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प्रश्न
उत्तर
\[\text{ Let I } = \int\frac{1}{4 \sin^2 x + 5 \cos^2 x}\text{ dx }\]
\[\text{Dividing numerator and denominator by} \cos^2 x\]
\[ \Rightarrow I = \int\frac{\sec^2 x}{4 \tan^2 x + 5}\text{ dx }\]
\[\text{ Let tan } x = t\]
\[ \Rightarrow \sec^2\text{ x }dx = dt\]
\[ \therefore I = \int \frac{dt}{4 t^2 + 5}\]
\[ = \frac{1}{4}\int \frac{dt}{t^2 + \frac{5}{4}}\]
\[ = \frac{1}{4}\int\frac{dt}{t^2 + \left( \frac{\sqrt{5}}{2} \right)^2}\]
\[ = \frac{1}{4} \times \frac{2}{\sqrt{5}} \text{ tan }^{- 1} \left( \frac{t}{\sqrt{5}} \times 2 \right) + C\]
\[ = \frac{1}{2\sqrt{5}} \text{ tan }^{- 1} \left( \frac{2 \tan x}{\sqrt{5}} \right) + C\]
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