Invariant differential form and compact Lie group actions

Question

Let $G$ be a compact Lie group acting on a manifold $M$. Denote the action by $\tau_g(p):=g\cdot p$. Let $dg=\nu$ be a left invariant volume form on $G$ such that $\int_G \nu =1$. Let $\alpha$ be a $k$-form on $M$. To simplify things, take $k=1$. Define $$ \widehat{\alpha} = \int_G [\tau_{g}^\ast \alpha]dg. $$ Concretely, for $X\in T_pM$, $\widehat{\alpha}_p(X)$ is defined by $$ \widehat{\alpha}_p(X)=\int_G \alpha_{g\cdot p}((\tau_g)_\ast X) dg. $$ How does one show that this is actually $G$-invariant? Let $f_{X,p}: G\rightarrow \mathbb{R}$ be the real valued smooth function on $G$ given by $$ f_{X,p}(g):=\alpha_{g\cdot p}((\tau_g)_\ast X). $$ Then we can rewrite $\widehat{\alpha}_p(X)$ as $$ \widehat{\alpha}_p(X) = \int_G f_{X,p}\nu $$ so that we're just integrating a real valued function on $G$ over $G$. $G$-invariance of $\widehat{\alpha}$ means $$ (\tau_h^\ast\widehat{\alpha})_p(X) =\widehat{\alpha}_{h\cdot p}((\tau_h)_\ast X)=\widehat{\alpha}_p(X). $$ To rephrase the question, how does one show $$ \int_G f_{(\tau_h)_\ast X,h\cdot p}\nu = \int_G f_{X,p}\nu? $$ The functions $f_{(\tau_h)_\ast X,h\cdot p}$ and $f_{X,p}$ are related by $$ (R_h)^\ast f_{X,p}= f_{(\tau_h)_\ast X,h\cdot p} $$ where $R_h:G \rightarrow G$ is right translation by $h\in G$. If $f_{(\tau_h)_\ast X,h\cdot p}$ and $f_{X,p}$ were related by a left translation, then we would be done, since we can use the fact that $L_h^\ast \nu =\nu$ and the fact that integrals are invariant under orientation preserving diffeomorphisms. The way to fix the problem is to define $$ \widehat{\alpha} = \int_G [\tau_{g^{-1}}^\ast \alpha]dg. $$ In other words use the pullback by $\tau_{g^{-1}}$ not $\tau_g$. However, no one seems to do this which suggests that I'm making a mistake somewhere. Can someone point out where my reasoning is incorrect?