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

$${\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{1+\tan {}^5\mathrm{x}}$$

Answered by Olaf_Thorendsen last updated on 21/Aug/21

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{\tan }^{5}x}$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{5}x}{{\cos }^{5}x+{\sin }^{5}x}dx\left(1\right)$$$$\mathrm{Let}u=\frac{\pi }{2}-x:$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{5}u}{{\sin }^{5}u+{\cos }^{5}u}du\left(2\right)$$$$\frac{\left(1\right)+\left(2\right)}{2}:\mathrm{I}=\frac{\pi }{4}$$

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