1. Find the constants \( a \) and \( b \) so that \( f(x)=a x \cos (x)+b \sin (x) \) is an antiderivative of \( f(x)=x \sin (x) \).(lint: do NOT integrate)

To find the constants \( a \) and \( b \) such that \( f(x) = a x \cos(x) + b \sin(x) \) is an antiderivative of \( g(x) = x \sin(x) \), we need to differentiate \( f(x) \) and set it equal to \( g(x) \). First, we differentiate \( f(x) \): \[ f(x) = a x \cos(x) + b \sin(x) \] Using the product rule and the chain rule, we differentiate \( f(x) \): 1. Differentiate \( a x \cos(x) \): \[ \frac{d}{dx}(a x \cos(x)) = a \left( \cos(x) + x (-\sin(x)) \right) = a \left( \cos(x) - x \sin(x) \right) \] 2. Differentiate \( b \sin(x) \): \[ \frac{d}{dx}(b \sin(x)) = b \cos(x) \] Now, combine these results: \[ f'(x) = a \left( \cos(x) - x \sin(x) \right) + b \cos(x) \] Simplifying this, we have: \[ f'(x) = (a + b) \cos(x) - a x \sin(x) \] We want \( f'(x) \) to equal \( g(x) = x \sin(x) \): \[ (a + b) \cos(x) - a x \sin(x) = x \sin(x) \] Now, we can equate the coefficients of \( \cos(x) \) and \( \sin(x) \): 1. For the coefficient of \( \cos(x) \): \[ a + b = 0 \] 2. For the coefficient of \( \sin(x) \): \[ -a = 1 \] From the second equation, we find: \[ a = -1 \] Substituting \( a = -1 \) into the first equation: \[ -1 + b = 0 \implies b = 1 \] Thus, the constants are: \[ \boxed{a = -1} \quad \text{and} \quad \boxed{b = 1} \]