Question 697886
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,{x\,\arccos(x)\,dx}]


Use Integration by Parts:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,u\,dv\ =\ uv\ -\ \int\,v\,du]


Let *[tex \LARGE \ \ u\ =\ \arccos(x)]


Then from any standard table of derivatives: *[tex \LARGE \ \ \frac{du}{dx}\ =\ -\frac{1}{\sqrt{1\ -\ x^2}}]


So *[tex \LARGE \ \ du\ =\ -\frac{1}{\sqrt{1\ -\ x^2}}\,dx]


And *[tex \LARGE \ \ dv\ =\ x\,dx]


So *[tex \LARGE \ \ v\ =\ \int\,dv\ =\ \int\,x\,dx\ =\ \frac{x^2}{2}\ +\ C]


Then substituting into the Integration by Parts formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,{x\,\arccos(x)\,dx}\ =\ \frac{1}{2}x^2\,\arccos(x)\ +\ \frac{1}{2}\int\,\frac{x^2}{\sqrt{1\,-\,x^2}}\,dx]


To calculate the remaining integral, let *[tex \LARGE \ \ x\ =\ \sin\varphi]


Then *[tex \LARGE \ \ \frac{dx}{d\varphi}\ =\ \cos\varphi]


Then *[tex \LARGE \ \ dx\ =\ \cos\varphi\,d\varphi]


And then *[tex \LARGE \ \ \int\,\frac{x^2}{\sqrt{1\,-\,x^2}}\,dx\ =\ \int\,\frac{\sin^2\varphi\,\cos\varphi}{\sqrt{1\,-\,\sin^2\varphi}}\,d\varphi]


But since *[tex \LARGE \ \ \sqrt{1\ -\ \sin^2\theta}\ =\ \cos\theta],


*[tex \LARGE \ \ \int\,\frac{x^2}{\sqrt{1\,-\,x^2}}\,dx\ =\ \int\,\sin^2\varphi\,d\varphi]


Rewrite:  *[tex \LARGE \ \ \sin^2\varphi\ =\ \frac{1}{2}\left(1\ -\ \cos{2\varphi}\right)]



*[tex \LARGE \ \ \int\,\sin^2\varphi\,d\varphi\ =\ \frac{1}{2}\int\,\left(1\ -\ \cos{2\varphi}\right)\,d\varphi]


*[tex \LARGE \ \ =\ \frac{1}{2}\int\,d\varphi\ -\ \frac{1}{2}\int\,\cos{2\varphi}\,d\varphi\ =\ ]


Then since *[tex \LARGE \ \ x\ =\ sin\varphi], *[tex \LARGE \ \ \varphi\ =\ \arcsin(x)] and *[tex \LARGE \ \ \cos\varphi\ =\ \sqrt{1\ -\ \sin^2\varphi}\ =\ \sqrt{1\ -\ x^2}], so:


*[tex \LARGE \ \ \frac{1}{2}\int\,d\varphi\ -\ \frac{1}{2}\int\,2\sin\varphi\cos\varphi\,d\varphi\ =\ \frac{1}{2}\arcsin(x)\ -\ \frac{1}{2}x\sqrt{1\,-\,x^2}\ +\ C]


Finally, putting the pieces back together:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,{x\,\arccos(x)\,dx}\ =\ \frac{1}{2}x^2\arccos(x)\ +\ \frac{1}{4}\arcsin(x)\ -\ \frac{1}{4}x\sqrt{1\,-\,x^2}\ +\ C]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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