How would I go about solving the following integral?
$\displaystyle
\int {{1}\over{5-2\sin{x}}} dx
$
Make the u-substitution: $\displaystyle u=\tan\left(\frac{x}{2}\right)$
Then $\displaystyle dx=\frac{2\,du}{u^2+1}$, and $\displaystyle \sin{x}=\frac{2u}{u^2+1}$. Thus
$\displaystyle \int\frac{1}{5-2\sin{x}}\,dx=\int\frac{2}{5u^2-4u+5}\,du=\frac{2}{5}\int\frac{du}{u^2-\frac{4}{5}u+1}$
Completing the square will allow you to find the answer.
--Kevin C.
I was only able to make one further step and then I hit a roadblock. After completing the square I have:
$\displaystyle \int\frac{1}{(u-\frac{2}{5})^2+\frac{21}{25}} du$
However, I don't know which rule to then use to calculate the integral. Any guidance would be greatly appreciated.