The value of \(\int \frac{1}{3 \sin x-\cos x+3} d x\) is

Let \(I=\int \frac{d x}{3 \sin x-\cos x+3}\)
\(\left\{\begin{array}{l}\because \quad \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \\ \text { and } \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\end{array}\right\}\)
\(I=\int \frac{d x}{3\left\{\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right\}-\left\{\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right\}+3}\)
\(=\int \frac{\left(1+\tan ^{2} \frac{x}{2}\right) d x}{6 \tan \frac{x}{2}-1+\tan ^{2} \frac{x}{2}+3+3 \tan ^{2} \frac{x}{2}}\)
\(=\int \frac{\sec ^{2} \frac{x}{2}}{4 \tan ^{2} \frac{x}{2}+6 \tan \frac{x}{2}+2} d x\)
\(\left(\right.\) let \(\left.t=\tan \frac{x}{2}, d t=\frac{1}{2} \sec ^{2} \frac{x}{2} d x\right)\)
\(=\int \frac{d t}{2 t^{2}+3 t+1}\)
\(=\int \frac{d t}{(t+1)(2 t+1)}\)
\(=\int\left\{\frac{-1}{(t+1)}+\frac{2}{(2 t+1)}\right\} d t\) (by partial
fraction) \(=-\log (t+1)+\frac{2}{2} \log (2 t+1)+C\)
\(=\log \frac{(2 t+1)}{(t+1)}+C\)
\(=\log \left(\frac{2 \tan \frac{x}{2}+1}{\tan \frac{x}{2}+1}\right)+C\)