7.1 Integration By Parts/10: Difference between revisions

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(Created page with "<math> f'(x)= \int_{}^{}\sin^{-1}(x)\cdot dx </math> <br><br> <math>\int_{}^{}\sin^{-1}(x)dx</math> = <math>x\sin^{-1}(x)-\int_{}^{}\frac{x}{\sqrt{1-x^2}}dx</math> = <math>x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}\frac{1}{\sqrt{u}}</math> = <math>x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}u^{-\frac{1}{2}}du</math> = <math>{\frac{\pi}{6}}-\frac{1}{2}\int_{1}^{\frac{3}{4}}u^{-\frac{1}{2}}du</math>")
 
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<math> f'(x)= \int_{}^{}\sin^{-1}(x)\cdot dx </math> <br><br>
<math> f'(x)= \int_{}^{}\sin^{-1}(x)\cdot dx </math> <br><br>
<math>\int_{}^{}\sin^{-1}(x)dx</math> = <math>x\sin^{-1}(x)-\int_{}^{}\frac{x}{\sqrt{1-x^2}}dx</math> = <math>x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}\frac{1}{\sqrt{u}}</math> = <math>x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}u^{-\frac{1}{2}}du</math> = <math>{\frac{\pi}{6}}-\frac{1}{2}\int_{1}^{\frac{3}{4}}u^{-\frac{1}{2}}du</math>
<math>\int_{}^{}\sin^{-1}(x)dx</math> = <math>x\sin^{-1}(x)-\int_{}^{}\frac{x}{\sqrt{1-x^2}}dx</math> = <math>x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}\frac{1}{\sqrt{u}}du</math> = <math>x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}u^{-\frac{1}{2}}du</math> = <math>x\sin^{-1}(x)+\frac{1}{2}(2u^{frac{1}{2}}</math> = <math>x\sin^{-1}(x)+\sqrt{u}</math> = <math>x\sin^{-1}(x)+sqrt{1-x^2}</math>

Revision as of 01:46, 27 November 2022



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