# Thread: Basic (I think) integral involving sin

1. ## Basic (I think) integral involving sin

How would I go about solving the following integral?

$\displaystyle \int {{1}\over{5-2\sin{x}}} dx$

2. 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.

3. Wow. That was much more complicated than I expected. Not quite sure how you made the step to $\displaystyle \sin{x}=\frac{2u}{u^2+1}$, but I'll digest everything and try and re-attempt it. Thanks for the reply.

4. Originally Posted by drew.walker
Wow. That was much more complicated than I expected. Not quite sure how you made the step to $\displaystyle \sin{x}=\frac{2u}{u^2+1}$, but I'll digest everything and try and re-attempt it. Thanks for the reply.

5. 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.

6. Originally Posted by drew.walker
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.
The integrand can be put into the form $\displaystyle \frac{a}{w^2 + a^2}$ and the integral of this (with respect to w) is $\displaystyle \tan^{-1} \left( \frac{w}{a}\right) + C$.