# What is the value of the integral e^{ax }cos (bx)?

**Solution:**

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

Here's the detailed solution.

⇒ I = ∫ e^{ax }cos (bx) dx

Let, u = e^{ax }and v = cos(bx)

∫ uv dx = u ∫ v dx − ∫ u' (∫ v dx) dx

⇒ I = 1/b e^{ax}.sin bx - a/b ∫ e^{ax}.sin bx dx

Again applying part integration to ∫ e^{ax }sin bx dx, we get

I = 1/b e^{ax}.sin bx + a/b^{2 }e^{ax}.cos(bx) - a^{2}/b^{2 }∫ e^{ax}.cos (bx) dx

⇒ I = 1/b e^{ax}.sin bx + a/b^{2 }e^{ax}.cos(bx) - (a^{2}/b^{2}) I

On Solving LHS and RHS, we get

I (1 + a^{2}/b^{2}) = 1/b e^{ax}.sin bx + a/b^{2 }e^{ax}.cos(bx)

On solving this we get,

I = e^{ax }/ (a^{2 }+ b^{2}) {b sin bx + a cos bx} + c

### Thus, the value of the integral e^{ax }cos(bx) is e^{ax }/ (a^{2 }+ b^{2}) {a cos bx + b sin bx} + c

## What is the value of the integral e^{ax }cos (bx)?

**Summary:**

The value of the integral e^{ax }cos (bx) is e^{ax }/ (a^{2 }+ b^{2}) {a cos bx + b sin bx} + c

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