s2vunov

2022-09-06

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

Salma Baird

Beginner2022-09-07Added 8 answers

I prefer to the following method:

$$\begin{array}{rl}I:={\int}_{0}^{\pi}\frac{x}{({a}^{2}{\mathrm{cos}}^{2}x+{b}^{2}{\mathrm{sin}}^{2}x{)}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx& =\frac{\pi}{2}{\int}_{0}^{\pi}\frac{dx}{({a}^{2}{\mathrm{cos}}^{2}x+{b}^{2}{\mathrm{sin}}^{2}x{)}^{2}}\\ & =\pi {\int}_{0}^{\frac{\pi}{2}}\frac{dx}{({a}^{2}{\mathrm{cos}}^{2}x+{b}^{2}{\mathrm{sin}}^{2}x{)}^{2}}\\ & =\pi {\int}_{0}^{\frac{\pi}{2}}\frac{1+{\mathrm{tan}}^{2}x}{({a}^{2}+{b}^{2}{\mathrm{tan}}^{2}x{)}^{2}}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}dx.\end{array}$$

Now we make the substitution $b\mathrm{tan}x\mapsto a\mathrm{tan}x$. Then

$$\begin{array}{rl}I& =\frac{\pi}{(ab{)}^{3}}{\int}_{0}^{\frac{\pi}{2}}\frac{{b}^{2}+{a}^{2}{\mathrm{tan}}^{2}x}{(1+{\mathrm{tan}}^{2}x{)}^{2}}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}dx\\ & =\frac{\pi}{(ab{)}^{3}}{\int}_{0}^{\frac{\pi}{2}}({b}^{2}{\mathrm{cos}}^{2}x+{a}^{2}{\mathrm{sin}}^{2}x)\phantom{\rule{thinmathspace}{0ex}}dx\\ & =\frac{\pi}{(ab{)}^{3}}\cdot \frac{\pi}{4}({a}^{2}+{b}^{2})=\frac{{\pi}^{2}({a}^{2}+{b}^{2})}{4(ab{)}^{3}}.\end{array}$$

$$\begin{array}{rl}I:={\int}_{0}^{\pi}\frac{x}{({a}^{2}{\mathrm{cos}}^{2}x+{b}^{2}{\mathrm{sin}}^{2}x{)}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx& =\frac{\pi}{2}{\int}_{0}^{\pi}\frac{dx}{({a}^{2}{\mathrm{cos}}^{2}x+{b}^{2}{\mathrm{sin}}^{2}x{)}^{2}}\\ & =\pi {\int}_{0}^{\frac{\pi}{2}}\frac{dx}{({a}^{2}{\mathrm{cos}}^{2}x+{b}^{2}{\mathrm{sin}}^{2}x{)}^{2}}\\ & =\pi {\int}_{0}^{\frac{\pi}{2}}\frac{1+{\mathrm{tan}}^{2}x}{({a}^{2}+{b}^{2}{\mathrm{tan}}^{2}x{)}^{2}}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}dx.\end{array}$$

Now we make the substitution $b\mathrm{tan}x\mapsto a\mathrm{tan}x$. Then

$$\begin{array}{rl}I& =\frac{\pi}{(ab{)}^{3}}{\int}_{0}^{\frac{\pi}{2}}\frac{{b}^{2}+{a}^{2}{\mathrm{tan}}^{2}x}{(1+{\mathrm{tan}}^{2}x{)}^{2}}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{sec}}^{2}x\phantom{\rule{thinmathspace}{0ex}}dx\\ & =\frac{\pi}{(ab{)}^{3}}{\int}_{0}^{\frac{\pi}{2}}({b}^{2}{\mathrm{cos}}^{2}x+{a}^{2}{\mathrm{sin}}^{2}x)\phantom{\rule{thinmathspace}{0ex}}dx\\ & =\frac{\pi}{(ab{)}^{3}}\cdot \frac{\pi}{4}({a}^{2}+{b}^{2})=\frac{{\pi}^{2}({a}^{2}+{b}^{2})}{4(ab{)}^{3}}.\end{array}$$

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