How would I go about solving the following integral?

$\displaystyle

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

$

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- Aug 11th 2009, 05:01 PMdrew.walkerBasic (I think) integral involving sin
How would I go about solving the following integral?

$\displaystyle

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

$ - Aug 11th 2009, 05:46 PMTwistedOne151
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. - Aug 11th 2009, 05:54 PMdrew.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.

- Aug 11th 2009, 06:02 PMmr fantastic
- Aug 12th 2009, 02:20 AMdrew.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. - Aug 12th 2009, 02:28 AMmr fantastic