The problem is as follows:
Solve the initial value problem $$9(\sin(t)\frac{dy}{dt} + \cos(t)y) = (\cos(t))(\sin(t))^3, $$ for $ 0 < t < \pi$ and $y(\pi/2) = 9$.
So far I have this by dividing by $9\sin(t)$: $$\frac{dy}{dx} + cot(t)y = \frac{\cot(t)\sin^2(t)}{9}, $$
Then used the integrating factor of $e^{\cot(t)}$ in order to get: $$\frac{d}{dt}(e^{\cot(t)}y) = e^{\cot(t)} \frac{\cot(t)\sin^2(t)}{9}, $$ I do not know how to proceed from here as the integration itself is complex and my current process does not seem to be right.
First, when you divide through, the coefficient on $y$ is also divided by 9, and the rhs should have $\cos$ not $\cot$.
Second, the integrating factor is $e^{\int \cot t/9 \; dt}$, which, as Moo says, turns out to be $\sqrt[9]{\sin t}$. This makes the "hard" integral on the rhs into a fairly easy $u$-sub.