Integral

[stuart clark]

$$\int_{0}^{\pi}\frac{\sin^2 x}{a^2-2ab\cos +b^2}dx, a>b>0$$ and

$$\int_{0}^{\pi}\frac{\sin^2 x}{a^2-2ab\cos +b^2}dx, b>a>0$$

[LaçinCanAtış]

$$\int _0^{\pi }\frac{sin^2x}{\:a^2-2abcosx+b^2}dx=\int _0^{\pi }\frac{sin^2x}{\left(a^2+b^2\right)+2abcosx}dx$$
-->$$2I=\int _0^{\pi }\frac{sin^2x\left(a^2+b^2+2abcosx+a^2-2abcosx+b^2\right)}{\left(a^2+b^2\right)^2-4a^2b^2cos^2x}$$
$=2\left(a^2+b^2\right)\int _0^{\pi }\frac{sin^2x}{\left(a^2+b^2\right)^2-4a^2b^2cos^2x}dx$Therefore $I=\left(a^2+b^2\right)\cdot 2\int _0^{\frac{\pi }{2}}\frac{sin^2x}{\left(a^2+b^2\right)^2-4a^2b^2cos^2x}dx=2\left(a^2+b^2\right)\int _0^{\frac{\pi }{2}}\frac{1-cos^2x}{\left(a^2+b^2\right)^2-4a^2b^2cos^2x}dx=\frac{a^2+b^2}{2}\cdot \int _0^{\frac{\pi }{2}}\frac{\left(a^2+b^2\right)^2-4a^2b^2cos^2x+4a^2b^2-\left(a^2+b^2\right)^2}{\left(a^2+b^2\right)^2-4a^2b^2cos^2x}$
$=\frac{a^2+b^2}{2a^2b^2}\int _0^{\frac{\pi }{2}}dx-\frac{\left(a^2+b^2\right)}{2a^2b^2}\left(a^2-b^2\right)^2\int \:\frac{sec^2xdx}{\left(a^2+b^2\right)^2-4a^2b^2+\left(a^2+b^2\right)^2cos^2x}$
=$=\frac{\pi }{4a^2b^2}\left(a^2+b^2-\left|a^2-b^2\right|\right)$
gerisini okuyucuya bıraktım:D

[stuart clark]

Thanks  LaçinCanAtış

[LaçinCanAtış]

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