thanks FlorianK
any ideas on how to do the nasty integral? one of my mistakes was assuming |sin(x)cos(x)|sin(x)cos(x) is the same as sin(x)^2*cos(x)^2...that's the last thing i integrated before changing variables and getting the above erroneous answer
That took a while, hopefully there are not errors in there...
Anyways, the train of thought, try to work on the
to make it into something that we can doa trig substitution with. Do the trig sub, using the right substitution so that we get something else out that we can work with, here that is
)
. Use the double angle formula for cos to bring it down to something we can integrate, then expand it out and split it up, working on each part. Start subbing back through the variables, and when we encounter the cos inside the sin and such, draw out a triangle to find the equivalent expression. Then keep subbing back through the variables to finish it off.
^{2}-\left(\frac{1}{2}\right)^{2}\right)<br />\\ & =\frac{1}{4}-\left(x-\frac{1}{2}\right)^{2}<br />\\ \int\sqrt{x-x^{2}}dx & =\int\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^{2}}dx<br />\\ \mathrm{Let}\: u=x-\frac{1}{2}<br />\\ \frac{du}{dx}=1<br />\\ \int\sqrt{x-x^{2}}dx & =\int\sqrt{\frac{1}{4}-u^{2}}du<br />\\ \mathrm{Let\:}u=\frac{1}{2}\sin(a)<br />\\ \frac{du}{da}=\frac{1}{2}\cos(a)<br />\\ \sqrt{\frac{1}{4}-u^{2}} & =\sqrt{\frac{1}{4}-\frac{1}{4}\sin^{2}(a)}<br />\\ & =\frac{1}{2}\cos(a)<br />\\ \int\sqrt{x-x^{2}}dx & =\int\frac{1}{2}\cos(a)\times\frac{1}{2}\cos(a)da<br />\\ & =\frac{1}{4}\int\cos^{2}(a)da<br />\\ \mathrm{Using\:} & \cos^{2}(a)=\frac{1+\cos(2a)}{2}<br />\\ \int\sqrt{x-x^{2}}dx & =\frac{1}{4}\int\frac{1+\cos(2a)}{2}da<br />\\ & =\frac{1}{4}\left(\frac{1}{2}a+\frac{1}{4}\sin(2a)\right)+C<br />\\ a=\sin^{-1}\left(2u\right)<br />\\ & =\frac{1}{4}\left(\frac{1}{2}\sin^{-1}\left(2u\right)+\frac{1}{4}\sin(2\sin^{-1}\left(2u\right))\right)+C<br />\\ & =\frac{1}{8}\sin^{-1}\left(2u\right)+\frac{1}{16}\times2\sin\left(\sin^{-1}\left(2u\right)\right)\cos\left(\sin^{-1}\left(2u\right)\right)+C<br />\\ \mathrm{Draw\: the\: triagnle\: out\: with\: angle\:}a,\:\sin a=\frac{2u}{1},\:\mathrm{pythagoras\: theorem}<br />\\ & =\frac{1}{8}\sin^{-1}\left(2u\right)+\frac{1}{4}u\sqrt{1-4u^{2}}<br />\\ u=x-\frac{1}{2} & =\frac{1}{8}\sin^{-1}\left(2x-1\right)+\frac{1}{4}\left(\frac{2x-1}{2}\right)\sqrt{1-4\left(\frac{2x-1}{2}\right)^{2}}+C<br />\\ & =\frac{1}{8}\sin^{-1}\left(2x-1\right)+\frac{2x-1}{8}\sqrt{1-4x^{2}+4x-1}+C<br />\\ & =\frac{1}{8}\sin^{-1}\left(2x-1\right)+\frac{1}{8}\left(2x-1\right)\sqrt{4\left(x-x^{2}\right)}+C<br />\\ & =\frac{1}{8}\sin^{-1}\left(2x-1\right)+\frac{1}{4}\left(2x-1\right)\sqrt{x-x^{2}}+C<br />\end{alignedat}<br />)
There will be domain restrictions that arise when you manipulate the triangle aswell, but anyways... after typing all that out... Anyways, hope it helps
Also had a feeling that hyperbolic functions may have helped at one stage... but looking back it it maybe not.
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