Question

The reduction formula for \(\int e^{ax}\cos bx\,dx\) is:

• A. \(\dfrac{1}{a^2+b^2}e^{ax}(a\sin bx-b\cos bx)\)
• B. \(\dfrac{1}{a^2+b^2}e^{ax}(a\cos bx+b\sin bx)\)
• C. \(\dfrac{1}{\sqrt{a^2+b^2}}e^{ax}\cos\left(bx-\tan^{-1}\dfrac{b}{a}\right)\)
• D. \(\dfrac{1}{\sqrt{a^2+b^2}}e^{ax}\sin\left(bx-\tan^{-1}\dfrac{b}{a}\right)\)

Hint:

Remember \(\int e^{ax}\cos bx\,dx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)\).

Answer 1

The Correct Option is B

Concept:
The standard integral is: \[ \int e^{ax}\cos bx\,dx = \frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx) \]

Step 1: Use standard result. \[ I=\int e^{ax}\cos bx\,dx \] \[ I=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx) \]

Step 2: Verification by differentiation.
Differentiate: \[ \frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx) \] Using product rule, the result becomes: \[ e^{ax}\cos bx \] So the formula is correct. \[ \therefore \text{Correct Answer is (B)} \]