evaluate the integral using integration by parts.\n int 9e^{- 9\theta}cos9\theta d\theta \n int 9e^{…

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 $u = \cos(9\theta)$ and $\mathrm{d}v=9e^{-9\theta}\mathrm{d}\theta$. Then $\mathrm{d}u=- 9\sin(9\theta)\mathrm{d}\theta$ and $v=-e^{-9\theta}$.

Step2: Apply integration - by - parts

$\int9e^{-9\theta}\cos(9\theta)\mathrm{d}\theta=-e^{-9\theta}\cos(9\theta)-\int(-e^{-9\theta})(-9\sin(9\theta))\mathrm{d}\theta=-e^{-9\theta}\cos(9\theta)-\int9e^{-9\theta}\sin(9\theta)\mathrm{d}\theta$.

Step3: Apply integration - by - parts again on $\int9e^{-9\theta}\sin(9\theta)\mathrm{d}\theta$

Let $u = \sin(9\theta)$ and $\mathrm{d}v=9e^{-9\theta}\mathrm{d}\theta$. Then $\mathrm{d}u = 9\cos(9\theta)\mathrm{d}\theta$ and $v=-e^{-9\theta}$. So $\int9e^{-9\theta}\sin(9\theta)\mathrm{d}\theta=-e^{-9\theta}\sin(9\theta)-\int(-e^{-9\theta})(9\cos(9\theta))\mathrm{d}\theta=-e^{-9\theta}\sin(9\theta)+\int9e^{-9\theta}\cos(9\theta)\mathrm{d}\theta$.

Step4: Substitute the result of Step 3 into Step 2

$\int9e^{-9\theta}\cos(9\theta)\mathrm{d}\theta=-e^{-9\theta}\cos(9\theta)-\left(-e^{-9\theta}\sin(9\theta)+\int9e^{-9\theta}\cos(9\theta)\mathrm{d}\theta\right)$. $\int9e^{-9\theta}\cos(9\theta)\mathrm{d}\theta=-e^{-9\theta}\cos(9\theta)+e^{-9\theta}\sin(9\theta)-\int9e^{-9\theta}\cos(9\theta)\mathrm{d}\theta$. Add $\int9e^{-9\theta}\cos(9\theta)\mathrm{d}\theta$ to both sides: $2\int9e^{-9\theta}\cos(9\theta)\mathrm{d}\theta=-e^{-9\theta}\cos(9\theta)+e^{-9\theta}\sin(9\theta)$.

Step5: Solve for $\int9e^{-9\theta}\cos(9\theta)\mathrm{d}\theta$

$\int9e^{-9\theta}\cos(9\theta)\mathrm{d}\theta=\frac{1}{2}e^{-9\theta}(\sin(9\theta)-\cos(9\theta))+C$

Answer:

$\frac{1}{2}e^{-9\theta}(\sin(9\theta)-\cos(9\theta))+C$