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

Bonus du Mardi 12/07/2022 I= ∫_0 ^(Π/2) ((sin^2 (x))/(1+cos^2 (x)))dx = ? J = ∫_0 ^(Π/2) (dx/(1+cos^2 (x))) I = ∫_0 ^(Π/2) ((2−(1+cos^2 (x))/(1+cos^2 (x)))dx = 2∫_0 ^(Π/2) (dx/(1+cos^2 (x)))−∫_0 ^(Π/2) dx . = 2J−(Π/2) determinant (((J=∫_0 ^(Π/2) (dx/(1+cos^2 (x))))),((I=2J−(Π/2)))) e

$$\:\:\: \\ $$$$\:\:{Bonus}\:{du}\:{Mardi}\:\mathrm{12}/\mathrm{07}/\mathrm{2022} \\ $$$$\:\:\:\:{I}=\:\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \:\frac{{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)}{dx}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:{J}\:\:=\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)} \\ $$$$\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{\mathrm{2}−\left(\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)\right.}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)}−\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \:{dx}\:.\:\: \\ $$$$=\:\mathrm{2}{J}−\frac{\Pi}{\mathrm{2}} \\ $$$$ \\ $$$$\begin{array}{|c|c|}{{J}=\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)}}\\{{I}=\mathrm{2}{J}−\frac{\Pi}{\mathrm{2}}}\\\hline\end{array} \\ $$$$\:{e}\: \\ $$

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