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Question
\[\int\frac{1}{1 + 3 \sin^2 x} \text{ dx }\]
Sum
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Solution
\[\text{ Let I }= \int \frac{1}{1 + 3 \sin^2 x}\text{ dx }\]
\[\text{Dividing numerator and denominator by} \cos^2 x\]
\[ \Rightarrow I = \int \frac{\sec^2 x}{\sec^2 x + 3 \tan^2 x}dx\]
\[ = \int \frac{\sec^2 x}{1 + \tan^2 x + 3 \tan^2 x}dx\]
\[ = \int \frac{\sec^2 x}{1 + 4 \tan^2 x}dx\]
\[ = \int \frac{\sec^2 x}{1 + \left( 2 \tan x \right)^2}dx\]
\[\text{ Let 2 }\tan x = t\]
\[ \Rightarrow 2 \sec^2 x \text{ dx } = dt\]
\[ \Rightarrow \sec^2 x \text{ dx } = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int \frac{dt}{1 + t^2}\]
\[ = \frac{1}{2} \text{ tan }^{- 1} \left( t \right) + C\]
\[ = \frac{1}{2} \text{ tan }^{- 1} \left( 2 \tan x \right) + C\]
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