7.1 Integration By Parts/27: Difference between revisions

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<math> f'(x)= \int_{0}^{\frac{1}{2}}\cos^{-1}(x)\cdot dx </math> <br><br>
<math> f'(x)= \int_{0}^{\frac{1}{2}}\cos^{-1}(x)\cdot dx </math> <br><br>
<math>\int_{0}^{\frac{1}{2}}\cos^{-1}(x)\cdot dx</math> = <math>\int_{0}^{\pi}e^{cost}(2sintcost)dt</math> = <math>{-2}\int_{cos0}^{cos\pi}e^{u}{u}du</math> = <math>{-2}{u}e^{u}\bigg|_{cos0}^{cos\pi}-(-2)\int_{cos0}^{cos\pi}e^{u}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>{-2}{u}e^{u}\bigg|_{cos0}^{cos\pi}+{2}e^{u}\bigg|_{cos0}^{cos\pi} du</math> = <math>{2}{u}e^{u}\bigg|_{cos\pi}^{cos0}-{2}e^{u}\bigg|_{cos\pi}^{cos0}du</math> = <math>{2}{cos(0)}e^{cos(0)}-{2}{cos(\pi)}e^{cos(\pi)}-{2}e^{cos(0)}+{2}e^{cos(\pi)}</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}(2-\sqrt{3})</math> = <math>\frac{\pi}{6}-\frac{\sqrt{3}}{2}+1</math> = <math>\frac{1}{6}(\pi+6-3\sqrt{3})</math>


= <math>{2}(1)e^{1}-{2}(-1)e^{-1}-{2}e^{1}+{2}e^{-1}</math> = <math>{2}e^{1}+{2}e^{-1}-{2}e^{1}+{2}e^{-1}</math> = <math>{2}e^{-1}+{2}e^{-1}</math> = <math> {4}e^{-1} </math>
 
<math>{u}</math> = <math>{1-x^2}</math>
 
<math>{du}</math> = <math>{-2x}</math>
 
<math>{-\frac{1}{2}du}</math> = <math>{x}dx</math>
 
 
<math>{u}</math> = <math>{\cos^{-1}(x)}</math> ,  <math>{dv}</math> = <math>dx</math>
 
<math>{du}</math> = <math>{-\frac{1}{\sqrt{1-x^2}}dx}</math> ,  <math>{v}</math> = <math>{x}</math>

Latest revision as of 01:25, 27 November 2022



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