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Question Number 151423 by peter frank last updated on 21/Aug/21

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{5}} \mathrm{x}} \\$$

Answered by Olaf_Thorendsen last updated on 21/Aug/21

$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\mathrm{tan}^{\mathrm{5}} {x}} \\$$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}^{\mathrm{5}} {x}}{\mathrm{cos}^{\mathrm{5}} {x}+\mathrm{sin}^{\mathrm{5}} {x}}\:{dx}\:\:\:\:\left(\mathrm{1}\right) \\$$$$\mathrm{Let}\:{u}\:=\:\frac{\pi}{\mathrm{2}}−{x}\:: \\$$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{\mathrm{5}} {u}}{\mathrm{sin}^{\mathrm{5}} {u}+\mathrm{cos}^{\mathrm{5}} {u}}\:{du}\:\:\:\:\left(\mathrm{2}\right) \\$$$$\frac{\left(\mathrm{1}\right)+\left(\mathrm{2}\right)}{\mathrm{2}}\::\:\mathrm{I}\:=\:\frac{\pi}{\mathrm{4}} \\$$

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