7.1 Integration By Parts/27: Difference between revisions
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<math>\int_{0}^{\frac{1}{2}}\cos^{-1}(x)dx</math> = <math>x\cos^{-1}(x)\bigg|_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}}\frac{x}{\sqrt{1-x^2}}dx</math> = <math>(\frac{1}{2}\cdot\frac{\pi}{3}){-\frac{1}{2}}\int_{1}^{\frac{3}{4}}\frac{1}{\sqrt{u}}du</math> = <math>{\frac{\pi}{6}}-\frac{1}{2}\int_{1}^{\frac{3}{4}}u^{-\frac{1}{2}}du</math> | <math>\int_{0}^{\frac{1}{2}}\cos^{-1}(x)dx</math> = <math>x\cos^{-1}(x)\bigg|_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}}\frac{x}{\sqrt{1-x^2}}dx</math> = <math>(\frac{1}{2}\cdot\frac{\pi}{3}){-\frac{1}{2}}\int_{1}^{\frac{3}{4}}\frac{1}{\sqrt{u}}du</math> = <math>{\frac{\pi}{6}}-\frac{1}{2}\int_{1}^{\frac{3}{4}}u^{-\frac{1}{2}}du</math> | ||
= <math>{\frac{\pi}{6}}+{(\frac{1}{2}}(2u^{\frac{1}{2}}))\bigg|_{\frac{3}{4}}^{1}</math> | = [change the sign and flip the limits] <math>{\frac{\pi}{6}}+{(\frac{1}{2}}(2u^{\frac{1}{2}}))\bigg|_{\frac{3}{4}}^{1}</math> = <math>\frac{\pi}{6}-\frac{1}{2}(\sqrt{3}-1)</math> = <math>\frac{\pi}{6}-\frac{\sqrt{3}}{2}+1</math> = <math>\frac{1}{6}(\pi+6-3\sqrt{3})</math> | ||
Revision as of 01:19, 27 November 2022
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