Question Number 185471 by liuxinnan last updated on 22/Jan/23
$$\int_{\:\mathrm{0}} ^{\:{r}} \frac{\mathrm{1}}{\:\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{dx}=? \\ $$
Answered by hmr last updated on 22/Jan/23
![= [sin^(−1) ((x/r))]_( 0) ^( r) = (sin^(−1) ((r/r))) − (sin^(−1) ((0/r))) = (π/2) − 0 = (π/2)](https://www.tinkutara.com/question/Q185476.png)
$$=\:\left[{sin}^{−\mathrm{1}} \:\left(\frac{{x}}{{r}}\right)\underset{\:\:\mathrm{0}} {\overset{\:\:{r}} {\right]}} \\ $$$$=\:\left({sin}^{−\mathrm{1}} \:\left(\frac{{r}}{{r}}\right)\right)\:−\:\left({sin}^{−\mathrm{1}} \:\left(\frac{\mathrm{0}}{{r}}\right)\right) \\ $$$$=\:\frac{\pi}{\mathrm{2}}\:−\:\mathrm{0}\:=\:\frac{\pi}{\mathrm{2}} \\ $$
Answered by mr W last updated on 22/Jan/23
$${let}\:{x}={r}\:\mathrm{sin}\:\theta \\ $$$${dx}={r}\:\mathrm{cos}\:\theta\:{d}\theta \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{r}\:\mathrm{cos}\:\theta\:{d}\theta}{{r}\:\mathrm{cos}\:\theta}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {d}\theta=\frac{\pi}{\mathrm{2}} \\ $$