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∫ 1 3 + 2 Cos 2 X D X - Mathematics

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प्रश्न

\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
योग

उत्तर

\[\text{ Let I } = \int \frac{1}{3 + 2 \cos^2 x}dx\]
\[\text{Dividing numerator and denominator by} \cos^2 x\]
\[ \Rightarrow I = \int \frac{\sec^2 x}{3 \sec^2 x + 2} dx\]
\[ = \int \frac{\sec^2 x}{3 \left( 1 + \tan^2 x \right) + 2}dx\]
\[ = \int \frac{\sec^2 x}{3 \tan^2 x + 5}dx\]
\[ = \int \frac{\sec^2 x}{\left( \sqrt{5} \right)^2 + \left( \sqrt{3} \tan x \right)^2}dx\]
\[\text{ Let }\sqrt{3} \tan x = t\]
\[ \Rightarrow \sqrt{3} \text{ sec}^2 x \text{ dx } = dt\]
\[ \Rightarrow \sec^2 x \text{ dx } = \frac{dt}{\sqrt{3}}\]
\[ \therefore I = \frac{1}{\sqrt{3}}\int \frac{dt}{\left( \sqrt{5} \right)^2 + t^2}\]
\[ = \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{5}} \text{ tan }^{- 1} \left( \frac{t}{\sqrt{5}} \right) + C\]
\[ = \frac{1}{\sqrt{15}} \text{ tan }^{- 1} \left( \frac{\sqrt{3} \tan x}{\sqrt{5}} \right) + C\]

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अध्याय 19: Indefinite Integrals - Exercise 19.22 [पृष्ठ ११४]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.22 | Q 6 | पृष्ठ ११४

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