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]

Question

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]

BlackTom AI · Accepted Answer

First, restating the problem in my own words: we want the indefinite integral ∫ e^(−x) sin(x) dx to be written in the form cos(x)/(A e^x) + sin(x)/(B e^x) + C, where A and B are integers, and we are told to use integration by parts twice. In this particular instance, the given answer for A and B are both −2.

Now, let's work through the integration step by step and keep track of how the coefficients emerge.

Step 1: Do the first integration by parts. Let u = sin x and dv = e^(−x) dx. Then du = cos x dx and v = −e^(−x). The integration by parts formula gives:
I = ∫ e^(−x) sin x dx = u v − ∫ v du
= (sin x)(−e^(−x)) − ∫ (−e^(−x))(cos x) dx
= − e^(−x) sin x + ∫ e^(−x) cos x dx.

Step 2: Set up the second integral J = ∫ e^(−x) cos x dx and apply integration by parts again. Take u = cos x and dv = e^(−x) dx (so v = −e^(−x), du = − sin x dx). Then:
J = ∫ e^(−x) cos x dx = u v − ∫ v du
= (cos x)(−e^(−x)) − ∫ (−e^(−x))(− sin x) dx
= − e^(−x) cos x − ∫ e^(−x) sin x dx
= − e^(−x) cos x − I.

Step 3: Combine the two results. From Step 1 we have I = − e^(−x) sin x + J. Substitute J from Step 2:
I = − e^(−x) sin x + (− e^(−x) cos x − I).
Thus I = − e^(−x) sin x − e^(−x) cos x − I.
Add I to both sides to collect like terms: 2I = − e^(−x)(sin x + cos x).
Divide by 2: I = − (1/2) e^(−x)(sin x + cos x) + C.

Step 4: rewrite in the requested form. Distribute the −1/2 and factor e^x in the denominator:
I = −(sin x)/(2 e^x) − (cos x)/(2 e^x) + C
= cos(x)/(A e^x) + sin(x)/(B e^x) + C, with A and B chosen so that 1/A = −1/2 and 1/B = −1/2.
This gives A = −2 and B = −2.

Edge note: If you plug A = −2 and B = −2 back into the expression, you recover I = (cos x)/(−2 e^x) + (sin x)/(−2 e^x) + C = −(cos x + sin x)/(2 e^x) + C, which matches the result obtained above. This confirms the values are consistent.

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