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

tutorial exercise evaluate the integral. ∫e^7θsin(8θ)dθ step 1 we will begin by letting u = sin(8θ) and dv = e^7θdθ. then du = 8cos(8θ)dθ and v = 1/7e^7θ. step 2 after integration by parts we have ∫e^7θsin(8θ)dθ = sin(8θ)1/7e^7θ - 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 = -8sin(8θ)dθ and v = 1/7e^7θ. step 4 after integration by parts we have ∫e^7θcos(8θ)dθ = 1/7e^7θcos(8θ) - 8/7×∫e^7θsin(8θ)dθ.

tutorial exercise evaluate the integral. ∫e^7θsin(8θ)dθ step 1 we will begin by letting u = sin(8θ) and dv = e^7θdθ. then du = 8cos(8θ)dθ and v = 1/7e^7θ. step 2 after integration by parts we have ∫e^7θsin(8θ)dθ = sin(8θ)1/7e^7θ - 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 = -8sin(8θ)dθ and v = 1/7e^7θ. step 4 after integration by parts we have ∫e^7θcos(8θ)dθ = 1/7e^7θcos(8θ) - 8/7×∫e^7θsin(8θ)dθ.

Answer

Explanation:

Step1: Recall integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$.

Step2: First application of integration by parts

Let $u = \sin(8\theta)$ and $\mathrm{d}v=e^{7\theta}\mathrm{d}\theta$. Then $\mathrm{d}u = 8\cos(8\theta)\mathrm{d}\theta$ and $v=\frac{1}{7}e^{7\theta}$. So, $\int e^{7\theta}\sin(8\theta)\mathrm{d}\theta=\sin(8\theta)\frac{1}{7}e^{7\theta}-\frac{8}{7}\int e^{7\theta}\cos(8\theta)\mathrm{d}\theta$.

Step3: Second application of integration by parts

For $\int e^{7\theta}\cos(8\theta)\mathrm{d}\theta$, let $U = \cos(8\theta)$ and $\mathrm{d}V=e^{7\theta}\mathrm{d}\theta$. Then $\mathrm{d}U=- 8\sin(8\theta)\mathrm{d}\theta$ and $V = \frac{1}{7}e^{7\theta}$. So, $\int e^{7\theta}\cos(8\theta)\mathrm{d}\theta=\frac{1}{7}e^{7\theta}\cos(8\theta)+\frac{8}{7}\int e^{7\theta}\sin(8\theta)\mathrm{d}\theta$.

Step4: Substitute and solve for $\int e^{7\theta}\sin(8\theta)\mathrm{d}\theta$

Substitute $\int e^{7\theta}\cos(8\theta)\mathrm{d}\theta$ into the result of Step 2: [ \begin{align*} \int e^{7\theta}\sin(8\theta)\mathrm{d}\theta&=\sin(8\theta)\frac{1}{7}e^{7\theta}-\frac{8}{7}\left(\frac{1}{7}e^{7\theta}\cos(8\theta)+\frac{8}{7}\int e^{7\theta}\sin(8\theta)\mathrm{d}\theta\right)\ \int e^{7\theta}\sin(8\theta)\mathrm{d}\theta&=\frac{1}{7}e^{7\theta}\sin(8\theta)-\frac{8}{49}e^{7\theta}\cos(8\theta)-\frac{64}{49}\int e^{7\theta}\sin(8\theta)\mathrm{d}\theta\ \int e^{7\theta}\sin(8\theta)\mathrm{d}\theta+\frac{64}{49}\int e^{7\theta}\sin(8\theta)\mathrm{d}\theta&=\frac{1}{7}e^{7\theta}\sin(8\theta)-\frac{8}{49}e^{7\theta}\cos(8\theta)\ \frac{49 + 64}{49}\int e^{7\theta}\sin(8\theta)\mathrm{d}\theta&=\frac{1}{7}e^{7\theta}\sin(8\theta)-\frac{8}{49}e^{7\theta}\cos(8\theta)\ \frac{113}{49}\int e^{7\theta}\sin(8\theta)\mathrm{d}\theta&=\frac{7e^{7\theta}\sin(8\theta)-8e^{7\theta}\cos(8\theta)}{49}\ \int e^{7\theta}\sin(8\theta)\mathrm{d}\theta&=\frac{e^{7\theta}(7\sin(8\theta)-8\cos(8\theta))}{113}+C \end{align*} ]

Answer:

$\frac{e^{7\theta}(7\sin(8\theta)-8\cos(8\theta))}{113}+C$