Question

Evaluate $\int \frac{1}{3\cos x+4\sin x+6}dx$

Answer 1

$\int \frac{dx}{3\cos x+4\sin x+6}$

Take $\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$ and $\sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$

$\therefore \int \frac{dx}{3\cos x+4\sin x+6}$

$=\int \frac{dx}{3\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+4\left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+6}$

$=\int \frac{(1+{\tan }^2\frac{x}{2})dx}{3(1-{\tan }^2\frac{x}{2})+4\left(2\tan \frac{x}{2}\right)+6(1+{\tan }^2\frac{x}{2})}$

$=\int \frac{{\sec }^2\frac{x}{2}dx}{1-2{\tan }^2\frac{x}{2}+8\tan \frac{x}{2}+6+6{\tan }^2\frac{x}{2}}$ where $1+{\tan }^2\frac{x}{2}={\sec }^2\frac{x}{2}$

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

$=\int \frac{2dt}{3-3t^2+8t+6+6t^2}$

$=2\int \frac{dt}{3t^2+8t+9}$

$=\frac{2}{3}\int \frac{dt}{t^2+\frac{8}{3}t+3}$

$=\frac{2}{3}\int \frac{dt}{t^2+2\times \frac{4}{3}t+{\left(\frac{4}{3}\right)}^2-{\left(\frac{4}{3}\right)}^2+3}$

$=\frac{2}{3}\int \frac{dt}{{(t+\frac{4}{3})}^2+\frac{11}{9}}$

$=\frac{2}{3}\int \frac{dt}{{(t+\frac{4}{3})}^2+{\left(\frac{\sqrt{11}}{3}\right)}^2}$ where $\int \frac{dt}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}$

$=\frac{2}{3}\frac{3}{\sqrt{11}}{\tan }^{-1}\left(\frac{t+\frac{4}{3}}{\frac{\sqrt{11}}{3}}\right)$

$=\frac{2}{\sqrt{11}}{\tan }^{-1}\left(\frac{3t+4}{\sqrt{11}}\right)+c$ where $c$ is the constant of integration

$=\frac{2}{\sqrt{11}}{\tan }^{-1}\left(\frac{3\tan \frac{x}{2}+4}{\sqrt{11}}\right)+c$ where $t=\tan \frac{x}{2}$