1. ## Difficult integral

I have been asked to integrate the folowing function:

sin(x)/(1+9cos^2(x))

and am having considerable difficulties.

Could anyone please tell me how to progress - I don't want the answer, just a method for solving this indefinate integral.

Thanks in advance for any help.

Regards,

Kirstie

2. Let $u=3cos(x), \;\ \frac{-du}{3}=sin(x)dx$.

Then you should see an arctan somewhere.

3. Originally Posted by kirstie
I have been asked to integrate the folowing function:

sin(x)/(1+9cos^2(x))

and am having considerable difficulties.

Could anyone please tell me how to progress - I don't want the answer, just a method for solving this indefinate integral.

Thanks in advance for any help.

Regards,

Kirstie
let $u=3\cos(x) \to du=-3\sin(x)dx$

$\int \frac{\sin(u)}{1+9\cos(u)}dx=-\frac{1}{3}\int\frac{du}{1+u^2}$

This is in the form of the arctangent so we get

$-\frac{1}{3}\int\frac{du}{1+u^2}=-\frac{1}{3}\tan^{-1}(u)+C=-\frac{1}{3}\tan^{-1}\left( 3\cos(x)\right)+C$

4. Thanks for your prompt replies.

However shouldn't the final answer be

1/3tan^-1(3cos(x))+c

since u = 3cos(x) not cos(x)/3

Kirstie

5. Yes it should. I fixed the above post.