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Can this: $$\frac{\cos x}{4 + \sin^2 x}$$

Be re-written using the fact that: $$\cot(t) = \frac{\cos (t)}{\sin (t)} = \frac{1}{\tan (t)}$$

I'm not good with algebra, but I'm getting there. I'm trying to simplify this expression, it's an integration by substitution task. I just don't see how I can separate $\cos x$ and $\sin x$ from the original equation.

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    If it's integration by substitution, you should substitute $u=\sin x$. You don't need $\cot$.2011-04-17
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    That would give me; 1/(4+u^2) which I still can't integrate. :(2011-04-17
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    [Are you sure about that?](http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Derivatives_of_inverse_trigonometric_functions)2011-04-17
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    Evil ponnies! The 1/a^2+x^2 rule! Thanks. If you put that in a reply I'll give you the creds!2011-04-17

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$$\int \frac{\cos x}{4 + \sin^2 x} dx = \int \frac{1}{4 + u^2} du \; .$$

Substituting $u=\sin x$.

The result is

$$\int \frac{1}{4 + u^2} du = \int \frac{1}{2(1 + (u/2)^2)} d(u/2)=\frac{1}{2}\arctan\left(\frac{u}{2}\right)+c \; .$$

Substituting back $u=\sin x$ we get the final result:

$$\int \frac{\cos x}{4 + \sin^2 x} dx = \frac{1}{2}\arctan\left(\frac{\sin x}{2}\right)+c \; .$$