Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 239: 46

$\sin 2x+\cos 3x+C$

As we are given that $ 2 \int (\cos 2x ) dx - 3 \int (\sin 3x) dx$ Thus, we have $2 \int (\cos 2x ) dx - 3 \int (\sin 3x) dx=(2)(\dfrac{1}{2}) \sin 2x -(3)(\dfrac{1}{3})(\cos 3x)+C$ or, $=\sin 2x+\cos 3x+C$