If $\int(\sin(2020x))(\sin^{2018} x) dx$ is equal to $\frac{(\sin(a)).(\sin x)^b}{c} + k$ (where $k$ is integration constant) then $\frac{a+b+c}{3} = $
Step-by-Step Solution
Key Concept: Decompose $\sin(2019x)$ using angle addition formulas to create a reducible structure with the power $\sin^{2018}x$.
Using product-to-sum technique, rewrite $\sin(2019x)\sin^{2018}x$ by expressing $\sin(2019x) = \sin((2019x+x)-x) = \sin(2019x+x)\cos x - \cos(2019x+x)\sin x$. This splits the integral into two parts: $I = \int(\sin 2019x)\sin^{2018}x\cos x\,dx + \int(\cos 2019x)\sin^{2019}x\,dx$. The first integral evaluates using substitution $u = \sin x$, giving $\frac{(\sin 2019x)(\sin^{2019}x)}{2019}$. The second integral recycles back into $I$, allowing algebraic cancellation to yield the final answer $I = \frac{(\sin 2019x)(\sin^{2019}x)}{2019} + c$.
Correct Answer: 0