Question

$\int \frac{1}{4{\sin }^{2}x+5{\cos }^{2}x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{4{\sin }^{2}x+5{\cos }^{2}x}dx \\ \mathrm{Dividing}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}{\cos }^{2}x \\ \Rightarrow I=\int \frac{{\sec }^{2}x}{4{\tan }^{2}x+5}dx \\ \mathrm{Let}\tan x=t \\ \Rightarrow {\sec }^{2}xdx=dt \\ \therefore I=\int \frac{dt}{4t^2+5} \\ =\frac{1}{4}\int \frac{dt}{t^2+{\displaystyle \frac{5}{4}}} \\ =\frac{1}{4}\int \frac{dt}{t^2+{\left({\displaystyle \frac{\sqrt{5}}{2}}\right)}^2} \\ =\frac{1}{4}\times \frac{2}{\sqrt{5}}{\tan }^{-1}\left(\frac{t}{\sqrt{5}}\times 2\right)+C \\ =\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{2\tan x}{\sqrt{5}}\right)+C \end{gathered}$$