Question Number 215775 by ajfour last updated on 17/Jan/25
$$\:\:\underset{\mathrm{0}} {\overset{\:\:{s}} {\int}}\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\left(\mathrm{1}+\sqrt{\mathrm{1}−\left(\frac{{x}}{{s}}\right)^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$
Commented by ajfour last updated on 18/Jan/25
https://youtu.be/ZcHJ_FnQ6Fk?si=r5hLYfUC_lLg_ccq
Answered by mr W last updated on 18/Jan/25
![∫(√(1−x^2 ))dx=(1/2)(sin^(−1) x+x(√(1−x^2 )))+C ∫_0 ^s (√(1−x^2 ))dx=(1/2)(sin^(−1) s+s(√(1−s^2 ))) ∫_0 ^s (√(1−((x/s))^2 ))dx=s∫_0 ^1 (√(1−t^2 ))dt=(s/2)((π/2))=((sπ)/4) ∫_0 ^( s) (√(1−s^2 ))(1+(√(1−((x/s))^2 ))−(√(1−x^2 )))dx =(√(1−s^2 ))[s+((sπ)/4)−(1/2)(sin^(−1) s+s(√(1−s^2 )))] ✓](Q215795.png)
$$\int\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}^{−\mathrm{1}} {x}+{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)+{C} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{{s}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}^{−\mathrm{1}} {s}+{s}\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\right) \\ $$$$\int_{\mathrm{0}} ^{{s}} \sqrt{\mathrm{1}−\left(\frac{{x}}{{s}}\right)^{\mathrm{2}} }{dx}={s}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}=\frac{{s}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\right)=\frac{{s}\pi}{\mathrm{4}} \\ $$$$\:\:\underset{\mathrm{0}} {\overset{\:\:{s}} {\int}}\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\left(\mathrm{1}+\sqrt{\mathrm{1}−\left(\frac{{x}}{{s}}\right)^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$$$=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\left[{s}+\frac{{s}\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}^{−\mathrm{1}} {s}+{s}\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\right)\right]\:\checkmark \\ $$