Is the following working correct?

\[ \int_0^\infty \sin x \, dx = \int_0^\pi \sin x \, dx +\int_\pi^\infty \sin x \, dx \]

The first integral in RHS is equal to \( [ -\cos x ]_0^\pi = \cos 0 - \cos \pi = 2 \).

For the second integral in RHS, we use a substitution \(y = x - \pi\), then \(x = y + \pi \). So \(\sin x = \sin(y + \pi) = -\sin y\). We have

\[\begin{eqnarray} \int_0^\infty \sin x \, dx &=& 2 - \int_0^{\infty} \sin y \, dy \\ &=& 2 - \int_0^{\infty} \sin x \, dx \\ 2 \int_0^\infty \sin x \, dx &=& 2 \\ \int_0^\infty \sin x \, dx &=& 1. \\ \end{eqnarray}\]

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