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Question Number 36892 by anik last updated on 06/Jun/18

$$2.\int \left[\sqrt{(1-x^2)/(1+x^2)}\right]dx=?$$

Commented by math khazana by abdo last updated on 11/Jun/18

$$letI=\int \sqrt{\frac{1-x^2}{1+x^2}}dxchangementx=sintgive$$$$I=\int \frac{cost}{\sqrt{2{cos}^{2}t-1}}costdt$$$$=\int \frac{{cos}^{2}t}{\sqrt{2{cos}^{2}t-1}}dt=\int \frac{1}{(1+{tan}^{2}t)\sqrt{2\frac{1}{1+{tan}^{2}t}-1}}dt$$$$=\int \frac{dt}{(1+{tan}^{2}t)\sqrt{2-1-{tan}^{2}t}}\sqrt{1+{tan}^{2}t}dt$$$$=\int \frac{dt}{\sqrt{1+{tan}^{2}t}\sqrt{1-{tan}^{2}t}}changement$$$$tant=ugive$$$$I=\int \frac{1}{\sqrt{1+u^2}\sqrt{1-u^2}}\frac{du}{1+u^2}$$$$=\int \frac{du}{(1+u^2)\sqrt{1-u^4}}....becontinued....$$$$$$

Answered by anik last updated on 06/Jun/18

Commented by anik last updated on 06/Jun/18

$$\mathrm{Indrajit}\mathrm{Karmakar}\mathrm{have}\mathrm{solved}\mathrm{this}.$$

Commented by MJS last updated on 06/Jun/18

$$\int \frac{{\cos }^2\theta }{\sqrt{2-{\cos }^2\theta }}d\theta =\int \frac{d\theta }{\sec \theta \sqrt{{2\sec }^2\theta -1}}$$$$\mathrm{so}\mathrm{something}\mathrm{went}\mathrm{wrong}...$$

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