Question

If $\int \frac{1}{5+4\cos 2\theta }d\theta =A{\tan }^{-1}\left(B\tan \theta \right)+c$ then $(A,B)=$

Answer 1

$\begin{array}{c}\int \frac{d\theta }{5+4\cos 2\theta }=\int \frac{d\theta }{5+\frac{4\left(1-\tan \theta \right)}{1+{\tan }^2\theta }}\\ =\int \frac{{\sec }^2\theta d\theta }{9+{\tan }^2\theta }\\ Let\tan \theta =u\\ \frac{du}{d\theta }={\sec }^2\theta \\ Now,\\ =\int \frac{du}{9+u^2}=\frac{1}{3}{\tan }^{-1}\left(\frac{\tan \theta }{3}\right)+C\\ \left(A,B\right)=\left(\frac{1}{3},\frac{1}{3}\right)\end{array}$