Question

$\int \frac{1}{4{\cos }^{2}x+9{\sin }^{2}x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{4{\cos }^{2}x+9{\sin }^{2}x}dx \\ \mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}{\cos }^{2}x \\ \Rightarrow I=\int \frac{{\displaystyle \frac{1}{{\cos }^{2}x}}}{4+9{\tan }^{2}x}dx \\ =\int \frac{{\sec }^{2}x}{4+9{\tan }^{2}x}dx \\ \mathrm{Let}\tan x=t \\ \Rightarrow {\sec }^{2}xdx=dt \\ \therefore I=\int \frac{dt}{4+9t^2} \\ =\frac{1}{9}\int \frac{dt}{{\displaystyle \frac{4}{9}}+t^2} \\ =\frac{1}{9}\int \frac{dt}{{\left({\displaystyle \frac{2}{3}}\right)}^2+t^2} \\ =\frac{1}{9}\times \frac{3}{2}{\tan }^{-1}\left(\frac{t}{{\displaystyle \frac{2}{3}}}\right)+C \\ =\frac{1}{6}{\tan }^{-1}\left(\frac{3t}{2}\right)+C \\ =\frac{1}{6}{\tan }^{-1}\left(\frac{3\tan x}{2}\right)+C \end{gathered}$$