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Question Number 34283 by math khazana by abdo last updated on 03/May/18

$$calculate{\int }_0^{\frac{\pi }{2}}\frac{dx}{{cos}^{4}x+{sin}^{4}x}$$

Commented by math khazana by abdo last updated on 08/May/18

$$letputI={\int }_0^{\frac{\pi }{2}}\frac{dx}{{cos}^{4}x+{sin}^{4}x}$$$$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{{({cos}^{2}x+{sin}^{2}x)}^2-2{cos}^{2}x{sin}^{2}x}$$$$={\int }_0^{\frac{\pi }{2}}\frac{dx}{1^2-2{cos}^{2}x.{sin}^{2}x}$$$$={\int }_0^{\frac{\pi }{2}}\frac{dx}{1-2\frac{1+cos\left(2x\right)}{2}.\frac{1-cos\left(2x\right)}{2}}$$$$={\int }_0^{\frac{\pi }{2}}\frac{dx}{1-\frac{1}{2}(1-{cos}^2\left(2x\right))}$$$$={\int }_0^{\frac{\pi }{2}}\frac{2dx}{2-1+{cos}^2\left(2x\right)}$$$$={\int }_0^{\frac{\pi }{2}}\frac{2dx}{1+\frac{1+cos\left(4x\right)}{2}}={\int }_0^{\frac{\pi }{2}}\frac{4dx}{3+\cos \left(4\mathrm{x}\right)}$$$$={}_{4\mathrm{x}=\mathrm{t}}{\int }_0^{2\pi }\frac{dt}{3+cost}changement{e}^{it}=zgive$$$$I={\int }_{\mid z\mid =1}\frac{1}{3+\frac{z+z^{-1}}{2}}\frac{dz}{iz}$$$$={\int }_{\mid z\mid =1}\frac{2dz}{iz(6+z+z^{-1})}={\int }_{\mid z\mid =1}\frac{-2i}{6z+z^2+1}dz$$$$letintroducethecomplexfunction$$$$\phi \left(z\right)=\frac{-2i}{z^2+6z+1}.polesof\phi ?$$$$$$

Commented by math khazana by abdo last updated on 08/May/18

$${\mathrm{\Delta }}^{{}^{\prime }}=3^2-1=8\Rightarrow z_1=-3+2\sqrt{2}$$$$z_2=-3-2\sqrt{2}$$$$\mid z_1\mid -1=2\sqrt{2}-3-1=2\sqrt{2}-4<0$$$$\mid z_2\mid -1=3+2\sqrt{2}-1=2+2\sqrt{2}>0(toeliminate$$$$fromresidus)$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_1)$$$$Res(\phi ,z_1)=\frac{-2i}{(z_1-z_2)}=\frac{-2i}{4\sqrt{2}}=\frac{-i}{2\sqrt{2}}$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi .\frac{-i}{2\sqrt{2}}=\frac{\pi }{\sqrt{2}}$$$$I=\frac{\pi }{\sqrt{2}}.$$

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