# Basic (I think) integral involving sin

• Aug 11th 2009, 05:01 PM
drew.walker
Basic (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 PM
TwistedOne151
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 PM
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.
• Aug 11th 2009, 06:02 PM
mr fantastic
Quote:

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.

• Aug 12th 2009, 02:20 AM
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.
• Aug 12th 2009, 02:28 AM
mr fantastic
Quote:

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 $\displaystyle \tan^{-1} \left( \frac{w}{a}\right) + C$.