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| \begin{align} | | \begin{align} |
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| g(x)=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{6}{\sqrt{1-t^2}} dt \\[2ex]
| | \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] |
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| g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg|_{1/2}^{\sqrt{3}/2}
| | &=6\arcsin{(x)}\bigg|_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \\[2ex] |
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| =6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2)) | | &=\left[6\arcsin\left(\frac{\sqrt{3}}{2}\right)\right]-\left[6\arcsin{\left(\frac{1}{2}\right)}\right] \\[2ex] |
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| =\pi | | &=\left[6\cdot\frac{\pi}{3}\right]-\left[6\cdot\frac{\pi}{6}\right] = 2\pi-\pi \\[2ex] |
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| | &=\pi |
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| | \end{align} |
| </math> | | </math> |
Revision as of 21:58, 6 September 2022
![{\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)