what is the volume of rotation?

between x=0 and x=6?

$$\\V=\pi \int_0^6\;(5(sin(4x))^2dx\\\\ V=25\pi \int_0^6\;sin^2(4x)dx\\\\ --------------------\\\\ cos(8x)=cos^2(4x)-sin^2(4x)\\\\ cos(8x)=1-sin^2(4x)-sin^2(4x)\\\\ cos(8x)=1-2sin^2(4x)\\\\ (cos(8x)-1)/-2=sin^2(4x)\\\\ sin^2(4x)=\frac{1-cos(8x)}{2}\\\\ --------------------\\\\$$

$$\\SO\\ V=25\pi \int_0^6\;sin^2(4x)dx\\\\ V=25\pi \int_0^6\;\frac{1-cos(8x)}{2}dx\\\\ V=12.5\pi \int_0^6\;1-cos(8x)dx\\\\ V=12.5\pi\left [\;x-\frac{sin(8x)}{8}\right]_0^6\\\\ V=12.5\pi\left [\left(\;6-\frac{sin(48)}{8}\right)-\left(\;0-\frac{sin(0)}{8}\right)\right]\\\\ V=12.5\pi \left(\;6-\frac{sin(48)}{8}\right)\\\\$$

sin(48)=-0.768254661     (remember that it is 48 radians)

$${\mathtt{12.5}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}\left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.768\: \!254\: \!661}}}{{\mathtt{8}}}}\right) = {\mathtt{239.390\: \!610\: \!267\: \!802\: \!800\: \!8}}$$

$$So I get $ 239\;units^3\qquad $(To the nearest whole number)$$ $