1. Să se calculeze integralele:
a) ∫ d x 1 − x 2 , x ∈ [ − 1 , 1 ] ; {\displaystyle \int {\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}\;,\;x\in [-1,1];} b) ∫ x 1 − x 2 d x , x ∈ [ − 1 , 1 ] . {\displaystyle \int {\frac {x}{\sqrt {1-x^{2}}}}\mathrm {d} x\;,\;x\in [-1,1].}
R. a) Se face schimbarea de variabilă: x = cos t , {\displaystyle x=\cos t,} deci d x = − sin t d t . {\displaystyle \mathrm {d} x=-\sin t\mathrm {d} t.}
Mai departe:
S-a ţinut cont că:
b) Se face schimbarea de variabilă: 1 − x 2 = t . {\displaystyle {\sqrt {1-x^{2}}}=t.}
Avem:
{\displaystyle } {\displaystyle } {\displaystyle } {\displaystyle }
![{\displaystyle \int {\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}\;,\;x\in [-1,1];}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3af65a070b33b4a16581df17b9064aa34f75c71e)
b) ![{\displaystyle \int {\frac {x}{\sqrt {1-x^{2}}}}\mathrm {d} x\;,\;x\in [-1,1].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1719b28f2e4a2f0060d0708e7cc111c46a804e5a)
deci 
![{\displaystyle \arcsin x+\arccos x={\frac {\pi }{2}}\;,\;\forall x\in [-1,1].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e90998e849f643ae527e45de1e4f724a3c41be46)
