Question

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

Short answer

\(\int e^{a x} \sin b x d x = \frac{e^{ax} [a \sin(bx) - b \cos(bx)]}{a^{2} + b^{2}} + C\).

Step-by-step solution

Set up the integral by parts formula

In our integral, we set \(u = \sin(bx)\) and \(dv = e^{ax} dx\). To find the other parts of the integration by parts formula, we calculate \(du/dx\) and \(v\), which gives us \(du = b\cos(bx) dx\) and \(v = \int e^{ax} dx = \frac{1}{a} e^{ax}\).

Apply the integration by parts formula

Applying the integration by parts formula, we get \(\int e^{a x} \sin b x d x = u*v - \int v du = \sin(bx) \frac{1}{a} e^{ax} - \frac{1}{a} \int e^{ax} b \cos(bx) dx\).

Apply the integration by parts again to the second term

Now, we apply the integration by parts again with \(u = \cos(bx)\), \(dv = b e^{ax} dx\), which gives us \(du = -b\sin(bx) dx\), \(v = \frac{1}{a} e^{ax}\). This results in \(-\frac{1}{a} \int e^{ax} b \cos(bx) dx = -\cos(bx) \frac{b}{a^{2}} e^{ax} + \frac{b}{a^{2}} \int e^{ax} b \sin(bx) dx\). This simplifies to \(\frac{e^{ax}}{a^{2}} [\sin(bx)a - \cos(bx)b] + C\).

Combine all steps into final result

The final result after steps 2 and 3 is \( \int e^{a x} \sin b x d x = \frac{e^{ax} \sin(bx)}{a} - \frac{e^{ax}}{a^{2}} [\sin(bx)a - \cos(bx)b] + C\). which simplifies further to: \(\frac{e^{ax} [a \sin(bx) - b \cos(bx)]}{a^{2} + b^{2}} + C\).