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Question Number 56471 by harish 12@g last updated on 17/Mar/19

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Mar/19

$${\int }_0^{\pi }\frac{xdx}{a^2{cos}^{2}x+b^2{sin}^{2}x}dx=I$$$$I={\int }_0^{\pi }\frac{(\pi -x)}{a^2{cos}^2(\pi -x)+b^2{sin}^2(\pi -x)}dx$$$$={\int }_0^{\pi }\frac{\pi -x}{a^2{cos}^{2}x+b^2{sin}^{2}x}dx$$$$2I={\int }_0^{\pi }\frac{\pi -x+x}{a^2{cos}^{2}x+b^2{sin}^{2}x}dx$$$$\frac{2I}{\pi }={\int }_0^{\pi }\frac{dx}{{cos}^{2}x(a^2+b^2{tan}^{2}x)}dx$$$$\frac{2I}{\pi }={\int }_0^{\pi }\frac{{sec}^{2}xdx}{a^2+b^2{tan}^{2}x}dx$$$$\frac{2I}{\pi }=\frac{1}{b^2}{\int }_0^{\pi }\frac{{sec}^{2}xdx}{{\left(\frac{a}{b}\right)}^2+{tan}^{2}x}dx$$$$now{\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dxwhenf(2a-x)=f\left(x\right)$$$$\frac{2I}{\pi }=\frac{2}{b^2}{\int }_0^{\frac{\pi }{2}}\frac{{sec}^{2}xdx}{{\left(\frac{a}{b}\right)}^2+{tan}^{2}x}dx$$$$tanx=\frac{a}{b}tanp$$$${sec}^{2}xdx=\frac{a}{b}{sec}^{2}pdp$$$$\frac{b^{2}I}{\pi }={\int }_0^{\frac{\pi }{2}}\frac{\frac{a}{b}{sec}^{2}p}{\left(\frac{a^2}{b^2}\right)(1+{tan}^{2}p)}dp$$$$\frac{b^{2}I}{\pi }=\frac{b}{a}{\int }_0^{\frac{\pi }{2}}dp$$$$\frac{abI}{\pi }=\mid p{\mid }_0^{\frac{\pi }{2}}$$$$\frac{abI}{\pi }=\frac{\pi }{2}$$$$soI=\frac{{\pi }^2}{2ab}proved$$

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