Right choice is (c) -cos2x+3x+C
Best explanation: To find ∫ 2 sin2x+3 dx
\(\int \,2 \,sin2x+3 \,dx=\int \,2 \,sin2x \,dx + \int \,3 \,dx\)
\(\int \,2 \,sin2x+3 \,dx=2\int \,sin2x \,dx+3\int \,dx\)
\(\int \,2 \,sin2x+3 \,dx=\frac{-2 cos2x}{2}+3x\)
∴∫2 sin2x+3 dx=-cos2x+3x+C