Question

If an antiderivative of \(f(x)\) is \(e^x\) and an antiderivative of \(g(x)\) is \(cosx\), then \(\int f(x)cosx\,dx+\int g(x)e^x\,dx =\)

• A. \(e^xsinx+c\)
• B. \(e^x(f(x)+g(x))+c\)
• C. \(e^xcosx+c\)
• D. \(e^x+c\)

Hint:

The two integrands are the parts of d(e^x cos x).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Let \(F(x)=e^x\) with \(F'=f\), and \(G(x)=\cos x\) with \(G'=g\). Then \(f=e^x\) and \(g=-\sin x\).

Step 2: Rewrite the integrals:
\(\int f\cos x\,dx=\int F'G\,dx\). And \(\int g\,e^x\,dx=\int G'F\,dx\).

Step 3: Add:
\[ \int(F'G+FG')dx=\int(FG)'dx=FG+c \]

Step 4: Substitute:
\(FG=e^x\cos x\). So the sum is \(e^x\cos x+c\).

Step 5: Verify directly:
\(\int e^x\cos x\,dx+\int(-\sin x)e^x\,dx\). The first is \(\tfrac{e^x}{2}(\cos x+\sin x)\) and the second is \(-\tfrac{e^x}{2}(\sin x-\cos x)\). The sum is \(e^x\cos x\). Option (C).

Final Answer:
The sum is e^x cos x + c. \[ \boxed{e^x\cos x+c} \]