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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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    $\begingroup$ If it's integration by substitution, you should substitute $u=\sin x$. You don't need $\cot$. $\endgroup$ – Raskolnikov Apr 17 '11 at 14:55
  • $\begingroup$ That would give me; 1/(4+u^2) which I still can't integrate. :( $\endgroup$ – Algific Apr 17 '11 at 15:06
  • $\begingroup$ Are you sure about that? $\endgroup$ – Raskolnikov Apr 17 '11 at 15:10
  • $\begingroup$ Evil ponnies! The 1/a^2+x^2 rule! Thanks. If you put that in a reply I'll give you the creds! $\endgroup$ – Algific Apr 17 '11 at 15:12
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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:

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

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