Question textThe indefinite integral\displaystyle \int e^{-x}\sin(x)\,dxcan be written in the form\dfrac{\cos(x)}{Ae^x} + \dfrac{\sin(x)}{Be^x} + Cwhere A and B are integers, and C is a constant of integration.Use integration by parts (twice) to find A and B.Fill in the spaces with the correct responses.A= Answer 1 Question 27[input]B= Answer 2 Question 27[input]多项填空题
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Question textThe indefinite integral\displaystyle \int e^{-x}\sin(x)\,dxcan be written in the form\dfrac{\cos(x)}{Ae^x} + \dfrac{\sin(x)}{Be^x} + Cwhere A and B are integers, and C is a constant of integration.Use integration by parts (twice) to find A and B.Fill in the spaces with the correct responses.A= Answer 1 Question 27[input]B= Answer 2 Question 27[input]多项填空题

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BlackTom AI · Accepted Answer

We need to evaluate the integral ∫ e^{-x} sin x dx and express it in the form cos(x)/(A e^x) + sin(x)/(B e^x) + C with integers A and B.

Approach idea: use integration by parts twice, or recall the standard formula for integrals of e^{ax} sin(bx) or e^{ax} cos(bx). Here a = -1 and b = 1, so the standard result applies and avoids long algebra: ∫ e^{ax} sin(bx) dx = e^{ax} (a sin(bx) - b cos(bx)) / (a^2 + b^2) + C.

Plugging a = -1 and b = 1 gives:
∫ e^{-x} sin x dx = e^{-x} [(-1) sin x - 1 cos x] / ((-1)^2 + 1^2) + C
= e^{-x} [ -sin x - cos x ] / 2 + C
= -(e^{-x} sin x)/2 - (e^{-x} cos x)/2 + C.

Now rewrite e^{-x} as 1/e^{x} to match the requested form:
= cos(x)/(A e^x) + sin(x)/(B e^x) + C, with A and B integers.
Comparing, we have: -(e^{-x} cos x)/2 = cos(x)/(A e^x) and similarly -(e^{-x} sin x)/2 = sin(x)/(B e^x).
This gives A = -2 and B = -2, since cos(x)/( -2 e^x) = -(cos x)/(2 e^x) and sin(x)/( -2 e^x) = -(sin x)/(2 e^x).

Thus, the values that fit the required form are A = -2 and B = -2.

Why other potential approaches would fail: attempting to force different A or B would lead to mismatched coefficients in front of cos x or sin x after expressing the integral in terms of e^{-x} times those trig functions. The standard closed form for this specific combination yields exactly the equal negative half-coefficients for both sine and cosine terms, i.e., both A and B equal -2.