# Thread: Another Trig Integration by Parts Problem

1. ## Another Trig Integration by Parts Problem

$\displaystyle \int9x\cos8x = [9x][\dfrac{1}{8}\sin x] - \int [\dfrac{1}{8}\sin x][9]$

$\displaystyle u = 9x$

$\displaystyle dv = \cos 8x dx$

$\displaystyle v = \dfrac{1}{8}\sin x$

$\displaystyle du = 9 dx$

$\displaystyle \dfrac{9}{8}\sin^{2}x - \int\dfrac{9}{8} \sin x$

$\displaystyle \dfrac{9}{8}\sin^{2}x + \dfrac{9}{8} \cos x$ - Online homework says this is wrong. ??

2. ## Re: Another Trig Integration by Parts Problem

\displaystyle \begin{align*} v = \frac{1}{8}\sin{(8x)} \end{align*}, not \displaystyle \begin{align*} \frac{1}{8}\sin{(x)} \end{align*}.

3. ## Re: Another Trig Integration by Parts Problem

$\displaystyle \int9x\cos8x = [9x][\dfrac{1}{8}\sin 8x] - \int [\dfrac{1}{8}\sin 8x][9] dx$

$\displaystyle u = 9x$

$\displaystyle dv = \cos 8x dx$

$\displaystyle v = \dfrac{1}{8}\sin 8x$

$\displaystyle du = 9 dx$

$\displaystyle = [\dfrac{9}{8}\sin^{2} 8x] - \int [\dfrac{9}{8}\sin 8x]$

$\displaystyle = [\dfrac{9}{8}\sin^{2} 8x] + [\dfrac{9}{64}\cos 8x] + C$ ???

4. ## Re: Another Trig Integration by Parts Problem

NO! \displaystyle \begin{align*} 9x\cdot \frac{1}{8}\sin{(8x)} \end{align*} is NOT \displaystyle \begin{align*} \frac{9}{8}\sin^2{(8x)} \end{align*}, it's \displaystyle \begin{align*} \frac{9}{8}x\sin{(8x)} \end{align*}.

Also \displaystyle \begin{align*} \int{\frac{9}{8}\sin{(8x)}\,\mathrm{d}x} = -\frac{1}{8}\cdot \frac{9}{8}\cos{(8x)} = -\frac{9}{64}\cos{(8x)} \end{align*}, NOT \displaystyle \begin{align*} \frac{72}{8}\cos{(8x)} \end{align*}.

5. ## Re: Another Trig Integration by Parts Problem

The answer came out right on homework using ProveIt's suggestion.