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Question Number 91892 by john santu last updated on 03/May/20

$$\underset{0}{\stackrel{\pi }{\int }}\frac{dx}{a^2\cos {}^{2}x+b^2\sin {}^{2}x}?$$

Commented by john santu last updated on 03/May/20

$$\int \frac{dx}{(b^2-a^2)\sin {}^{2}x+a^2}$$$$say{a}^2=p\wedge b^2-a^2=q$$$$\int \frac{dx}{q\sin {}^{2}x+p}=$$$$W\mathrm{ierstrass}\mathrm{let}\tan \left(\frac{x}{2}\right)=\mathrm{t}$$$$dx=\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$\mathrm{tobe}\mathrm{continue}$$

Commented by Prithwish Sen 1 last updated on 03/May/20

$${\int }_0^{\pi }\frac{{\sec }^2\mathrm{x}\mathrm{dx}}{{\mathrm{a}}^2+{\mathrm{b}}^2\tan {}^2\mathrm{x}}=\frac{1}{\mathrm{ab}}{\left[{\tan }^{-1}\left(\frac{\mathrm{btan}\mathrm{x}}{\mathrm{a}}\right)\right]}_0^{\pi }$$

Commented by mathmax by abdo last updated on 03/May/20

$$I={\int }_0^{\pi }\frac{dx}{a^2{cos}^{2}x+b^2{sin}^{2}x}\Rightarrow I={\int }_0^{\pi }\frac{2dx}{a^2(1+cos\left(2x\right))+b^2(1-cos\left(2x\right))}$$$$={\int }_0^{\pi }\frac{2dx}{a^2+b^2+(a^2-b^2)cos\left(2x\right)}{=}_{2x=t}{\int }_0^{2\pi }\frac{dt}{a^2+b^2+(a^2-b^2)cost}$$$${=}_{e^{it}=z}{\int }_{\mid z\mid =1}\frac{dz}{iz(a^2+b^2+(a^2-b^2)\times \frac{z+z^{-1}}{2})}$$$$={\int }_{\mid z\mid =1}\frac{2dz}{iz\{2(a^2+b^2)+(a^2-b^2)(z+z^{-1})\}}$$$$=\int \frac{-2idz}{2(a^2+b^2)z+(a^2-b^2)z^2+a^2-b^2}let$$$$\phi \left(z\right)=\frac{-2i}{(a^2-b^2)z^2+2(a^2+b^2)z+a^2-b^2}polesof\phi ?$$$${\mathrm{\Delta }}^{{}^{\prime }}={(a^2+b^2)}^2-{(a^2-b^2)}^2=a^4+b^4+2a^2{b}^2-a^4-b^4+2a^2{b}^2$$$$=4a^2{b}^2\Rightarrow z_1=\frac{-(a^2+b^2)+2\mid ab\mid }{a^2-b^2}=-\frac{{(\mid a\mid -\mid b\mid )}^2}{\mid a{\mid }^2-\mid b{\mid }^2}=-\frac{\mid a\mid -\mid b\mid }{\mid a\mid +\mid b\mid }(wesupposea\ne b)$$$$z_2=\frac{-(a^2+b^2)-2\mid ab\mid }{a^2-b^2}=-\frac{\mid a\mid +\mid b\mid }{\mid a\mid -\mid b\mid }=\frac{1}{z_1}$$$$\mid z_1\mid -1=\frac{\mid \mid a\mid -\mid b\mid \mid }{\mid a\mid +\mid b\mid }-1<0because\mid ....\mid ⩽\mid a\mid +\mid b\mid \Rightarrow \mid z_2\mid -1>0\Rightarrow$$$$\left(z_{2}isoutofcircle\right)$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_1)$$$$\phi \left(z\right)=\frac{-2i}{(a^2-b^2)(z-z_1)(z-z_2)}\Rightarrow Res(\phi ,z_1)=\frac{-2i}{(a^2-b^2)(z_1-z_2)}$$$$=\frac{-2i}{(a^2-b^2)\times \frac{4\mid ab\mid }{a^2-b^2}}=\frac{-i}{2\mid ab\mid }\Rightarrow {\int }_{\mid z\mid =1}\phi \left(z\right)dz=\frac{2i\pi \times (-i)}{2\mid ab\mid }$$$$=\frac{\pi }{\mid ab\mid }\Rightarrow I=\frac{\pi }{\mid a\mid \times \mid b\mid }(ab\ne 0)$$$$ifa=b\ne 0\Rightarrow I={\int }_0^{\pi }\frac{dx}{a^2}=\frac{\pi }{a^2}$$

Commented by john santu last updated on 04/May/20

$$\mathrm{it}\mathrm{does}\mathrm{the}\mathrm{result}\mathrm{equal}\mathrm{to}\mathrm{zero}?$$

Commented by john santu last updated on 04/May/20

$$\mathrm{waw}\mathrm{via}\mathrm{complex}\mathrm{method}$$

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