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Question Number 87335 by naka3546 last updated on 04/Apr/20

Answered by TANMAY PANACEA. last updated on 04/Apr/20

$${\int }_0^{\frac{\pi }{2}}\frac{{sin}^{a}x}{{sin}^{a}x+{cos}^{a}x}dx=\frac{I\pi }{4}$$$$using{\int }_0^{b}f\left(x\right)dx={\int }_0^{b}f(b-x)dx$$$$\frac{I\pi }{4}={\int }_0^{\frac{\pi }{2}}\frac{{sin}^a(\frac{\pi }{2}-x)}{{sin}^a(\frac{\pi }{2}-x)+{cos}^a(\frac{\pi }{2}-x)}dx$$$$\frac{I\pi }{4}={\int }_0^{\frac{\pi }{2}}\frac{{cos}^{a}x}{{cos}^{a}x+{sin}^{a}x}dx$$$$2\times \frac{I\pi }{4}={\int }_0^{\frac{\pi }{2}}(\frac{{sin}^{a}x}{{cos}^{a}x+{sin}^{a}x}+\frac{{cos}^{a}x}{{sin}^{a}x+{cos}^{a}x})dx$$$$\frac{I\pi }{2}={\int }_0^{\frac{\pi }{2}}dx$$$$I=1$$

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