7.1 Integration By Parts/10 - Revision history

Kattieh70488@students.laalliance.org at 01:49, 27 November 2022

2022-11-27T01:49:55Z

← Older revision		Revision as of 01:49, 27 November 2022
Line 2:		Line 2:
	<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>\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}+C</math>		= <math>x\sin^{-1}(x)+\sqrt{u}</math> = <math>x\sin^{-1}(x)+\sqrt{1-x^2}+C</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>{\sin^{-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>

Kattieh70488@students.laalliance.org at 01:48, 27 November 2022

2022-11-27T01:48:59Z

← Older revision		Revision as of 01:48, 27 November 2022
Line 1:		Line 1:
	<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}}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}+C</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}+C</math>

Kattieh70488@students.laalliance.org at 01:48, 27 November 2022

2022-11-27T01:48:44Z

← Older revision		Revision as of 01:48, 27 November 2022
Line 1:		Line 1:
	<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}}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}+C</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}+C</math>

Kattieh70488@students.laalliance.org at 01:48, 27 November 2022

2022-11-27T01:48:20Z

← Older revision		Revision as of 01:48, 27 November 2022
Line 1:		Line 1:
	<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}}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>		<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}+C</math>

Kattieh70488@students.laalliance.org at 01:47, 27 November 2022

2022-11-27T01:47:21Z

← Older revision		Revision as of 01:47, 27 November 2022
Line 1:		Line 1:
	<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}}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>		<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>

Kattieh70488@students.laalliance.org at 01:46, 27 November 2022

2022-11-27T01:46:42Z

← Older revision		Revision as of 01:46, 27 November 2022
Line 1:		Line 1:
	<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>

Kattieh70488@students.laalliance.org: Created page with " $f'(x)= \int_{}^{}\sin^{-1}(x)\cdot dx$

$\int_{}^{}\sin^{-1}(x)dx$ = $x\sin^{-1}(x)-\int_{}^{}\frac{x}{\sqrt{1-x^2}}dx$ = $x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}\frac{1}{\sqrt{u}}$ = $x\sin^{-1}(x)+\frac{1}{2}\int_{}^{}u^{-\frac{1}{2}}du$ = ${\frac{\pi}{6}}-\frac{1}{2}\int_{1}^{\frac{3}{4}}u^{-\frac{1}{2}}du$ "

2022-11-27T01:43:52Z

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>"

New page