Question

In Exercises \(17-20,\) use parts and solve for the unknown integral. $$\int e^{-x} \sin 2 x d x$$

Short answer

\(\int e^{-x}\sin 2x \,dx = -\frac{2}{3}e^{-x}\cos 2x -\frac{2}{3} e^{-x}\sin 2x\)

Step-by-step solution

Identify the functions

First, identify the functions that 'u' and 'dv' should represent. We can let \(u = e^{-x}\) and \(dv = \sin 2x\, dx\). This is because the derivative of an exponential function is simpler, and the integral of the sine function is easily calculated.

Apply the formula of integration by parts

Next, calculate the derivatives and integrals necessary, \(du = -e^{-x} \, dx\) and \(v = -\frac{1}{2}\cos 2x\). Then apply the integration by parts formula: \(\int u\,dv = uv - \int v\,du\). Substituting 'u', 'v', 'du' and 'dv' into the formula gives \(e^{-x} \cdot -\frac{1}{2}\cos 2x - \int -\frac{1}{2}\cos 2x \cdot -e^{-x} dx\). After simplification this gives \(-\frac{1}{2}e^{-x}\cos 2x + \frac{1}{2}\int e^{-x}\cos 2x \,dx\).

Repeat Step 1 and Step 2

We have another integral similar to the beginning, but this time it involves cosine. Similarly as before, we can let \(u = e^{-x}\), \(dv = \cos 2x \, dx\), \(du = -e^{-x} \, dx\) and \(v = \frac{1}{2}\sin 2x\). This then gives \(-\frac{1}{2}\int e^{-x}\cos 2x \, dx = -\frac{1}{2} e^{-x}\sin 2x + \frac{1}{4}\int e^{-x}\sin 2x \, dx\).

Substitute integral into original equation

Substitute \(-\frac{1}{2}\int e^{-x}\cos 2x \, dx\) from Step 3 into \(-\frac{1}{2}e^{-x}\cos 2x + \frac{1}{2}\int e^{-x}\cos 2x \, dx\) from Step 2. This gives \(-\frac{1}{2}e^{-x}\cos 2x -\frac{1}{2} e^{-x}\sin 2x + \frac{1}{4}\int e^{-x}\sin 2x \, dx\). When simplified, \(0.75 \int e^{-x}\sin 2x \,dx = -\frac{1}{2}e^{-x}\cos 2x -\frac{1}{2} e^{-x}\sin 2x \).

Solve for the unknown integral

Multiply \(0.75 \int e^{-x}\sin 2x \,dx = -\frac{1}{2}e^{-x}\cos 2x -\frac{1}{2} e^{-x}\sin 2x\) by \(\frac{4}{3}\) to isolate the integral. It gives \(\int e^{-x}\sin 2x \,dx = -\frac{2}{3}e^{-x}\cos 2x -\frac{2}{3} e^{-x}\sin 2x\).