Question

Solve: $\int \frac{dx}{13+3cosx+4sinx}$

Answer 1

$\int \frac{dx}{13+3\cos x+4\sin x}$

$=\int \frac{dx}{13+3\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+4\left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}$

$=\frac{(1+{\tan }^2\frac{x}{2})dx}{13+13{\tan }^2\frac{x}{2}+3-3{\tan }^2\frac{x}{2}+8\tan \frac{x}{2}}$

$=\frac{{\sec }^2\frac{x}{2}dx}{16+10{\tan }^2\frac{x}{2}+8\tan \frac{x}{2}}$

Let $t=\tan \frac{x}{2}\Rightarrow dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx$

$\Rightarrow 2dt={\sec }^2\frac{x}{2}dx$

$=\int \frac{2dt}{16+10t^2+8t}$

$=\int \frac{dt}{8+5t^2+4t}$

$=\int \frac{dt}{5(t^2+\frac{4t}{5}+\frac{8}{5})}$

$=\frac{1}{5}\int \frac{dt}{t^2+2\times \frac{2t}{5}+{\left(\frac{2}{5}\right)}^2-{\left(\frac{2}{5}\right)}^2+\frac{8}{5}}$

$=\frac{1}{5}\int \frac{dt}{{(t+\frac{2}{5})}^2-\frac{4}{25}+\frac{8}{5}}$

$=\frac{1}{5}\int \frac{dt}{{(t+\frac{2}{5})}^2+\frac{-4+40}{25}}$

$=\frac{1}{5}\int \frac{dt}{{(t+\frac{2}{5})}^2+\frac{36}{25}}$

$=\frac{1}{5}\int \frac{dt}{t^2+2\times \frac{2t}{5}+{\left(\frac{2}{5}\right)}^2-{\left(\frac{2}{5}\right)}^2+\frac{8}{5}}$

$=\frac{1}{5}\int \frac{dt}{{(t+\frac{2}{5})}^2-\frac{4}{25}+\frac{8}{5}}$

$=\frac{1}{5}\int \frac{dt}{{(t+\frac{2}{5})}^2+\frac{-4+40}{25}}$

$=\frac{1}{5}\int \frac{dt}{{(t+\frac{2}{5})}^2+\frac{36}{25}}$

$=\frac{1}{5}\int \frac{dt}{{(t+\frac{2}{5})}^2+{\left(\frac{6}{5}\right)}^2}$

$=\frac{1}{5}\times \frac{5}{6}{\tan }^{-1}\left(\frac{t+\frac{2}{5}}{\frac{6}{5}}\right)$

$=\frac{1}{6}{\tan }^{-1}\left(\frac{5t+2}{6}\right)+c$

$=\frac{1}{6}{\tan }^{-1}\left(\frac{5\tan \frac{x}{2}+2}{6}\right)+c$ where $t=\tan \frac{x}{2}$