Question

Use integration by parts to verify the formula. (For Exercises \(57-60\), assume that \(n\) is a positive integer.) $$ \int e^{a x} \cos b x d x=\frac{e^{a x}(a \cos b x+b \sin b x)}{a^{2}+b^{2}}+C $$

Short answer

After solving the exercise via the integration by parts method, the given formula \(\int e^{a x} \cos b x dx=\frac{e^{a x}(a \cos b x+b \sin b x)}{a^{2}+b^{2}}+C\) is found to be correct.

Step-by-step solution

Identify the factors to be used in the Integration By Parts formula.

The formula for integration by parts is \(\int u dv = u v - \int v du \). The challenge is choosing what \(u\) and \(dv\) should be. In this exercise, consider \(u=\cos(bx)\) and \(dv = e^{a x} dx\).

Compute for \(du\) and \(v\).

Taking the differential of \(u\), we can find \(du = -b \sin(bx) dx\). Next, by integrating \(dv\), obtain \(v = \frac{1}{a} e^{ax}\).

Apply the integration by parts formula.

Substitute \(u\), \(v\), and \(du\) into the integration by parts formula: \( \int e^{a x} \cos b x dx = u v - \int v du = \cos(bx) \cdot \frac{1}{a} e^{ax} - \int \frac{1}{a} e^{ax} (-b \sin(bx)) dx\).

Simplify the equation.

The equation simplifies to \(\frac{e^{ax} \cos(bx)}{a} - \frac{-b}{a} \int e^{ax} \sin(bx) dx\). This integral can be solved by another round of integration by parts where \(u=\sin(bx)\), \(dv = e^{ax} dx\), \(du=b\cos(bx) dx\), and \(v=\frac{1}{a} e^{ax}\).

Use integration by parts for the remaining integral.

Perform the integration by parts on the second integral: \(\int e^{ax} \sin(bx) dx = \sin(bx) \cdot \frac{1}{a} e^{ax} - \frac{1}{a} \int e^{ax} b \cos(bx)dx\). The integral appears like the original problem again and can be expressed using this same formula. The equation then simplifies to \(\frac{e^{ax} \cos(bx)}{a} + \frac{b}{a^2} \cdot e^{ax} \sin(bx) - \frac{b}{a^2} \int e^{ax} \cos(bx) dx\).

Move the recurring integral to the left side of the equation.

After moving the remaining integral to the left side, obtain \(\left(1+\frac{b^2}{a^2}\right) \int e^{ax} \cos(bx) dx = \frac{e^{ax} \cos(bx)}{a} + \frac{b}{a^2} \cdot e^{ax} \sin(bx)\).

Solve the left side for the original integral.

Solve for \(\int e^{ax} \cos(bx) dx\), to get \(\int e^{ax} \cos(bx) dx = \frac{e^{ax} (a \cos(bx) + b \sin(bx))}{a^2 + b^2} + C\) where \(C\) is the constant of integration.