Question:
If \(\int \mathrm{e}^{2x}(\cos x+7\sin x)\,\mathrm{d}x=\mathrm{e}^{2x}g(x)+\mathrm{c}\), where c is a constant of integration, then \(g(0)+g\left(\frac{\pi}{2}\right)\) is equal to ________.
If \(\int \mathrm{e}^{2x}(\cos x+7\sin x)\,\mathrm{d}x=\mathrm{e}^{2x}g(x)+\mathrm{c}\), where c is a constant of integration, then \(g(0)+g\left(\frac{\pi}{2}\right)\) is equal to ________.
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Differentiate \(e^{2x}(A\cos x+B\sin x)\) and match coefficients with \(e^{2x}(\cos x+7\sin x)\).
Updated On: Nov 12, 2026
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Correct Answer: 2
Solution and Explanation
Let \(g(x)=A\cos x+B\sin x\). Then \(\frac{d}{dx}\left(e^{2x}g(x)\right)=e^{2x}\left[(2A+B)\cos x+(2B-A)\sin x\right]\).
Matching with \(\cos x+7\sin x\): \(2A+B=1\) and \(2B-A=7\).
Solving: \(A=-1,\ B=3\), so \(g(x)=-\cos x+3\sin x\).
\(g(0)=-1\) and \(g\left(\frac{\pi}{2}\right)=3\), so \(g(0)+g\left(\frac{\pi}{2}\right)=2\).
Answer: 2
Matching with \(\cos x+7\sin x\): \(2A+B=1\) and \(2B-A=7\).
Solving: \(A=-1,\ B=3\), so \(g(x)=-\cos x+3\sin x\).
\(g(0)=-1\) and \(g\left(\frac{\pi}{2}\right)=3\), so \(g(0)+g\left(\frac{\pi}{2}\right)=2\).
Answer: 2
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