Using the substitution u=3x+32sin2xu = 3x + \frac{3}{2}\sin 2xu=3x+23sin2x, show that
∫0π2e3x+32sin2xcos2x dx=16(e3π2−1) \int_0^{\frac{\pi}{2}} e^{3x + \frac{3}{2}\sin 2x} \cos^2 x \, dx = \frac{1}{6}(e^{\frac{3\pi}{2}} - 1) ∫02πe3x+23sin2xcos2xdx=61(e23π−1)The design of a high-performance aerodynamic component involves a surface generated by rotating a region RRR through 2π2\pi2π radians about the xxx-axis. The region RRR is bounded by the curve with equation
y=18e32x+34sin2xcosx y = \sqrt{18} e^{\frac{3}{2}x + \frac{3}{4}\sin 2x} \cos x y=18e23x+43sin2xcosxand the coordinate axes in the first quadrant.
Use the result from part (a) to find the volume of the component, giving your answer in simplest form.