$\int \frac{\cos 5 x + 5 \cos 3 x + 10 cosx}{\cos 6 x + 6 \cos 4 x + 15 \cos 2 x + 10} dx =$

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$\frac{1}{4} \log \tan (\frac{π}{6} + \frac{x}{2}) + C$

$\frac{1}{4} \log \tan (\frac{π}{6} - \frac{x}{2}) + C$

$\frac{1}{2} logtan (\frac{π}{4} + \frac{x}{2}) + C$

$\frac{1}{3} \log (\frac{π}{6} + \frac{x}{2}) + C$

answer is B.

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Detailed Solution

$\int \frac{\cos 5 x + 5 \cos 3 x + 10 \cos x}{\cos 6 x + 6 \cos 4 x + 15 \cos 2 x + 10} d x \begin{array}{l} = \int \frac{\cos 5 x + 5 \cos 3 x + 10 \cos x}{(\cos 6 x + \cos 4 x) + (5 \cos 4 x + 5 \cos 2 x) + (10 \cos 2 x + 10)} d x \\ = \int \frac{\cos 5 x + 5 \cos 3 x + 10 \cos x}{(\cos 6 x + \cos 4 x) + 5 (\cos 4 x + \cos 2 x) + 10 (\cos 2 x + 1)} d x \end{array}$
$\begin{array}{l} = \int \frac{\cos 5 x + 5 \cos 3 x + 10 \cos x}{2 (\cos 5 x \cos x) + 5 (2 \cos 3 x \cos x) + 10 (\cos 2 x + 1)} d x \\ = \int \frac{\cos 5 x + 5 \cos 3 x + 10 \cos x}{2 (\cos 5 xcos x) + 5 (2 \cos 3 xcos x) + 10 (2 \cos^{2} x)} d x \begin{aligned} = \int \frac{\cos 5 x + 5 \cos 3 x + 10 \cos x}{2 \cos x (\cos 5 x + 5 \cos 3 x + 10 \cos x)} d x \end{aligned} \end{array}$
$= \frac{1}{2} \int \sec xdx = \frac{1}{2} \log (\tan \frac{π}{4} + \frac{x}{2}) + C$