If $\int_0^1 \frac{\sin t}{1+t} dt = a$, then the value of $\int_{4\pi-2}^{4\pi} \frac{\sin t}{4\pi + 2 - t} dt$ is:
Step-by-Step Solution
Key Concept: Use the substitution property ∫ₐᵇ f(x)dx = ∫ₐᵇ f(a+b-x)dx to transform the integral ∫₄π₋₂⁴π sin(t)/(4π+2-t) dt into a form comparable with ∫₀¹ sin(t)/(1+t) dt by setting u = 4π - t, which reveals that the second integral equals -a due to the sign change from the substitution limits.
Let $I = \int_{4π-2}^{4π} \frac{\sin(t/2)}{4π+2-t}dt$. Using the property $\int_a^b f(x)dx = \int_a^b f(b-a+x)dx$ with substitution $u = 4π - t$, this becomes $I = 2\int_0^π \frac{\sin(t-1)dt}{(4-2t)}$, which equals $2\int_0^π \frac{\sin(t-1)dt}{4-2t}$.
Correct Answer: 2