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Question-189293




Question Number 189293 by Rupesh123 last updated on 14/Mar/23
Answered by Sutrisno last updated on 14/Mar/23
∫_0 ^(π/2) ((sin^2 x)/((1/(sin^2 x))+((cos^2 x)/(sin^2 x))))dx  ∫_0 ^(π/2) ((sin^4 x)/(1+cos^2 x))dx  ∫_0 ^(π/2) (((1−cos^2 x)^2 )/(1+cos^2 x))dx  ∫_0 ^(π/2) ((cos^4 x−2cos^2 x+1)/(1+cos^2 x))dx  ∫_0 ^(π/2) cos^2 x−3+(4/(cos^2 x+1))dx  ∫_0 ^(π/2) (1/2)cos2x−(5/2)+(4/(cos^2 x+1))dx  (1/4)sin2x−(5/2)x+2(√2)tan^(−1) (((tanx)/( (√2))))∣_0 ^(π/2)   (0−((5π)/4)+2(√2).(π/2))−0  (π/4)(4(√2)−5)  (((4(5+4(√2)))/7)).(π/4)(4(√2)−5)  (π/7)(32−25)  π
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}} {x}}{\frac{\mathrm{1}}{{sin}^{\mathrm{2}} {x}}+\frac{{cos}^{\mathrm{2}} {x}}{{sin}^{\mathrm{2}} {x}}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{4}} {x}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{4}} {x}−\mathrm{2}{cos}^{\mathrm{2}} {x}+\mathrm{1}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} {x}−\mathrm{3}+\frac{\mathrm{4}}{{cos}^{\mathrm{2}} {x}+\mathrm{1}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{2}{x}−\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{4}}{{cos}^{\mathrm{2}} {x}+\mathrm{1}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{2}{x}−\frac{\mathrm{5}}{\mathrm{2}}{x}+\mathrm{2}\sqrt{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\frac{{tanx}}{\:\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\left(\mathrm{0}−\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{2}\sqrt{\mathrm{2}}.\frac{\pi}{\mathrm{2}}\right)−\mathrm{0} \\ $$$$\frac{\pi}{\mathrm{4}}\left(\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{5}\right) \\ $$$$\left(\frac{\mathrm{4}\left(\mathrm{5}+\mathrm{4}\sqrt{\mathrm{2}}\right)}{\mathrm{7}}\right).\frac{\pi}{\mathrm{4}}\left(\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{5}\right) \\ $$$$\frac{\pi}{\mathrm{7}}\left(\mathrm{32}−\mathrm{25}\right) \\ $$$$\pi \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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