Question Number 127355 by bemath last updated on 29/Dec/20
$$\:\int\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=?\: \\ $$
Answered by liberty last updated on 29/Dec/20
![I=∫ (dx/((1+x^2 )(√(1−x^2 )))) ; [ x = sin h ] I =∫ (dh/(1+sin^2 h)) = ∫ (dh/(sin^2 h(1+csc^2 h))) I= ∫ ((csc^2 h )/(cot^2 h +2)) dh = −(1/( (√2))) tan^(−1) (((cot h)/( (√2))))+c I= −(1/( (√2))) tan^(−1) (((√(1−x^2 ))/(x(√2)))) + c](https://www.tinkutara.com/question/Q127356.png)
$$\:{I}=\int\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:;\:\left[\:{x}\:=\:\mathrm{sin}\:{h}\:\right]\: \\ $$$$\:{I}\:=\int\:\frac{{dh}}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {h}}\:=\:\int\:\frac{{dh}}{\mathrm{sin}\:^{\mathrm{2}} {h}\left(\mathrm{1}+\mathrm{csc}^{\mathrm{2}} \:{h}\right)} \\ $$$$\:{I}=\:\int\:\frac{\mathrm{csc}^{\mathrm{2}} \:{h}\:}{\mathrm{cot}\:^{\mathrm{2}} {h}\:+\mathrm{2}}\:{dh}\:=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{cot}\:{h}}{\:\sqrt{\mathrm{2}}}\right)+{c} \\ $$$$\:{I}=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}\sqrt{\mathrm{2}}}\right)\:+\:{c}\: \\ $$