evaluate the integral using integration by parts.\n int 64e^{4\theta}sin 4\theta d\theta \n int…

Explanation:

Step1: Recall integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = \sin(4\theta)$ and $\mathrm{d}v = 64e^{4\theta}\mathrm{d}\theta$. Then $\mathrm{d}u=4\cos(4\theta)\mathrm{d}\theta$ and $v = 16e^{4\theta}$.

Step2: Apply integration - by - parts

$\int64e^{4\theta}\sin(4\theta)\mathrm{d}\theta=16e^{4\theta}\sin(4\theta)-\int16e^{4\theta}\times4\cos(4\theta)\mathrm{d}\theta=16e^{4\theta}\sin(4\theta)-64\int e^{4\theta}\cos(4\theta)\mathrm{d}\theta$.

Step3: Apply integration - by - parts again on $\int e^{4\theta}\cos(4\theta)\mathrm{d}\theta$

Let $u=\cos(4\theta)$ and $\mathrm{d}v = e^{4\theta}\mathrm{d}\theta$. Then $\mathrm{d}u=- 4\sin(4\theta)\mathrm{d}\theta$ and $v=\frac{1}{4}e^{4\theta}$. So $\int e^{4\theta}\cos(4\theta)\mathrm{d}\theta=\frac{1}{4}e^{4\theta}\cos(4\theta)+\int\frac{1}{4}e^{4\theta}\times4\sin(4\theta)\mathrm{d}\theta=\frac{1}{4}e^{4\theta}\cos(4\theta)+\int e^{4\theta}\sin(4\theta)\mathrm{d}\theta$.

Step4: Substitute the result of Step 3 into Step 2

$\int64e^{4\theta}\sin(4\theta)\mathrm{d}\theta=16e^{4\theta}\sin(4\theta)-64\left(\frac{1}{4}e^{4\theta}\cos(4\theta)+\int e^{4\theta}\sin(4\theta)\mathrm{d}\theta\right)$ $=16e^{4\theta}\sin(4\theta)-16e^{4\theta}\cos(4\theta)-64\int e^{4\theta}\sin(4\theta)\mathrm{d}\theta$. Let $I = \int64e^{4\theta}\sin(4\theta)\mathrm{d}\theta$. Then $I = 16e^{4\theta}\sin(4\theta)-16e^{4\theta}\cos(4\theta)-64\times\frac{I}{64}$.

Step5: Solve for $I$

$I = 16e^{4\theta}\sin(4\theta)-16e^{4\theta}\cos(4\theta)-I$. $2I=16e^{4\theta}(\sin(4\theta)-\cos(4\theta))$. $I = 8e^{4\theta}(\sin(4\theta)-\cos(4\theta))+C$.

Answer:

$8e^{4\theta}(\sin(4\theta)-\cos(4\theta))+C$