 1 \left(x\right) + \int \setminus \frac{x}{\sqrt{1 - {x}^{2}}} \setminus \mathrm{dx}$

This integral can be cracked using a u-substitution with $u = 1 - {x}^{2}$. The derivative of $u$ is then $- 2 x$, so we divide through by that to integrate with respect to $u$:
$\int \setminus \frac{\cancel{x}}{- 2 \cancel{x} \sqrt{u}} \setminus \mathrm{du} = - \int \setminus \frac{1}{2 \sqrt{u}} \setminus \mathrm{du} =$

This is the common derivative of $\sqrt{u}$:
$= - \sqrt{u} + C$

Plugging back into the original expression and resubstituting, we get:
$x {\cos}^{-} 1 \left(x\right) - \sqrt{1 - {x}^{2}} + C$

Now we can plug in the limits of integration:
${\int}_{0}^{\frac{1}{2}} {\cos}^{-} 1 \left(x\right) \setminus \mathrm{dx} = {\left[x {\cos}^{-} 1 \left(x\right) - \sqrt{1 - {x}^{2}}\right]}_{0}^{\frac{1}{2}} =$

$= \frac{1}{2} {\cos}^{-} 1 \left(\frac{1}{2}\right) - \sqrt{1 - {\left(\frac{1}{2}\right)}^{2}} - \left(0 {\cos}^{-} 1 \left(0\right) - \sqrt{1}\right) =$

$= \frac{1}{2} \cdot \frac{\pi}{3} - \sqrt{1 - \frac{1}{4}} + 1 = \frac{\pi}{6} - \sqrt{\frac{3}{4}} + 1 =$

$= \frac{\pi}{6} - \frac{\sqrt{3}}{2} + 1 = \frac{\pi}{6} - \frac{3 \sqrt{3}}{6} + \frac{6}{6} = \frac{\pi + 6 - 3 \sqrt{3}}{6}$

Jan 2, 2018

${\int}_{0}^{\frac{1}{2}} \arccos \left(x\right) \mathrm{dx} = \frac{\pi}{6} - \frac{\sqrt{3}}{2} + 1 = \approx 0.657573$

#### Explanation:

Let's first have a look at the following integral (that we are going to need later on):

 int x / sqrt(1 - x^2) dx .

To solve this integral, we substitute $x = \cos u$ and $\mathrm{dx} = - \sin \left(u\right) \mathrm{du}$, which leads to

 [ - int cos u / sqrt(1 - cos^2 u) sin u du ]_{u = arccos x} =

 = [ - int cos u du ]_{u = arccos x} =

 = [ - sin u + C]_{u = arccos x} =

 = [ - sqrt(1 - cos^2 u) + C]_{u = arccos x} =

 = - sqrt(1 - x^2) + C .

Now we are ready to evaluate the integral that we are actually interested in:

 int arccos(x) dx = int \underbrace{1}_{u'} * \underbrace{arccos(x)}_v dx.

Using integration by parts, and remembering that

 d / dx arccos x = - 1 / sqrt(1 - x^2) ,

this yields

 x arccos x + int x / sqrt(1 - x^2) dx = x arccos x - sqrt(1 - x^2) + C .

Finally, we get

 int_0^{1/2} arccos(x) dx = [ x arccos x - sqrt(1 - x^2) ]_0^{1/2} =

 = pi/6 - sqrt(3) / 2 + 1 = approx 0.657573 .