s2vunov

2022-09-06

Definite integration evaluation of ${\int }_{0}^{\pi }\frac{x}{\left({a}^{2}{\mathrm{cos}}^{2}x+{b}^{2}{\mathrm{sin}}^{2}x{\right)}^{2}}dx$

Salma Baird

I prefer to the following method:
$\begin{array}{rl}I:={\int }_{0}^{\pi }\frac{x}{\left({a}^{2}{\mathrm{cos}}^{2}x+{b}^{2}{\mathrm{sin}}^{2}x{\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx& =\frac{\pi }{2}{\int }_{0}^{\pi }\frac{dx}{\left({a}^{2}{\mathrm{cos}}^{2}x+{b}^{2}{\mathrm{sin}}^{2}x{\right)}^{2}}\\ & =\pi {\int }_{0}^{\frac{\pi }{2}}\frac{dx}{\left({a}^{2}{\mathrm{cos}}^{2}x+{b}^{2}{\mathrm{sin}}^{2}x{\right)}^{2}}\\ & =\pi {\int }_{0}^{\frac{\pi }{2}}\frac{1+{\mathrm{tan}}^{2}x}{\left({a}^{2}+{b}^{2}{\mathrm{tan}}^{2}x{\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}dx.\end{array}$
Now we make the substitution $b\mathrm{tan}x↦a\mathrm{tan}x$. Then
$\begin{array}{rl}I& =\frac{\pi }{\left(ab{\right)}^{3}}{\int }_{0}^{\frac{\pi }{2}}\frac{{b}^{2}+{a}^{2}{\mathrm{tan}}^{2}x}{\left(1+{\mathrm{tan}}^{2}x{\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}dx\\ & =\frac{\pi }{\left(ab{\right)}^{3}}{\int }_{0}^{\frac{\pi }{2}}\left({b}^{2}{\mathrm{cos}}^{2}x+{a}^{2}{\mathrm{sin}}^{2}x\right)\phantom{\rule{thinmathspace}{0ex}}dx\\ & =\frac{\pi }{\left(ab{\right)}^{3}}\cdot \frac{\pi }{4}\left({a}^{2}+{b}^{2}\right)=\frac{{\pi }^{2}\left({a}^{2}+{b}^{2}\right)}{4\left(ab{\right)}^{3}}.\end{array}$

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