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>{-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> | |||
= <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> | |||
Revision as of 23:11, 25 November 2022
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