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Question Number 42593 by aseerimad last updated on 28/Aug/18

Commented by maxmathsup by imad last updated on 28/Aug/18

$$wehaveI={\int }_0^{\frac{\pi }{2}}\frac{{cos}^{4}x+{sin}^{4}x-{sin}^{4}x}{{sin}^{4}x+{cos}^{4}x}dx=\frac{\pi }{2}-{\int }_0^{\frac{\pi }{2}}\frac{{sin}^{4}x}{{sin}^{4}x+{cos}^{4}x}dxbut$$$$changementx=\frac{\pi }{2}-tgive{\int }_0^{\frac{\pi }{2}}\frac{{sin}^{4}x}{{sin}^{4}x+{cos}^{4}x}dx={\int }_0^{\frac{\pi }{2}}\frac{{cos}^{4}t}{{cos}^{4}x+{sin}^{4}x}dt$$$$=I\Rightarrow 2I=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Aug/18

$$I={\int }_0^{\frac{\mathrm{\Pi }}{2}}f\left(x\right)dx={\int }_0^{\frac{\mathrm{\Pi }}{4}}f(\frac{\mathrm{\Pi }}{2}-x)dx$$$$I={\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{{cos}^{4}x}{{sin}^{4}x+{cos}^{4}x}dx={\int }_0^{\frac{\mathrm{\Pi }}{4}}\frac{{cos}^4(\frac{\mathrm{\Pi }}{2}-x)}{{sin}^4(\frac{\mathrm{\Pi }}{2}-x)+{cos}^4(\frac{\mathrm{\Pi }}{2}-x)}dx$$$$I={\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{{sin}^{4}x}{{cos}^{4}x+{sin}^{4}x}dx$$$$2I={\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{{cos}^{4}x+{sin}^{4}x}{{cos}^{4}x+{sin}^{4}x}dx={\int }_0^{\frac{\mathrm{\Pi }}{2}}dx$$$$2I=\mid x{\mid }_0^{\frac{\mathrm{\Pi }}{2}}=\frac{\mathrm{\Pi }}{2}$$$$I=\frac{\mathrm{\Pi }}{4}$$$$$$

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