Question

\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{  dx }\]

Answer 1

\[\text{ Let I } = \int \frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{ dx }\]
\[\text{Dividing numerator and denominator by} \cos^2 x\]

\[ \Rightarrow I = \int \frac{\frac{1}{\cos^2 x}}{4 + 9 \tan^2 x}dx\]
\[ = \int \frac{\sec^2 x}{4 + 9 \tan^2 x}dx\]
\[\text{ Let tan } x = t\]
` ⇒  sec^2  x   dx = dt `
\[ \therefore I = \int \frac{dt}{4 + 9 t^2}\]
\[ = \frac{1}{9}\int \frac{dt}{\frac{4}{9} + t^2}\]
\[ = \frac{1}{9}\int \frac{dt}{\left( \frac{2}{3} \right)^2 + t^2}\]
\[ = \frac{1}{9} \times \frac{3}{2} \text[\text{  tan }^{- 1} \left( \frac{t}{\frac{2}{3}} \right) + C\]
\[ = \frac{1}{6} \text{ tan }^{- 1} \left( \frac{3t}{2} \right) + C\]
\[ = \frac{1}{6} \text{ tan }^{- 1} \left( \frac{3 \tan x}{2} \right) + C\]