I need help with a calculus integration problem

Determine the integral I=cosx/9+sin^2(x)

Help please?

Please use parantheses next time!! Do you mean cos(x)/9 + sin²(x) or (cos(x))/(9 + sin²(x))? I'll do both for this case only.

∫cos(x)/9 + sin²(x) dx = 1/9*sin(x) + 1/2*[x - 1/2*sin(2x)] + C

∫(cos(x))/(9 + sin²(x)) dx

u = sin(x)

du = cos(x) dx

∫du/(u² + 9) = 1/3*arctan(u/3) + C

= 1/3*arctan(sin(x)/3) + C

u = sin(x)
du = cos(x) * dx

du / (9 + u^2)

Let u = 3 * tan(t)
du = 3 * sec(t)^2 * dt

3 * sec(t)^2 * dt / (9 + 9tan(t)^2) =>
3 * sec(t)^2 * dt / (9 * sec(t)^2) =>
dt / 3

Integrate

(1/3) * t + C
(1/3) * arctan(u/3) + C
(1/3) * arctan(sin(x) / 3) + C

∫sin² x dx = ½ x − ¼ sin 2x + C

and ∫cos x/9 = sin x/9

answer: sin x/9 + ½ x − ¼ sin 2x + C