Question Number 46657 by Necxx last updated on 29/Oct/18
$${integrte}\:\mathrm{sin}^{−\mathrm{1}} {x} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18
![∫sin^(−1) x sin^(−1) x×∫dx−∫[((dsin^(−1) x)/dx)∫dx]dx xsin^(−1) x−∫(x/(√(1−x^2 )))dx t^2 =1−x^2 2tdt=−2xdx so∫((xdx)/(√(1−x^2 ))) ∫((−tdt)/t).=−t=−(√(1−x^2 )) ans is xsin^(−1) x+(√(1−x^2 ))](Q46662.png)
$$\int{sin}^{−\mathrm{1}} {x} \\ $$$${sin}^{−\mathrm{1}} {x}×\int{dx}−\int\left[\frac{{dsin}^{−\mathrm{1}} {x}}{{dx}}\int{dx}\right]{dx} \\ $$$${xsin}^{−\mathrm{1}} {x}−\int\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${t}^{\mathrm{2}} =\mathrm{1}−{x}^{\mathrm{2}} \:\:\:\mathrm{2}{tdt}=−\mathrm{2}{xdx} \\ $$$${so}\int\frac{{xdx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\int\frac{−{tdt}}{{t}}.=−{t}=−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\: \\ $$$${ans}\:{is} \\ $$$${xsin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\: \\ $$$$ \\ $$
Commented by Necxx last updated on 29/Oct/18
$${thanks}\:{boss} \\ $$