Question:medium

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

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Remember \(\int e^{ax}\cos bx\,dx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)\).
Updated On: May 19, 2026
  • \(\dfrac{1}{a^2+b^2}e^{ax}(a\sin bx-b\cos bx)\)
  • \(\dfrac{1}{a^2+b^2}e^{ax}(a\cos bx+b\sin bx)\)
  • \(\dfrac{1}{\sqrt{a^2+b^2}}e^{ax}\cos\left(bx-\tan^{-1}\dfrac{b}{a}\right)\)
  • \(\dfrac{1}{\sqrt{a^2+b^2}}e^{ax}\sin\left(bx-\tan^{-1}\dfrac{b}{a}\right)\)
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The Correct Option is B

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