Integrating the difference of brownian motion

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I'm reading the solutions to an exercise where it is stated that $$\int_t^T\Big(W(u) - W(t)\Big)du = \int_t^T (T-u)dW(u).$$ But can someone enlighten me to what theorem/rule can be used to show this?

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we can start from the right hand side.

$$\int_t^T ( T - u ) \, dW_u = \int_t^T T \, dW_u - \int_t^T u \, dW_u = T\big( W_T - W_t \big) - \int_t^T u \, dW_u \,\, . $$

Apply Ito formula to $T W_T$, we have

$$TW_T = t W_t + \int_t^T u \, dW_u + \int_t^T W_u \, du \,\, , $$

then we have

$$\int_t^T ( T - u ) \, dW_u = T\big( W_T - W_t \big) - \left[\,\, TW_T - t W_t - \int_t^T W_u \, du \right] = ( - T + t ) W_t + \int_t^T W_u \, du $$

while the left side is equal to

$$\int_t^T \big( W_u - W_t \big)\, du = \int_t^T W_u\, du - \int_t^T W_t\, du = \int_t^T W_u\, du - (T - t)W_t $$

Q.E.D.