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Question Number 132610 by liberty last updated on 15/Feb/21

$$\mathrm{\Omega }=\int \frac{\sin {}^2\left(\mathrm{x}\right)}{1+\sin {}^2\left(\mathrm{x}\right)}\mathrm{dx}$$

Answered by Dwaipayan Shikari last updated on 15/Feb/21

$$\int \frac{{sin}^{2}x}{1+{sin}^{2}x}dx=x-\int \frac{1}{1+{sin}^{2}x}dx$$$$=x-\int \frac{{cosec}^{2}x}{{cot}^{2}x+2}dx=x-\frac{1}{\sqrt{2}}{tan}^{-1}\frac{cotx}{\sqrt{2}}+C$$

Answered by mathmax by abdo last updated on 15/Feb/21

$$\mathrm{\Phi }=\int \frac{{\sin }^2\mathrm{x}}{1+{\sin }^2\mathrm{x}}\mathrm{dx}\Rightarrow \mathrm{\Phi }=\int \frac{\frac{1-\cos \left(2\mathrm{x}\right)}{2}}{1+\frac{1-\cos \left(2\mathrm{x}\right)}{2}}\mathrm{dx}$$$$=\int \frac{1-\cos \left(2\mathrm{x}\right)}{3-\cos \left(2\mathrm{x}\right)}\mathrm{dx}{=}_{2\mathrm{x}=\mathrm{t}}\frac{1}{2}\int \frac{1-\mathrm{cost}}{3-\mathrm{cost}}\mathrm{dt}$$$${=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{y}}\frac{1}{2}\int \frac{1-\frac{1-{\mathrm{y}}^2}{1+{\mathrm{y}}^2}}{3-\frac{1-{\mathrm{y}}^2}{1+{\mathrm{y}}^2}}\frac{2\mathrm{dy}}{1+{\mathrm{y}}^2}=\int \frac{1+{\mathrm{y}}^2-1+{\mathrm{y}}^2}{3+{3\mathrm{y}}^2-1+{\mathrm{y}}^2}\mathrm{dy}$$$$=\int \frac{{2\mathrm{y}}^2}{2+{4\mathrm{y}}^2}\mathrm{dy}=\int \frac{{\mathrm{y}}^2}{1+{2\mathrm{y}}^2}\mathrm{dy}=\frac{1}{2}\int \frac{{2\mathrm{y}}^2+1-1}{1+{2\mathrm{y}}^2}\mathrm{dy}$$$$=\frac{\mathrm{y}}{2}-\frac{1}{2}\int \frac{\mathrm{dy}}{1+{2\mathrm{y}}^2}(\to \mathrm{t}=\sqrt{2}\mathrm{y})$$$$=\frac{\mathrm{y}}{2}-\frac{1}{2}\int \frac{\mathrm{dt}}{\sqrt{2}(1+{\mathrm{t}}^2)}=\frac{\mathrm{y}}{2}-\frac{1}{2\sqrt{2}}\arctan \left(\sqrt{2}\mathrm{y}\right)+\mathrm{C}$$$$=\frac{1}{2}\tan \left(\mathrm{x}\right)-\frac{1}{2\sqrt{2}}\arctan \left(\sqrt{2}\tan \left(\mathrm{x}\right)\right)+\mathrm{C}$$

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