Question

$\int \frac{dx}{13+3cosx+4sinx}$ is equal to

• A. $\frac{1}{6}co{t}^{-1}\left(5tan\left(\frac{x}{2}\right)\right)+C$
• B. $\frac{1}{6}\left(\frac{ta{n}^{-1}x+5}{6}\right)+C$
• C. $\frac{1}{6}co{t}^{-1}\left(\frac{5tan\left(\frac{x}{2}\right)+2}{6}\right)+C$
• D. $\frac{1}{12}{\tan }^{-1}\left(\frac{5\tan \frac{x}{2}+2}{6}\right)+C$

Answer 1

Answer: (D)

$\frac{1}{12}{\tan }^{-1}\left(\frac{5\tan \frac{x}{2}+2}{6}\right)+C$

$\int \frac{1}{13+3\cos x+4\sin x}dx$

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

$\Rightarrow dx=\frac{dt}{1+t^2}$

We have $\cos t=\frac{1-t^2}{1+t^2}$ and $\sin x=\frac{2t}{1+t^2}$

$\int \frac{1}{13+3\cos x+4\sin x}dx$

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

$=\int \frac{\frac{dt}{1+t^2}}{13+3\frac{1-t^2}{1+t^2}+4\frac{2t}{1+t^2}}$

$=\int \frac{dt}{13(1+t^2)+3(1-t^2)+8t}$

$=\frac{1}{10}\int \frac{dt}{t^2+\frac{4}{5}t+8}$

$=\frac{1}{10}\int \frac{dt}{t^2+2\times \frac{2t}{5}+{\left(\frac{2}{5}\right)}^2-{\left(\frac{2}{5}\right)}^2+\frac{8}{5}}$ by completing the square method

$=\frac{1}{10}\int \frac{dt}{{(t+\frac{2}{5})}^2-\frac{4}{25}+\frac{8}{5}}$

$=\frac{1}{10}\int \frac{dt}{{(t+\frac{2}{5})}^2+\frac{-4+40}{25}}$

$=\frac{1}{10}\int \frac{dt}{{(t+\frac{2}{5})}^2+\frac{36}{25}}$

$=\frac{1}{10}\int \frac{dt}{{(t+\frac{2}{5})}^2+{\left(\frac{6}{5}\right)}^2}$

$=\frac{1}{10}\frac{1}{\frac{6}{5}}{\tan }^{-1}\left(\frac{t+\frac{2}{5}}{\frac{6}{5}}\right)$

$=\frac{5}{10\times 6}{\tan }^{-1}\left(\frac{5t+2}{6}\right)+c$ where $c$ is the constant of integration

$=\frac{1}{12}{\tan }^{-1}\left(\frac{5\tan \frac{x}{2}+2}{6}\right)+c$ where $t=\tan \frac{x}{2}$