\(\int \frac{1}{1+3 \sin ^2 x+8 \cos ^2 x} d x\) is equals to

Let \(I=\int \frac{1}{1+3 \sin ^2 x+8 \cos ^2 x} d x\)
Dividing the numerator and denominator by
\(\cos ^2 x\), we get
\(\begin{aligned} & \Rightarrow I=\int \frac{\sec ^2 x}{\sec ^2 x+3 \tan ^2 x+8} d x \\ & \Rightarrow I=\int \frac{\sec ^2 x}{1+\tan ^2 x+3 \tan ^2 x+8} d x \\ & \Rightarrow I=\int \frac{\sec ^2 x}{4 \tan ^2 x+9} d x\end{aligned}\)
Putting, \(\tan x=t \Rightarrow \sec ^2 x d x=d t\), we get
\(\begin{gathered}I=\int \frac{d t}{4 t^2+9}=\frac{1}{4} \int \frac{d t}{t^2+\left(\frac{3}{2}\right)^2}+C \\ I=\frac{1}{4} \times \frac{1}{3 / 2} \tan ^{-1}\left(\frac{t}{3 / 2}\right)+C \\ \Rightarrow \quad I=\frac{1}{6} \tan ^{-1}\left(\frac{2 t}{3}\right)+C=\frac{1}{6} \tan ^{-1}\left(\frac{2 \tan x}{3}\right)+C\end{gathered}\)