Question

Evaluate:$\int \frac{dt}{16+6\sin t}$

Answer 1

$\int \frac{dt}{16+6\sin t}$

$=\frac{1}{2}\int \frac{dt}{8+3\sin t}$

Let $\sin t=\frac{1-{\tan }^2\frac{t}{2}}{1+{\tan }^2\frac{t}{2}}$

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

$=\frac{1}{2}\int \frac{{\sec }^2\frac{t}{2}}{8+8{\tan }^2\frac{t}{2}+3-3{\tan }^2\frac{t}{2}}dt$

Let $u=\tan \frac{t}{2}\Rightarrow du=\frac{1}{2}{\sec }^2\frac{t}{2}dt$

$\Rightarrow 2du={\sec }^2\frac{t}{2}dt$

$=\frac{1}{2}\int \frac{2du}{11+5u^2}$

$=\frac{1}{2}\int \frac{2du}{\frac{1}{5}({\left(\sqrt{55}\right)}^2+u^2)}$

$=\frac{5}{\sqrt{55}}{\tan }^{-1}\left(\frac{u}{\sqrt{55}}\right)+c$ where $c$ is the constant of integration.

$=\frac{5}{\sqrt{55}}{\tan }^{-1}\left(\frac{\tan \frac{t}{2}}{\sqrt{55}}\right)+c$ where $u=\tan \frac{t}{2}$