Here's one way to do it.

int{sinycosy/(9 + cos^4(y)}dy

= int{sinycosy/[9 + (cos^2(y))^2]}dy

Now we proceed by substitution:

Let u = cos^2y

=> du = -2cosysiny dy

=> (-1/2)du = cosysiny dy

So our integral becomes

(-1/2)*int{1/(9 + u^2)}du

= (-1/2)*int{1/(3^2 + u^2)}du

= (-1/2)*int{(1/9)*(1/[1 + (u^2)/9])}du

= (-1/2)(1/9)*int{1/(1 + (u/3)^2)}du

Substitution again:

Let v = u/3

=> dv = 1/3 du

=> 3dv = du

So our new integral is:

(-1/18)(3)int{1/(1 + v^2)}dv

= (-1/6)int{1/(1 + v^2)}dv

= (-1/6)arctan(v) + C

= (-1/6)arctan(cos^2(y)/3) + C

This method is correct, but there are other ways to do it that will end up with answers that look different. If you type this function in the "integrator" for instance, the answer looks different from what we have here. But any difference in the answer can be resolved by algebraic manipulation, if not, the difference is absorbed in the constant.