$\int e^{x}\sin x\,dx$ equals
A$\dfrac{e^{x}}{2}(\sin x-\cos x)+C$
B$\dfrac{e^{x}}{2}(\sin x+\cos x)+C$
C$e^{x}(\sin x-\cos x)+C$
D$-e^{x}\cos x+C$
Answer & Solution
Correct answer: A. $\dfrac{e^{x}}{2}(\sin x-\cos x)+C$
1. Let $I=\int e^{x}\sin x\,dx$. By parts ($u=e^{x}$): $I=-e^{x}\cos x+\int e^{x}\cos x\,dx$.
2. For $\int e^{x}\cos x\,dx$, by parts: $e^{x}\sin x-\int e^{x}\sin x\,dx=e^{x}\sin x-I$.
3. So $I=-e^{x}\cos x+e^{x}\sin x-I$, giving $2I=e^{x}(\sin x-\cos x)$.
4. Hence $I=\dfrac{e^{x}}{2}(\sin x-\cos x)+C$. Option B has the wrong inner sign.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.327_
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