Evaluate the following integrals: _ 0 ^ 2 1 a^2 ^2x+b^2 ^2x dx - Vidyadip

Question

Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$

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Answer

We have,$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$$\int_{0}^\limits{\frac{\pi}{2}}\begin{pmatrix}\frac{\frac{1}{\cos^2\text{x}}}{\text{a}^2\frac{\sin^2\text{x}}{\cos^2\text{x}}+\text{b}^2\frac{\cos^2\text{x}}{\cos^2\text{x}}} \end{pmatrix}\text{dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\Big(\frac{\sec^2\text{x}}{\text{a}^2\tan^{2}\text{x}+\text{b}^2}\Big)\text{dx}$
$=\frac{1}{\text{a}^2}\int_{0}^\limits{\frac{\pi}{2}}\Bigg(\frac{\sec^2\text{x}}{\tan^{2}\text{x}+\big(\frac{\text{b}}{\text{a}}\big)^2}\Bigg)\text{dx}$
Let $\tan\text{x}=\text{t}$ Differentiating w.r.t. x, we get$\sec^2\text{xdx}=\text{dt}$
When $\text{x}=0\Rightarrow\text{t}=0$$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=\infty$
$\therefore\ \frac{1}{\text{a}^2}\int_{0}^\limits{\frac{\pi}{2}}\Bigg(\frac{\sec^2\text{x}}{\tan^{2}\text{x}+\big(\frac{\text{b}}{\text{a}}\big)^2}\Bigg)\text{dx}$
$=\frac{1}{\text{a}^2}\int_{0}^\limits{\infty}\frac{\text{dt}}{\big(\frac{\text{b}}{\text{a}}\big)^2+\text{t}^2}$
$=\frac{1}{\text{a}^2}\Big[\frac{\text{a}}{\text{b}}\tan^{-1}\frac{\text{at}}{\text{b}}\Big]^{\infty}_0$
$=\frac{1}{\text{a}^2}\frac{\text{a}}{\text{b}}\big[\tan^{-1}\infty-\tan^{-1}0\big]$
$=\frac{1}{\text{ab}}\Big[\tan^{-1}\tan\frac{\pi}{2}\Big]$
$=\frac{\pi}{2\text{ab}}$

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