1. ## Integration by Parts

Hello,

$\int_{ 0}^{1/2} {cos^{-1}(x)dx}$

let
$u = cos^{-1}x$ and $dv=dx$

$du = \frac{-1}{\sqrt{1-x^2}}dx$ and $v=x$

Then:

$= [xarccos(x)]_{0}^{1/2} - \int_{0}^{1/2} \frac{-x}{\sqrt{1-x^2}}dx$

However at this point I would use u-sub to find the value of the second integral, but the answer has changed the limits of integration? I'm confused. Any help would be appreciated.

2. ## Re: Integration by Parts

then use the u-sub to find the antiderivative, then back sub ...

$x\arccos{x} - \frac{1}{2} \int \frac{-2x}{\sqrt{1-x^2}} \, dx$

$\left[x\arccos{x} - \sqrt{1-x^2}\right]_0^{\frac{1}{2}}$

$\left(\frac{\pi}{6} - \frac{\sqrt{3}}{2}\right) - (-1)$

3. ## Re: Integration by Parts

Originally Posted by skeeter
then use the u-sub to find the antiderivative, then back sub ...

$x\arccos{x} - \frac{1}{2} \int \frac{-2x}{\sqrt{1-x^2}} \, dx$

$\left[x\arccos{x} - \sqrt{1-x^2}\right]_0^{\frac{1}{2}}$

$\left(\frac{\pi}{6} - \frac{\sqrt{3}}{2}\right) - (-1)$
The answer in the solution manual is:

$= [xarccos(x)]_{0}^{1/2} + \int_{1}^{3/4} t^{\frac{-1}{2}}[\frac{-1}{2}dt]$

Is it wrong? -- notice when it did the u-sub it changed the limits of integration...

4. ## Re: Integration by Parts

Originally Posted by sjmiller
The answer in the solution manual is:

$= [xarccos(x)]_{0}^{1/2} + \int_{1}^{3/4} t^{\frac{-1}{2}}[\frac{-1}{2}dt]$

Is it wrong? -- notice when it did the u-sub it changed the limits of integration...
no, it's not wrong.

$t = 1-x^2$

lower limit, $x = 0$

reset lower limit, $t = 1$

upper limit, $x = \frac{1}{2}$

reset upper limit, $t = \frac{3}{4}$