91Ó°ÊÓ

Choose \(u\) and \(dv\) as follows: $$u = \sin 6 \theta$$ $$dv = e^{-2 \theta} d\theta$$

Differentiate \(u\) and integrate \(dv\) to obtain \(du\) and \(v\): $$du = \frac{d(\sin 6 \theta)}{d\theta} d\theta = 6\cos 6 \theta d\theta$$ $$v = \int e^{-2 \theta} d\theta = -\frac{1}{2} e^{-2 \theta}$$

Using the integration by parts formula, \(\int u dv = uv - \int v du\), we can write the given integral as follows: $$\int e^{-2 \theta} \sin 6 \theta d \theta = \left[-\frac{1}{2} e^{-2 \theta}\sin 6\theta\right] - \int -\frac{1}{2} e^{-2 \theta}(6\cos 6 \theta) d\theta$$

Simplify the integral and apply integration by parts again: $$\int e^{-2 \theta} \sin 6 \theta d\theta = -\frac{1}{2} e^{-2 \theta}\sin 6\theta + 3\int e^{-2 \theta}\cos 6\theta d\theta$$ Now, let \(u_2 = \cos 6\theta\) and \(dv_2 = e^{-2\theta} d\theta\): $$du_2 = -6\sin 6\theta d\theta$$ $$v_2 = -\frac{1}{2} e^{-2 \theta}$$ Apply the integration by parts formula again: $$3\int e^{-2 \theta}\cos 6\theta d\theta = 3\left[-\frac{1}{2} e^{-2\theta}\cos 6\theta - \int -\frac{1}{2} e^{-2 \theta}(-6\sin6\theta) d\theta\right]$$

Simplify the integral and combine the results: $$\int e^{-2 \theta} \sin 6 \theta d \theta = -\frac{1}{2} e^{-2 \theta}\sin 6\theta - \frac{3}{2} e^{-2\theta}\cos 6\theta - 9\int e^{-2\theta} \sin 6\theta d\theta$$ Notice that the integral on the right-hand side is the same as the integral we started with. Let \(I = \int e^{-2 \theta} \sin 6 \theta d \theta\). Then we have: $$I = -\frac{1}{2} e^{-2 \theta}\sin 6\theta - \frac{3}{2} e^{-2\theta}\cos 6\theta - 9I$$ Combine the terms involving I: $$10I = -\frac{1}{2} e^{-2 \theta}\sin 6\theta - \frac{3}{2} e^{-2\theta}\cos 6\theta$$ Finally, divide by 10 to obtain the result for the integral: $$I = \int e^{-2 \theta} \sin 6 \theta d \theta = -\frac{1}{20} e^{-2 \theta}\sin 6\theta - \frac{3}{20} e^{-2\theta}\cos 6\theta + C$$ where \(C\) is the constant of integration.