could someone please help me with the integration of cos-1x and tan-1x?
i know that we have to draw the cos and tan graphs and somehow calculate the areas through the method of integration by parts
thankyouuuu !
For example: $\displaystyle u=\arccos x\Longrightarrow\,u'=-\frac{1}{\sqrt{1-x^2}}\,,\,\,v'=1\Longrightarrow\,v=x$ , so integrating by parts:
$\displaystyle \int\arccos x\,dx=x\arccos x+\int\frac{x}{\sqrt{1-x^2}}\,dx=x\arccos x-\frac{1}{2}\int\frac{-2x}{\sqrt{1-x^2}}\,dx=$ $\displaystyle x\arccos x-\sqrt{1-x^2}+C$
Now you do the other one following the lines above.
Tonio