int frac{dx}{a^2 sin^2 x + b^2 cos^2 x} is eq | vedclass

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Question

(A) To evaluate the integral $I = \int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x}$, divide both the numerator and the denominator by $\cos^2 x$:$I = \int

Vedclass · Accepted Answer

$\frac{1}{ab} 	an^{-1}\left(\frac{a 	an x}{b}ight) + C$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x}$, divide both the numerator and the denominator by $\cos^2 x$:$I = \int \frac{\sec^2 x dx}{a^2 	an^2 x + b^2}$Now, let $u = 	an x$, then $du = \sec^2 x dx$:$I = \int \frac{du}{a^2 u^2 + b^2} = \frac{1}{a^2} \int \frac{du}{u^2 + (b/a)^2}$Using the standard integral formula $\int \frac{dx}{x^2 + k^2} = \frac{1}{k} 	an^{-1}(\frac{x}{k}) + C$:$I = \frac{1}{a^2} \cdot \frac{1}{b/a} 	an^{-1}\left(\frac{u}{b/a}ight) + C$$I = \frac{1}{ab} 	an^{-1}\left(\frac{a 	an x}{b}ight) + C$

$\int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x}$ is equal to

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