Question

Solve:
${\int }_0^{\pi }\frac{xdx}{(a^{2}co{s}^{2}x+b^{2}si{n}^{2}x)}$

Answer 1

$\begin{array}{c}\int \frac{dx}{a^2{\cos }^2\left(x\right)+b^2{\sin }^2\left(x\right)}=\int \frac{dx}{a^2{\cos }^2\left(x\right)\left(1+\frac{b^2}{a^2}{\tan }^{2}x\right)}\\ Thisgivesus\\ \frac{1}{a^2}\int \frac{{\sec }^{2}xdx}{1+\frac{b^2}{a^2}{\tan }^{2}x}\\ y=\frac{b}{a}\tan \left(x\right).du\\ =\frac{b}{a}{\sec }^{2}xdx\\ or,{\sec }^{2}xdx=\frac{a}{b}\int \frac{du}{1+u^2}\\ weknowthat,\\ u=\frac{b}{a}\tan \left(x\right)\\ weget,\\ \int \frac{dx}{a^2{\cos }^2\left(x\right)+b^2{\cos }^2\left(x\right)}\\ =\frac{1}{ab}arc\tan \left(\frac{b}{a}\tan \left(x\right)\right)+C\end{array}$