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Question Number 61465 by arcana last updated on 02/Jun/19

$${\int }_0^{2\pi }\frac{1}{a^2{cos}^2\left(t\right)+b^2{sin}^2\left(t\right)}dt=\frac{2\pi }{ab}?$$

Commented by maxmathsup by imad last updated on 03/Jun/19

$$letA={\int }_0^{2\pi }\frac{dt}{a^2{cos}^{2}t+b^2{sin}^{2}t}\Rightarrow A={\int }_0^{2\pi }\frac{2dt}{a^2(1+cos\left(2t\right))+b^2(1-cos\left(2t\right)}$$$$={\int }_0^{2\pi }\frac{2dt}{a^2+b^2+(a^2-b^2)cos\left(2t\right)}{=}_{2t=x}{\int }_0^{4\pi }\frac{dx}{a^2+b^2+(a^2-b^2)cos\left(x\right)}$$$$={\int }_0^{2\pi }\frac{dx}{a^2+b^2+(a^2-b^2)cos\left(x\right)}+{\int }_{2\pi }^{4\pi }\frac{dx}{a^2+b^2+(a^2-b^2)cos\left(x\right)}=H+K$$$$H{=}_{e^{ix}=z}{\int }_{\mid z\mid =1}\frac{dz}{iz\{a^2+b^2+(a^2-b^2)\frac{z+z^{-1}}{2}\}}={\int }_{\mid z\mid =1}\frac{2dz}{iz\{2(a^2+b^2)+(a^2-b^2)(z+z^{-1})\}}$$$$={\int }_{\mid z\mid =1}\frac{2dz}{2i(a^2+b^2)z+i(a^2-b^2)z^2+i(a^2-b^2)}$$$$=\int \frac{-2idz}{(a^2-b^2)z^2+2(a^2+b^2)z+a^2-b^2}let\phi \left(z\right)=\frac{-2i}{(a^2-b^2)z^2+2(a^2+b^2)z+a^2-b^2}$$$${\mathrm{\Delta }}_d^{{}^{\prime }}=={(a^2+b^2)}^2-{(a^2-b^2)}^2=a^4+2a^2{b}^2+b^4-a^4+2a^2{b}^2-b^4=4a^2{b}^2\Rightarrow$$$$z_1=\frac{-a^2-b^2+2\mid ab\mid }{a^2-b^2}=-\frac{{(\mid a\mid -\mid b\mid )}^2}{\mid a{\mid }^2-\mid b{\mid }^2}=-\frac{\mid a\mid -\mid b\mid }{\mid a\mid +\mid b\mid }(a\ne b)$$$$z_2=\frac{-a^2-b^2-2\mid ab\mid }{a^2-b^2}=-\frac{{(\mid a\mid +\mid b\mid )}^2}{a^2-b^2}=-\frac{\mid a\mid +\mid b\mid }{\mid a\mid -\mid b\mid }(=\frac{1}{z_1})\mid z$$$$\mid z_1\mid -1=\frac{\mid \mid a\mid -\mid b\mid \mid }{\mid a\mid +\mid b\mid }-1=\frac{\mid \mid a\mid -\mid b\mid \mid -(\mid a\mid +\mid b\mid )}{(....)}<0$$$$\mid z_2\mid -1>0(toeliminatefromresidus\Rightarrow$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_1)wehave\phi \left(z\right)=\frac{-2i}{(a^2-b^2)(z-z_1)(z-z_2)}\nLeftrightarrow$$$$Res(\phi ,z_1)=\frac{-2i}{(a^2-b^2)(z_1-z_2)}=\frac{-2i}{(a^2-b^2)(\frac{\mid a\mid +\mid b\mid }{\mid a\mid -\mid b\mid }-\frac{\mid a\mid -\mid b\mid }{\mid a\mid +\mid b\mid })}=\frac{-2i}{{(\mid a\mid +\mid b\mid )}^2-{(\mid a\mid -\mid b\mid )}^2}$$$$=\frac{-2i}{4\mid ab\mid }=-\frac{i}{2\mid ab\mid }\Rightarrow {\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi \frac{-i}{2\mid ab\mid }=\frac{\pi }{\mid ab\mid }=H$$$$K{=}_{x=2\pi +t}{\int }_0^{2\pi }\frac{dt}{a^2+b^2+(a^2-b^2)cost}=H\Rightarrow$$$${\int }_0^{2\pi }\frac{dt}{a^2{cos}^{2}t+b^2{sin}^{2}t}=2H=\frac{2\pi }{\mid ab\mid }$$$$ifa=bweget{\int }_0^{2\pi }\frac{dt}{a^2{cos}^{2}t+b^2{sin}^{2}t}=\frac{1}{a^2}\left(2\pi \right)=\frac{2\pi }{a^2}.$$

Answered by tanmay last updated on 03/Jun/19

$${\int }_0^{2\pi }\frac{{sec}^{2}t}{a^2+b^2{tan}^{2}t}dt$$$$\frac{1}{b^2}{\int }_0^{2\pi }\frac{d\left(tant\right)}{{\left(\frac{a}{b^{}}\right)}^2+{\left(tant\right)}^2}dt$$$${\int }_0^{2a}f\left(t\right)dt=2{\int }_0^{a}f\left(t\right)dtwhenf(2a-t)=f\left(t\right)$$$$\frac{2}{b^2}{\int }_0^{\pi }\frac{d\left(tant\right)}{{\left(\frac{a}{b}\right)}^2+{\left(tant\right)}^2}$$$$=\frac{2}{b^2}\times 2{\int }_0^{\frac{\pi }{2}}\frac{d\left(tant\right)}{{\left(\frac{a}{b}\right)}^2+{\left(tant\right)}^2}$$$$=\frac{4}{b^2}\times \frac{1}{\left(\frac{a}{b}\right)}\times \mid {tan}^{-1}\left(\frac{tant}{\frac{a}{b}}\right){\mid }_0^{\frac{\pi }{2}}$$$$=\frac{4}{ab}\times [{tan}^{-1}\left(\mathrm{\infty }\right)-{tan}^{-1}\left(0\right)]$$$$=\frac{4}{ab}\times \frac{\pi }{2}=\frac{2\pi }{ab}$$

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