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Question Number 173497 by SIENSE last updated on 12/Jul/22

$$$$$$BonusduMardi12/07/2022$$$$I={\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{{sin}^2\left(x\right)}{1+{cos}^2\left(x\right)}dx=?$$$$J={\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{dx}{1+{cos}^2\left(x\right)}$$$$I={\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{2-(1+{cos}^2\left(x\right)}{1+{cos}^2\left(x\right)}dx$$$$$$$$=2{\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{dx}{1+{cos}^2\left(x\right)}-{\int }_0^{\frac{\mathrm{\Pi }}{2}}dx.$$$$=2J-\frac{\mathrm{\Pi }}{2}$$$$$$$$\overline{)\begin{array}{c}J={\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{dx}{1+{cos}^2\left(x\right)}\\ I=2J-\frac{\mathrm{\Pi }}{2}\end{array}}$$$$e$$

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