For any two continuous functions  $$f,g : \mathbb{R} \to \mathbb{R}$$​, define
​$$(f * g)(t) = \int_{0}^{t} f(s) g(t-s) \, ds.$$​​
Which of the following is the value of 
​$$(f * g)(t) \ when \ f(t) = e^{-t} \ and \ g(t) = \sin t ?$$​​

Correct option is A

​$$\text{We have} \\[10pt](f * g)(t) = \int_{0}^{t} e^{-s} \sin(t-s) \, ds. \\[10pt]\text{Using trigonometric identity:} \\[10pt]\sin (t - s) = \sin t \cos s - \cos t \sin s, \\[10pt]\text{we rewrite the integral as:} \\[10pt](f * g)(t) = \int_{0}^{t} e^{-s} (\sin t \cos s - \cos t \sin s) \, ds. \\[10pt]\text{Splitting the integral:} \\[10pt](f * g)(t) = \sin t \int_{0}^{t} e^{-s} \cos s \, ds - \cos t \int_{0}^{t} e^{-s} \sin s \, ds. \\[10pt]\text{Solving the integrals:} \\[10pt]\int_{0}^{x} e^{-s} \cos s \, ds = \frac{e^{-x}}{2} [-\cos x + \sin x] \\[10pt]\int_{0}^{x} e^{-s} \sin s \, ds = \frac{-e^{-x}}{2} [\sin x + \cos x] \\[10pt]\text{Substituting the limits, we get:} \\[10pt]\sin t \left[ \frac{e^{-t}}{2} (\sin t - \cos t) + \frac{1}{2} \right] - \cos t \left[ \frac{-e^{-t}}{2} (\sin t + \cos t) + \frac{1}{2} \right]. \\[10pt]\text{Expanding:} \\[10pt]\frac{e^{-t}}{2} \left[ \sin^2 t - \sin t \cos t + \cos t \sin t + \cos^2 t \right] + \frac{\sin t}{2} - \frac{\cos t}{2} \\[10pt]\text{Using } \sin^2 t + \cos^2 t = 1: \\[10pt]\frac{1}{2} \left[ e^{-t} + \sin t - \cos t \right] \\[10pt]\text{Thus,} \\[10pt](f * g)(t) = \frac{1}{2} \left[ e^{-t} + \sin t - \cos t \right]. \\[10pt]\textbf{Correct answer:} \text{Option (A).}$$​​