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$\int \frac{d x}{a^2 \sin ^2 x+b^2 \cos ^2 x}$ is equal to
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The correct answer is:
$\frac{1}{a b} \tan ^{-1}\left(\frac{a \tan x}{b}\right)+C$
$\begin{aligned} & \int \frac{d x}{a^2 \sin ^2 x+b^2 \cos ^2 x} \\ & =\frac{1}{a^2} \int \frac{\sec ^2 x d x}{\left(\frac{b}{a}\right)^2+\tan ^2 x} \\ & \Rightarrow \quad=\frac{1}{a b} \tan ^{-1}\left(\frac{a \tan x}{b}\right)+C\end{aligned}$
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