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/b² e^ax.cos(bx) - a²/b² ∫ e^ax.cos (bx) dx

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

On Solving LHS and RHS, we get

I (1 + a²/b²) = 1/b e^ax.sin bx + a/b² e^ax.cos(bx)

On solving this we get,

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

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