What Is The Value Of The Integral E^[ax] Cos(bx) Dx? [Solved] - Cuemath

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Solution:

Integration is exactly the reverse of differentiation. We will use integration by parts to solve this function.

Here's the detailed solution.

⇒ I = ∫ eax cos (bx) dx

Let, u = eax and v = cos(bx)

∫ uv dx = u ∫ v dx − ∫ u' (∫ v dx) dx

⇒ I = 1/b eax.sin bx - a/b ∫ eax.sin bx dx

Again applying part integration to ∫ eax sin bx dx, we get

I = 1/b eax.sin bx + a/b2 eax.cos(bx) - a2/b2 ∫ eax.cos (bx) dx

⇒ I = 1/b eax.sin bx + a/b2 eax.cos(bx) - (a2/b2) I

On Solving LHS and RHS, we get

I (1 + a2/b2) = 1/b eax.sin bx + a/b2 eax.cos(bx)

On solving this we get,

I = eax / (a2 + b2) {b sin bx + a cos bx} + c

Thus, the value of the integral eax cos(bx) is eax / (a2 + b2) {a cos bx + b sin bx} + c

What is the value of the integral eax cos (bx)?

Summary:

The value of the integral eax cos (bx) is eax / (a2 + b2) {a cos bx + b sin bx} + c

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