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Question Number 83824 by Power last updated on 06/Mar/20

Commented by mathmax by abdo last updated on 06/Mar/20

$$changementx=\frac{\pi }{2}-tgive$$$$\mathrm{\Omega }=-{\int }_0^{\frac{\pi }{2}}\frac{\pi (\frac{\pi }{2}-t)+{cos}^{4}t-{sin}^{4}t}{{cos}^{4}t+{sin}^{4}t}(-dt)$$$$=\frac{{\pi }^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{dt}{{cos}^{4}t+{sin}^{4}tdt}+{\int }_0^{\frac{\pi }{2}}\frac{-\pi t+{cos}^{4}t-{sin}^{4}t}{{cos}^{4}t+{sin}^{4}t}dt$$$$=\frac{{\pi }^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{dt}{{cos}^{4}t+{sin}^{4}t}-\mathrm{\Omega }\Rightarrow 2\mathrm{\Omega }=\frac{{\pi }^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{dt}{1-2{cos}^{2}t{sin}^{2}t}\Rightarrow$$$$\mathrm{\Omega }=\frac{{\pi }^2}{4}{\int }_0^{\frac{\pi }{2}}\frac{dt}{1-2{\left(\frac{sin\left(2t\right)}{2}\right)}^2}=\frac{{\pi }^2}{4}{\int }_0^{\frac{\pi }{2}}\frac{dt}{1-\frac{1}{2}{sin}^2\left(2t\right)}$$$$=\frac{{\pi }^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{dt}{2-\frac{1-cos\left(4t\right)}{2}}={\pi }^2{\int }_0^{\frac{\pi }{2}}\frac{dt}{3+cos\left(4t\right)}$$$${=}_{4t=u}{\pi }^2{\int }_0^{2\pi }\frac{1}{3+cosu}\frac{du}{4}=\frac{{\pi }^2}{4}{\int }_0^{2\pi }\frac{du}{3+cosu}changement{e}^{iu}=z\Rightarrow$$$${\int }_0^{2\pi }\frac{du}{3+cosu}={\int }_{\mid z\mid =1}\frac{1}{3+\frac{z+z^{-1}}{2}}\frac{dz}{iz}$$$$={\int }_{\mid z\mid =1}\frac{2dz}{iz(6+z+z^{-1})}={\int }_{\mid z\mid =1}\frac{-2idz}{6z+z^2+1}letW\left(z\right)=\frac{-2i}{z^2+6z+1}$$$$polesofW?$$$${\mathrm{\Delta }}^{{}^{\prime }}=9-1=8\Rightarrow z_1=-3+2\sqrt{2}and{z}_2=-3-2\sqrt{2}\Rightarrow W\left(z\right)=\frac{-2i}{(z-z_1)(z-z_2)}$$$$\mid z_1\mid -1=\mid -3+2\sqrt{2}\mid -1=3-2\sqrt{2}-1=2-2\sqrt{2}<0\Rightarrow \mid z_1\mid <1$$$$\mid z_2\mid =3+2\sqrt{2}-1=2+2\sqrt{2}>1residustheoremgive$$$${\int }_{\mid z\mid =1}W\left(z\right)dz=2i\pi Res(W,z_1)=2i\pi \times \frac{-2i}{z_1-z_2}=\frac{4\pi }{4\sqrt{2}}=\frac{\pi }{\sqrt{2}}\Rightarrow$$$$\mathrm{\Omega }=\frac{{\pi }^2}{4}\times \frac{\pi }{\sqrt{2}}=\frac{{\pi }^3}{4\sqrt{2}}$$

Answered by TANMAY PANACEA last updated on 06/Mar/20

$$I={\int }_0^{\frac{\pi }{2}}\frac{\pi (\frac{\pi }{2}-x)+{sin}^4(\frac{\pi }{2}-x)-{cos}^4(\frac{\pi }{2}-x)}{{sin}^4(\frac{\pi }{2}-x)+{cos}^4(\frac{\pi }{2}-x)}dx$$$$2I={\int }_0^{\frac{\pi }{2}}\frac{\pi \left(\frac{\pi }{2}\right)}{{cos}^{4}x+{sin}^{4}x}$$$$2I=\frac{{\pi }^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{{sec}^{2}x\times (1+{tan}^{2}x)}{1+{tan}^{4}x}dx$$$$I=\frac{{\pi }^2}{4}{\int }_0^{\mathrm{\infty }}\frac{1+t^2}{1+t^4}dt[t=tanx]$$$$I=\frac{{\pi }^2}{4}{\int }_0^{\mathrm{\infty }}\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt$$$$=\frac{{\pi }^2}{4}{\int }_0^{\mathrm{\infty }}\frac{d(t-\frac{1}{t})}{{(t-\frac{1}{t})}^2+2}$$$$\frac{{\pi }^2}{4}\times \frac{1}{\sqrt{2}}\mid {tan}^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right){\mid }_0^{\mathrm{\infty }}$$$$\frac{{\pi }^2}{4\sqrt{2}}\times \{({tan}^{-1}\mathrm{\infty }-{tan}^{-1}(-\mathrm{\infty })\}$$$$\frac{{\pi }^2}{4\sqrt{2}}\times (\frac{\pi }{2}+\frac{\pi }{2})=\frac{{\pi }^3}{4\sqrt{2}}$$$$\mathit{p}\mathit{l}\mathit{s}\mathit{c}\mathit{h}\mathit{e}\mathit{c}\mathit{k}$$

Commented by mathmax by abdo last updated on 06/Mar/20

$$youransweriscorrectsirtanmay$$

Commented by Power last updated on 06/Mar/20

$$\mathrm{thanks}$$

Commented by TANMAY PANACEA last updated on 06/Mar/20

$$thankyoubothofyousir$$

Answered by Kamel Kamel last updated on 07/Mar/20

$$Putt=\frac{\pi }{2}-x$$$$\therefore \mathrm{\Omega }={\int }_0^{\frac{\pi }{2}}\frac{\frac{{\pi }^2}{2}-\pi t+{cos}^4\left(t\right)-{sin}^4\left(t\right)}{{cos}^4\left(t\right)+{sin}^4\left(t\right)}dx$$$$\therefore 2\mathrm{\Omega }=\frac{{\pi }^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{dx}{{cos}^4\left(x\right)+{sin}^4\left(x\right)}=\frac{{\pi }^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{dx}{1-2{sin}^2\left(x\right){cos}^2\left(x\right)}$$$$=2{\pi }^2{\int }_0^{\frac{\pi }{2}}\frac{dx}{2-{sin}^2\left(2x\right)}\stackrel{t=2x}{=}{\pi }^2{\int }_0^{\pi }\frac{dt}{2-{sin}^2\left(t\right)}=2{\pi }^2{\int }_0^{\pi }\frac{dt}{cos\left(2t\right)+3}$$$$\stackrel{u=2t}{=}{\pi }^2{\int }_0^{2\pi }\frac{du}{cos\left(u\right)+3}={\pi }^2{\int }_{-\pi }^{\pi }\frac{du}{cos\left(u\right)+3}\stackrel{v=tg(\frac{u}{2}}{=}2{\pi }^2{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dv}{1-v^2+3+3v^2}$$$$=2{\pi }^2{\int }_0^{+\mathrm{\infty }}\frac{dv}{v^2+2}=\frac{{\pi }^2}{\sqrt{2}}Arctg\left(\frac{v}{\sqrt{2}}\right){\mid }_0^{+\mathrm{\infty }}=\frac{{\pi }^3}{2\sqrt{2}}\Rightarrow \mathrm{\Omega }=\frac{{\pi }^3}{4\sqrt{2}}$$$$\therefore {\int }_0^{\frac{\pi }{2}}\frac{\pi x+{sin}^4\left(x\right)-{cos}^4\left(x\right)}{{sin}^4\left(x\right)+{cos}^4\left(x\right)}dx=\frac{{\pi }^3}{4\sqrt{2}}$$$$\mathit{K}\mathit{A}\mathit{M}\mathit{E}\mathit{L}\mathit{B}\mathit{E}\mathit{N}\mathit{A}\mathit{I}\mathit{C}\mathit{H}\mathit{A}$$

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