To find $\gamma^∗ω$ given $\gamma(t) = (\cos t, \sin t)$.

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Consider the 1-form $w = \frac{−ydx+xdy}{x^2+y^2}$ defined on $R^2 − \{(0, 0)\}$. Let $\gamma : R → R^2 − \{(0, 0)\}$ be a curve given by $\gamma(t) = (\cos t, \sin t)$. To find $\gamma^∗ω$.

My Attempt:

$$\gamma^∗ω = \gamma^*(\frac{−ydx+xdy}{x^2+y^2}) = \frac{−(\gamma^*y)d(\gamma^*x)+(\gamma^*x)d(\gamma^*y)}{(\gamma^*x)^2+(\gamma^*y)^2} = \frac{ \sin^2t dt + \cos^2t dt}{\cos^2t+sin^2t} = dt$$

Is the solution correct?