Question:

<div>Write the Numerator = $\ell(\text{denominator}) + m(\text{d.c. of denominator}) + n$<br/>$\Rightarrow 2 + 3 \cos \theta = \ell(\sin\theta + 2\cos\theta + 3) + m(\cos\theta - 2\sin\theta) + n$.<br/>Comparing the coefficients of $\sin\theta, \cos\theta$ and constant terms, we get<br/>$3\ell + n = 2, 2\ell + m = 3, \ell - 2m = 0 \Rightarrow \ell = 6/5, m = 3/5$ and $n = -8/5$<br/>Hence $I = \int \frac{6}{5} d\theta + \frac{3}{5} \int \frac{\cos\theta - 2\sin\theta}{\sin\theta + 2\cos\theta + 3} d\theta - \frac{8}{5} \int \frac{d\theta}{\sin\theta + 2\cos\theta + 3}$<br/>$= \frac{6}{5}\theta + \frac{3}{5} \ln |\sin\theta + 2\cos\theta + 3| - \frac{8}{5} I_3$ where $I_3 = \int \frac{d\theta}{\sin\theta + 2\cos\theta + 3}$<br/>In $I_3$, put $\tan \frac{\theta}{2} = t \Rightarrow \sec^2 \frac{\theta}{2} d\theta = 2dt$<br/>$I_3 = 2 \int \frac{dt}{t^2 + 2t + 5} = 2 \int \frac{dt}{(t+1)^2 + 2^2} = 2 \cdot \frac{1}{2} \tan^{-1} \left( \frac{t+1}{2} \right) = \tan^{-1} \left( \frac{\tan \theta/2 + 1}{2} \right)$<br/>Hence $I = \frac{6\theta}{5} + \frac{3}{5} \ln |\sin\theta + 2\cos\theta + 3| - \frac{8}{5} \tan^{-1} \left( \frac{\tan \theta/2 + 1}{2} \right) + C$</div>