# Integration by parts for definite integrals?

## ${\int}_{0}^{\frac{1}{2}} \arccos \left(2 x\right) \mathrm{dx}$

Jan 22, 2018

The answer is $= \frac{1}{2}$

#### Explanation:

Perform the inegration by parts

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

Reminder :

$\left(\arccos x\right) ' = - \frac{1}{\sqrt{1 - {u}^{2}}}$

Let $u ' = 1$, $\implies$, $u = x$

$v = \arccos \left(2 x\right)$, $\implies$, $v ' = - \frac{2}{\sqrt{1 - 4 {x}^{2}}}$

Therefore,

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

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

So,

${\int}_{0}^{\frac{1}{2}} \arccos \left(2 x\right) \mathrm{dx} = {\left[x \arccos \left(2 x\right) - \frac{1}{2} \sqrt{1 - 4 {x}^{2}}\right]}_{0}^{\frac{1}{2}}$

$= \left(\frac{1}{2} \arccos \left(2 \cdot \frac{1}{2}\right) - \frac{1}{2} \sqrt{1 - 4 {\left(\frac{1}{2}\right)}^{2}}\right) - \left(0 \arccos \left(2 \cdot 0\right) - \frac{1}{2} \sqrt{1 - 4 {\left(0\right)}^{2}}\right)$

$= \frac{1}{2}$

Jan 22, 2018

${\int}_{0}^{\frac{1}{2}} \arccos \left(2 x\right) \mathrm{dx} = \frac{1}{2}$

#### Explanation:

${\int}_{0}^{\frac{1}{2}} \arccos \left(2 x\right) \mathrm{dx}$

Re write as:

${\int}_{0}^{\frac{1}{2}} 1 \cdot \arccos \left(2 x\right) \mathrm{dx}$

So $u = \arccos \left(2 x\right) \to u ' = - \frac{2}{\sqrt{1 - 4 {x}^{2}}}$

$v ' = 1 \to v = x$

Now:

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

${\int}_{0}^{\frac{1}{2}} 1 \cdot \arccos \left(2 x\right) \mathrm{dx} = {\left[x \arccos \left(2 x\right)\right]}_{0}^{\frac{1}{2}} - {\int}_{0}^{\frac{1}{2}} \frac{- 2}{\sqrt{1 - 4 {x}^{2}}} x \mathrm{dx}$

$= {\left[x \arccos \left(2 x\right)\right]}_{0}^{\frac{1}{2}} + {\int}_{0}^{\frac{1}{2}} \frac{2 x}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx}$

To evaluate this integral:

${\int}_{0}^{\frac{1}{2}} \frac{2 x}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx}$

Use the substitution:

$2 x = \sin \left(u\right) \to 2 \mathrm{dx} = \cos \left(u\right) \mathrm{du}$

Also note the new limits:
$x = 0 \to u = 0$
$x = \frac{1}{2} \to u = \frac{\pi}{2}$

Substituting these in, the integral now becomes:

${\int}_{0}^{\frac{\pi}{2}} \frac{\sin \left(u\right)}{\sqrt{1 - {\sin}^{2} \left(u\right)}} \cos \frac{u}{2} \mathrm{du}$

Now, using:

${\cos}^{2} u + {\sin}^{2} u = 1 \to {\cos}^{2} \left(u\right) = 1 - {\sin}^{2} \left(u\right)$

we can simplify the expression in the square root:

$= {\int}_{0}^{\frac{\pi}{2}} \frac{\sin \left(u\right)}{\sqrt{{\cos}^{2} u}} \cos \frac{u}{2} \mathrm{du}$

$= {\int}_{0}^{\frac{\pi}{2}} \frac{\sin \left(u\right)}{\cos} \left(u\right) \cos \frac{u}{2} \mathrm{du} = {\int}_{0}^{\frac{\pi}{2}} \sin \frac{u}{2} \mathrm{du}$

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

So:

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

Evaluating these limits:

${\left[x \arccos \left(2 x\right)\right]}_{0}^{\frac{1}{2}}$

$= \left\{\frac{1}{2} \arccos \left(2 \cdot \frac{1}{2}\right)\right\} - \left\{0 \cdot \arccos \left(\frac{1}{2} \cdot 0\right)\right\} = 0 - 0 = 0$

and

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

$= 0 + \frac{1}{2} = \frac{1}{2}$

So the final answer will be:

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

$= 0 + \frac{1}{2} = \frac{1}{2}$

Jan 22, 2018

This solution is not by parts, but may be of interest.

#### Explanation:

Note that

${\int}_{0}^{\frac{1}{2}} \arccos \left(2 x\right) \mathrm{dx} = \frac{1}{2} {\int}_{0}^{1} \arccos \left(u\right) \mathrm{du}$

and also note that

$y = \arccos \left(u\right)$ if and only if $\cos \left(u\right) = y$ and $0 < u < \pi$

If we turn our thinking ${90}^{\circ}$, we can integrate one quarter a period of the cosine.

graph{arccosx [-3.093, 3.067, -0.643, 2.438]}

The area below $y = \arccos \left(u\right)$ from $u = 0$ to $u = 1$ and above the $u$ axis is exactly the same as the area left of $u = \cos y$ and right of the $y$ axis from $y = 0$ to $y = \frac{\pi}{2}$.

That is a lot of words to explain that

$\frac{1}{2} {\int}_{0}^{1} \arccos \left(u\right) \mathrm{du} = \frac{1}{2} {\int}_{0}^{\frac{\pi}{2}} \cos y \mathrm{dy}$

$= \frac{1}{2} {\left[\sin y\right]}_{0}^{\frac{\pi}{2}} = \frac{1}{2} \left[1 - 0\right] = \frac{1}{2}$

Jan 22, 2018

${\int}_{0}^{\frac{1}{2}} \arccos \left(2 x\right) \mathrm{dx} = \frac{1}{2}$

#### Explanation:

Use Integration by Parts to find the indefinite integral:

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

Let:

$u = \arccos \left(2 x\right) \implies \mathrm{du} = - \frac{2}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx}$ (See proof below)

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

$- - - - - - - - - - - - - - - - - - - -$
color(blue)(y=arccos(2x)<=>cos(y)=2x

color(blue)(cos(y)=2x

Differentiate both sides implicitly (W.R.T $x$)

color(blue)(dy/dx*-sin(y)=2

Divide both sides by $- \sin \left(y\right)$

color(blue)(dy/dx=-2/cos(y)

Rewrite in terms of $x$

Since color(blue)(sin(y)=(2x)/1

Then color(blue)(cos(y)=sqrt(1-4x^2)/1=sqrt(1-4x^2)

So color(blue)(dy/dx=-2/cos(y)=-2/sqrt(1-4x^2)

:.color(blue)(d/dx[arccos(2x)]=-2/sqrt(1-4x^2)

$- - - - - - - - - - - - - - - - - - - -$

Plugging in into the formula:

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

Simplify:

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

Solving for $\int \frac{x}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx}$

Make a substitution:

Let $u = 1 - 4 {x}^{2} \implies \mathrm{du} = - 8 x$

$\int - \frac{1}{8 \sqrt{u}} \mathrm{du} \implies - \frac{1}{8} \int \frac{1}{\sqrt{u}} \mathrm{du} \implies - \frac{1}{8} \int {u}^{- \frac{1}{2}} \mathrm{du}$

Use: color(blue)(intx^adx=(x^(a+1))/(a+1)

$- \frac{1}{8} \cdot {u}^{- \frac{1}{2} + 1} / \left(- \frac{1}{2} + 1\right) = - \frac{1}{8} \cdot {u}^{\frac{1}{2}} / \left(\frac{1}{2}\right) = - \frac{2}{8} {u}^{\frac{1}{2}} = - \frac{1}{4} {u}^{\frac{1}{2}}$

Reverse the substitution:

$= - \frac{1}{4} {\left(1 - 4 {x}^{2}\right)}^{\frac{1}{2}} = - \frac{1}{4} \sqrt{1 - 4 {x}^{2}}$

Going back to the problem:

$= x \arccos \left(2 x\right) + 2 \left[- \frac{1}{4} \sqrt{1 - 4 {x}^{2}}\right]$

$= x \arccos \left(2 x\right) - \frac{2}{4} \sqrt{1 - 4 {x}^{2}}$

$= {\left[x \arccos \left(2 x\right) - \frac{1}{2} \sqrt{1 - 4 {x}^{2}}\right]}_{0}^{\frac{1}{2}}$

Evaluate the upper/lower limits

$= \left[\left(\textcolor{red}{\frac{1}{2}}\right) \arccos \left(2 \cdot \textcolor{red}{\frac{1}{2}}\right) - \frac{1}{2} \sqrt{1 - 4 {\left(\textcolor{red}{\frac{1}{2}}\right)}^{2}}\right] - \left[\left(\textcolor{red}{0}\right) \arccos \left(2 \cdot \textcolor{red}{0}\right) - \frac{1}{2} \sqrt{1 - 4 {\left(\textcolor{red}{0}\right)}^{2}}\right]$

$= \left[0\right] - \left[- \frac{1}{2}\right] = \frac{1}{2}$