1. ## Integration by Parts

Hello,

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

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

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

Then:

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

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 ...

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

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

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

3. ## Re: Integration by Parts

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

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

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

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

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

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:

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

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

$\displaystyle t = 1-x^2$

lower limit, $\displaystyle x = 0$

reset lower limit, $\displaystyle t = 1$

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

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