Question

If $u = \int e^x \cos x \, dx,\; v = \int e^x \sin x \, dx$, then $u + v =$

• A. $-u' + C$
• B. $u' + C$
• C. $-v' + C$
• D. $v' + C$
• E. $2v' + C$

Hint:

For integrals involving $e^x\sin x$ and $e^x\cos x$: - Try combining them - Check derivatives of $e^x\sin x$ or $e^x\cos x$ to simplify quickly

Answer 1

The Correct Option is D

Concept: Use differentiation of integrals: \[ \frac{d}{dx}\left(\int f(x)\,dx\right) = f(x) \] So, \[ u' = e^x \cos x, \quad v' = e^x \sin x \]

Step 1: Write expressions for $u'$ and $v'$.
\[ u' = e^x \cos x, \quad v' = e^x \sin x \]

Step 2: Add the two given integrals.
\[ u + v = \int e^x \cos x \, dx + \int e^x \sin x \, dx \] \[ = \int e^x (\cos x + \sin x)\, dx \]

Step 3: Observe relation with derivative.
\[ v' = e^x \sin x \] Also, \[ \frac{d}{dx}(e^x \sin x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x) \]

Step 4: Relate integral to derivative.
\[ \int e^x(\sin x + \cos x)\, dx = e^x \sin x + C \]

Step 5: Express in terms of $v'$.
\[ v' = e^x \sin x \] \[ \therefore u + v = v' + C \]