Question

If $\int \frac{dx}{5+4\cos x}=a{\tan }^{-1}\left(b\tan \frac{x}{2}\right)+c$, then calculate $a$ and $b$.

Answer 1

$\because \cos x=\frac{1-{\tan }^2\left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)}$

$\therefore \int \frac{dx}{5+4\cos x}=\int \frac{dx}{5+4\left(\frac{1-{\tan }^2\left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)}\right)}$

$=\int \frac{(1+{\tan }^2\left(\frac{x}{2}\right))dx}{5(1+{\tan }^2\left(\frac{x}{2}\right))+4(1-{\tan }^2\left(\frac{x}{2}\right))}$

$=\int \frac{{\sec }^2\left(\frac{x}{2}\right)dx}{(9+{\tan }^2\left(\frac{x}{2}\right))}$

Let $\tan \left(\frac{x}{2}\right)=y$ on differentiating this we get, ${\sec }^2\left(\frac{x}{2}\right)\times \frac{dx}{2}=dy$

Thus, $\int \frac{{\sec }^2\left(\frac{x}{2}\right)dx}{(9+{\tan }^2\left(\frac{x}{2}\right))}=\int \frac{2dy}{9+y^2}=2\int \frac{1}{y^2+3^2}dx=2\times \frac{1}{3}{\tan }^{-1}y+c$

Now putting in the value of y in above equation, we get,

$\int \frac{dx}{5+4\cos x}=\frac{2}{3}{\tan }^{-1}\left(\tan \frac{x}{2}\right)+c$

Then $a=\frac{2}{3}$ and $b=1$