∫ 1 2 3 2 6 1 − t 2 d t = 6 ∫ 1 2 3 2 1 1 − t 2 d t = 6 arcsin ( x ) | 1 2 3 2 = [ 6 arcsin ( 3 2 ) ] − [ 6 arcsin ( 1 2 ) ] = [ 6 ⋅ π 3 ] − [ 6 ⋅ π 6 ] = 2 π − π = π {\displaystyle {\begin{aligned}\int _{\frac {1}{2}}^{\frac {\sqrt {3}}{2}}{\frac {6}{\sqrt {1-t^{2}}}}\,dt&=6\int _{\frac {1}{2}}^{\frac {\sqrt {3}}{2}}{\frac {1}{\sqrt {1-t^{2}}}}\,dt\\[2ex]&=6\arcsin {(x)}{\bigg |}_{\frac {1}{2}}^{\frac {\sqrt {3}}{2}}\\[2ex]&=\left[6\arcsin \left({\frac {\sqrt {3}}{2}}\right)\right]-\left[6\arcsin {\left({\frac {1}{2}}\right)}\right]\\[2ex]&=\left[6\cdot {\frac {\pi }{3}}\right]-\left[6\cdot {\frac {\pi }{6}}\right]=2\pi -\pi \\[2ex]&=\pi \end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{\frac {1}{2}}^{\frac {\sqrt {3}}{2}}{\frac {6}{\sqrt {1-t^{2}}}}\,dt&=6\int _{\frac {1}{2}}^{\frac {\sqrt {3}}{2}}{\frac {1}{\sqrt {1-t^{2}}}}\,dt\\[2ex]&=6\arcsin {(x)}{\bigg |}_{\frac {1}{2}}^{\frac {\sqrt {3}}{2}}\\[2ex]&=\left[6\arcsin \left({\frac {\sqrt {3}}{2}}\right)\right]-\left[6\arcsin {\left({\frac {1}{2}}\right)}\right]\\[2ex]&=\left[6\cdot {\frac {\pi }{3}}\right]-\left[6\cdot {\frac {\pi }{6}}\right]=2\pi -\pi \\[2ex]&=\pi \end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/687cfb7880dbac82492500ce4b3b613ea2951264)