Problem Statement
Find:
$$\int \frac{dx}{5 + 4\cos x}$$
Verified Solution & Marking Scheme
Use half-angle identity for cos x
Express $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$.
$$\frac{1}{5 + 4\cos x} = \frac{1}{5 + 4\left( \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} \right)} = \frac{1 + \tan^2(x/2)}{5(1 + \tan^2(x/2)) + 4(1 - \tan^2(x/2))}$$
$$= \frac{\sec^2(x/2)}{5 + 5\tan^2(x/2) + 4 - 4\tan^2(x/2)} = \frac{\sec^2(x/2)}{\tan^2(x/2) + 9}$$
Substitute t = tan(x/2)
Let $t = \tan(x/2)$, then $dt = \frac{1}{2}\sec^2(x/2) \, dx \implies \sec^2(x/2) \, dx = 2\, dt$.
$$\int \frac{2\, dt}{t^2 + 3^2} = 2 \cdot \frac{1}{3} \tan^{-1}\left(\frac{t}{3}\right) + C = \frac{2}{3} \tan^{-1}\left( \frac{\tan(x/2)}{3} \right) + C$$