Examples of de Rham cohomology being easier to compute that singular cohomology

Question

De Rham's theorem states that for any smooth manifold $M$ the singular cohomology and de Rham cohomology of $M$ are isomorphic.

Are there any examples of manifolds for which it is easier to compute the de Rham cohomology?

Answer 1

I would say that de Rham cohomology is always easier to calculate than singular cohomology!
Singular cohomology is essentially impossible to calculate if you just use the definition: you need general theorems.
You don't believe me? I challenge you to calculate $H^2_{\operatorname {sing}}((0,1), \mathbb R)$ from the definition : good luck with your non-denumerable set of singular $2$-simplices in $(0,1)$ and the $\mathbb R$-vector space they generate!
By contrast $H^2_{\operatorname {de Rham}}((0,1), \mathbb R)=0$ is obvious because the only differential form of degree $2$ on $(0,1)$ is zero.
So you can make this "calculation" one picosecond after you have learned the definition of de Rham cohomology!