# Tricky Integral

Hey guys,

By what rules is the integral of

$$\frac{cos(t)}{1+9sin^2(t)}=\frac{1}{3}tan^{-1}(sin(3t))$$

I know this is right, but I have no idea why and can't find any trig identities to help. Thanks.

Let $$u = 3\sin t$$
and use the fact that $$\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$$