Basic (I think) integral involving sin

**drew.walker** · August 11th 2009, 05:01 PM

How would I go about solving the following integral?

$<br /> \int {{1}\over{5-2\sin{x}}} dx<br />$

**TwistedOne151** · August 11th 2009, 05:46 PM

Make the u-substitution: $u=\tan\left(\frac{x}{2}\right)$
Then $dx=\frac{2\,du}{u^2+1}$ , and $\sin{x}=\frac{2u}{u^2+1}$ . Thus
$\int\frac{1}{5-2\sin{x}}\,dx=\int\frac{2}{5u^2-4u+5}\,du=\frac{2}{5}\int\frac{du}{u^2-\frac{4}{5}u+1}$
Completing the square will allow you to find the answer.

--Kevin C.

**drew.walker** · August 11th 2009, 05:54 PM

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

**mr fantastic** · August 11th 2009, 06:02 PM

Originally Posted by drew.walker

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

Google Weierstrass substitution.

**drew.walker** · August 12th 2009, 02:20 AM

I was only able to make one further step and then I hit a roadblock. After completing the square I have:

$\int\frac{1}{(u-\frac{2}{5})^2+\frac{21}{25}} du$

However, I don't know which rule to then use to calculate the integral. Any guidance would be greatly appreciated.

**mr fantastic** · August 12th 2009, 02:28 AM

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:

$\int\frac{1}{(u-\frac{2}{5})^2+\frac{21}{25}} du$

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 $\frac{a}{w^2 + a^2}$ and the integral of this (with respect to w) is $\tan^{-1} \left( \frac{w}{a}\right) + C$ .

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