I am a bit rusty on my de Rham cohomology, and I'm hoping that someone here could help me.
I want to find the cohomology of $T^*\mathbb{CP}^n$ (seen as a real manifold). Now, this should be equal to the cohomology of $\mathbb{CP}^n$ since the two are homotopic (by homotopy of each fibre with a point), thus the problem reduces to the computation of $H^\bullet(\mathbb{CP}^n)$. How can I proceed to find it? Would something as the third possibility proposed in this answer work (by taking $G=\mathbb{C}^*$ acting on $\mathbb{C}^{n+1}$)?
Let me elaborate on @Aaron's comment. Let $\omega$ be the Fubini-Study symplectic form on $\mathbb{C}\mathbb{P}^n$. You can use Aaron's approach to conclude that $H^i(\mathbb{C}\mathbb{P}^n)$ is one-dimensional for even $i$ between $0$ and $2n$ and is trivial otherwise. To compute the ring structure, one can show $\omega^j$ is closed but not exact for $0\leq j\leq n$. Therefore, its class in cohomology generates $H^{2j}(\mathbb{C}\mathbb{P}^n)$ as a vector space for all such $j$. Therefore, the cohomology of $\mathbb{C}\mathbb{P}^n$ is generated by $[\omega]$ as an algebra. The only relation is that $[\omega]^{n+1}=0$. This gives you Aaron's answer.