Evaluate the following integrals:

• ∫ e^ax cos b x d x

To evaluate the integral ∫ e^(ax) cos(bx) dx, we can use the method of integration by parts or the technique of complex numbers. Here’s a straightforward approach using integration by parts.

Integration by Parts

We start by letting:

• u = cos(bx) (which we will differentiate)
• dv = e^(ax) dx (which we will integrate)

Now, we differentiate and integrate:

• du = -b sin(bx) dx
• v = (1/a)e^(ax)

Applying Integration by Parts

Using the integration by parts formula ∫u dv = uv - ∫v du, we have:

∫ e^(ax) cos(bx) dx = (1/a)e^(ax) cos(bx) - ∫ (1/a)e^(ax)(-b sin(bx)) dx

This simplifies to:

∫ e^(ax) cos(bx) dx = (1/a)e^(ax) cos(bx) + (b/a) ∫ e^(ax) sin(bx) dx

Second Integral

Now, we need to evaluate ∫ e^(ax) sin(bx) dx using a similar method:

• u = sin(bx)
• dv = e^(ax) dx

Following the same steps:

• du = b cos(bx) dx
• v = (1/a)e^(ax)

Final Steps

Applying integration by parts again:

∫ e^(ax) sin(bx) dx = (1/a)e^(ax) sin(bx) - (b/a) ∫ e^(ax) cos(bx) dx

Now, we have two equations:

• ∫ e^(ax) cos(bx) dx = (1/a)e^(ax) cos(bx) + (b/a) ∫ e^(ax) sin(bx) dx
• ∫ e^(ax) sin(bx) dx = (1/a)e^(ax) sin(bx) - (b/a) ∫ e^(ax) cos(bx) dx

Solving the System

Let’s denote:

• I_c = ∫ e^(ax) cos(bx) dx
• I_s = ∫ e^(ax) sin(bx) dx

Substituting I_s into the equation for I_c gives us a system of equations that can be solved simultaneously. After some algebra, we find:

I_c = (e^(ax) (a cos(bx) + b sin(bx))) / (a^2 + b^2) + C

Thus, the final result for the integral is:

∫ e^(ax) cos(bx) dx = (e^(ax) (a cos(bx) + b sin(bx))) / (a^2 + b^2) + C