evaluate the integral. ∫e^7θsin(8θ)dθ step 1 we will begin by letting u = sin(8θ) and dv = e^7θdθ. then du =…

evaluate the integral. ∫e^7θsin(8θ)dθ step 1 we will begin by letting u = sin(8θ) and dv = e^7θdθ. then du = 8cos8θ 8cos(8θ)dθ and v = 1/7e^7θ e^7θ/7. step 2 after integration by parts we have ∫e^7θsin(8θ)dθ = sin8θ 1/7e^7θ 1/7e^7θsin(8θ) - 8/7 8/7 ∫e^7θcos(8θ)dθ. step 3 well now apply the integration by parts procedure to the new integral ∫e^7θcos(8θ)dθ, letting u = cos(8θ) and dv = e^7θdθ. then du = -8sin8θ -8sin(8θ)dθ and v = 1/7e^7θ e^7θ/7. step 4 after integration by parts we have ∫e^7θcos(8θ)dθ = 1/7e^7θcos8θ - 8/7 × ∫e^7θsin(8θ)dθ.

evaluate the integral. ∫e^7θsin(8θ)dθ step 1 we will begin by letting u = sin(8θ) and dv = e^7θdθ. then du = 8cos8θ 8cos(8θ)dθ and v = 1/7e^7θ e^7θ/7. step 2 after integration by parts we have ∫e^7θsin(8θ)dθ = sin8θ 1/7e^7θ 1/7e^7θsin(8θ) - 8/7 8/7 ∫e^7θcos(8θ)dθ. step 3 well now apply the integration by parts procedure to the new integral ∫e^7θcos(8θ)dθ, letting u = cos(8θ) and dv = e^7θdθ. then du = -8sin8θ -8sin(8θ)dθ and v = 1/7e^7θ e^7θ/7. step 4 after integration by parts we have ∫e^7θcos(8θ)dθ = 1/7e^7θcos8θ - 8/7 × ∫e^7θsin(8θ)dθ.

Answer

Answer:

Let (I = \int e^{7\theta}\sin(8\theta)d\theta).

Using integration - by - parts: First, let (u=\sin(8\theta)) and (dv = e^{7\theta}d\theta). Then (du = 8\cos(8\theta)d\theta) and (v=\frac{1}{7}e^{7\theta}). By the integration - by - parts formula (\int u;dv=uv-\int v;du), we have: [ \begin{align*} I&=\frac{1}{7}e^{7\theta}\sin(8\theta)-\frac{8}{7}\int e^{7\theta}\cos(8\theta)d\theta \end{align*} ]

For (\int e^{7\theta}\cos(8\theta)d\theta), let (U = \cos(8\theta)) and (dV=e^{7\theta}d\theta). Then (dU=- 8\sin(8\theta)d\theta) and (V = \frac{1}{7}e^{7\theta}). By the integration - by - parts formula, (\int U;dV=UV-\int V;dU): [ \begin{align*} \int e^{7\theta}\cos(8\theta)d\theta&=\frac{1}{7}e^{7\theta}\cos(8\theta)+\frac{8}{7}\int e^{7\theta}\sin(8\theta)d\theta\ &=\frac{1}{7}e^{7\theta}\cos(8\theta)+\frac{8}{7}I \end{align*} ]

Substitute (\int e^{7\theta}\cos(8\theta)d\theta) into the first integration - by - parts result: [ \begin{align*} I&=\frac{1}{7}e^{7\theta}\sin(8\theta)-\frac{8}{7}\left(\frac{1}{7}e^{7\theta}\cos(8\theta)+\frac{8}{7}I\right)\ I&=\frac{1}{7}e^{7\theta}\sin(8\theta)-\frac{8}{49}e^{7\theta}\cos(8\theta)-\frac{64}{49}I\ I+\frac{64}{49}I&=\frac{1}{7}e^{7\theta}\sin(8\theta)-\frac{8}{49}e^{7\theta}\cos(8\theta)\ \frac{49I + 64I}{49}&=\frac{7e^{7\theta}\sin(8\theta)-8e^{7\theta}\cos(8\theta)}{49}\ \frac{113I}{49}&=\frac{e^{7\theta}(7\sin(8\theta)-8\cos(8\theta))}{49}\ I&=\frac{e^{7\theta}(7\sin(8\theta)-8\cos(8\theta))}{113}+C \end{align*} ]

Explanation:

Step1: First integration - by - parts

Let (u = \sin(8\theta)), (dv=e^{7\theta}d\theta), find (du) and (v). [u=\sin(8\theta),du = 8\cos(8\theta)d\theta,v=\frac{1}{7}e^{7\theta}] [I=\frac{1}{7}e^{7\theta}\sin(8\theta)-\frac{8}{7}\int e^{7\theta}\cos(8\theta)d\theta]

Step2: Second integration - by - parts

Let (U=\cos(8\theta)), (dV = e^{7\theta}d\theta), find (dU) and (V). [U=\cos(8\theta),dU=-8\sin(8\theta)d\theta,V=\frac{1}{7}e^{7\theta}] [\int e^{7\theta}\cos(8\theta)d\theta=\frac{1}{7}e^{7\theta}\cos(8\theta)+\frac{8}{7}I]

Step3: Substitute and solve for (I)

Substitute (\int e^{7\theta}\cos(8\theta)d\theta) into first result. [I=\frac{1}{7}e^{7\theta}\sin(8\theta)-\frac{8}{7}\left(\frac{1}{7}e^{7\theta}\cos(8\theta)+\frac{8}{7}I\right)] [I+\frac{64}{49}I=\frac{1}{7}e^{7\theta}\sin(8\theta)-\frac{8}{49}e^{7\theta}\cos(8\theta)] [I=\frac{e^{7\theta}(7\sin(8\theta)-8\cos(8\theta))}{113}+C]