Question

If $\int \frac{dx}{5+4cosx}=Kta{n}^{-1}\left(Mtan\frac{x}{2}\right)+C$, then

• A. $K=1$
• B. $K=2/3$
• C. $M=1/3$
• D. $M=2/3$

Answer 1

Answer: (B), (C)

The correct options are
B $M=1/3$
C $K=2/3$

$I=\int \frac{dx}{5+4\cos x}$
Put $\tan \frac{x}{2}=t$

$\therefore dt=\frac{{\sec }^2\frac{x}{2}}{2}dx$

We know that, $\cos x=1-\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$

$I=\int \frac{2dt}{(1+t^2)(5+\frac{4(1-t^2)}{1+t^2})}$

$=2\int \frac{dt}{5(1+t^2)+4(1-t^2)}$

$=2\int \frac{dt}{t^2+9}=\frac{2}{3}{\tan }^{-1}\frac{t}{3}+c$

$I=\frac{2}{3}{\tan }^{-1}\left(\frac{1}{3}\tan \frac{x}{2}\right)+c$

Hence $K=\frac{2}{3}$ and $M=\frac{1}{3}$