Evaluate: ^ _ 0 x dx a^ 2 ^ 2 x + b^ 2 ^ 2 x . - Vidyadip

Question

Evaluate: $\int\limits^{\pi}_{0}\frac{\text{x dx}}{\text{a}^{2}\cos^{2}\text{x + b}^{2}\sin^{2}\text{x}}.$

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Answer

$\text{I}=\int\limits^{\pi}_{0}\frac{\text{x dx}}{\text{a}^{2}\cos^{2}\text{x + b}^{2}\sin^{2}\text{x}}=\int\limits_{0}^{\pi}\frac{(\pi-\text{x) dx}}{\text{a}^{2}\cos^{2}\text{x + b}^{2}\sin^{2}\text{x}}$
$\therefore2\text{I}=\pi\int\limits^{\pi}_{0}\frac{\text{dx}}{\text{a}^{2}\cos^{2}\text{x + b}^{2}\sin^{2}\text{x}}=2\pi\int\limits_{0}^{\pi/2}\frac{\sec^{2}\text{x}}{\text{a}^{2}+\text{b}^{2}\text{ }\tan^{2}\text{x}}\text{dx}$
$ =2\pi\int\limits^{\infty}_{0}\frac{\text{dt}}{\text{a}^{2}+\text{b}^{2}\text{ t}^{2}}=\frac{2\pi}{\text{ab}}\Bigg[\tan^{-1}\frac{\text{bt}}{\text{a}}\Bigg]^{\infty}_{0}\text{where tan x = t}$
$\text{2I}=\frac{2\pi}{\text{ab}}\cdot\frac{\pi}{2}=\frac{\pi^{2}}{\text{ab}}$
$\Rightarrow\text{I}=\frac{\pi^{2}}{\text{2ab}}.$

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