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Let $M$ be a smooth manifold and let $\Delta _i$ for $i=1,...,k$ be distributions of $TM$ which are integrable such that $\Delta_i \cap \Delta _j$ is zero distribution for $i \neq j$. Suppose $\Delta = \Delta_1 \oplus ...\oplus \Delta _k$. Is $\Delta$ integrable?

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No. On the 3-sphere, take the usual metric and the 2-plane field $\Delta$ perpendicular to the fibers of the Hopf fibration. It splits as a sum of two line fields $\Delta_1$ and $\Delta_2$. I will write out the details when I have a moment.

Edit: some details. You think of $S^3$ as the unit quaternions, and so as a Lie group, with Maurer-Cartan form $\omega=\omega_1 i + \omega_2 j + \omega_3 k$ valued in the imaginary quaternions. (Or use $SO(3)$ if you prefer; it has the same Lie algebra). Then $\omega_1=0$ is a contact plane field $\Delta$, by the structure equations $d \omega_1 = -2 \omega_2 \wedge \omega_3$. But this is the sum of the plane fields $\Delta_1=(\omega_1=\omega_2=0)$ and $\Delta_2=(\omega_1=\omega_3=0)$.

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