Question

If $\int \frac{1}{5+4\cos 2\theta }d\theta =A{\tan }^{-1}\left(B\tan \theta \right)+C,$ then $(A,B)=$
(where $A,B$ are fixed constants and $C$ is integration constant)

• A. $(\frac{1}{2},\frac{1}{2})$
• B. $(\frac{1}{3},\frac{1}{3})$
• C. $(\frac{1}{2},3)$
• D. $(\frac{1}{3},2)$

Answer 1

The correct option is B $(\frac{1}{3},\frac{1}{3})$

$I=\int \frac{1}{5+4\cos 2\theta }d\theta \Rightarrow I=\int \frac{1+{\tan }^2\theta }{5(1+{\tan }^2\theta )+4(1-{\tan }^2\theta )}d\theta [\because \cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }]\Rightarrow I=\int \frac{{\sec }^2\theta }{9+{\tan }^2\theta }d\theta$

Put $\tan \theta =t$
$\Rightarrow {\sec }^2\theta d\theta =dt\Rightarrow I=\int \frac{dt}{(3{)}^2+(t{)}^2}\Rightarrow I=\frac{1}{3}{\tan }^{-1}\left(\frac{t}{3}\right)+C\Rightarrow I=\frac{1}{3}{\tan }^{-1}\left(\frac{\tan \theta }{3}\right)+C\therefore (A,B)\equiv (\frac{1}{3},\frac{1}{3})$