- (a) Using the substitution \(u = 4 x + 2 \sin 2 x\), or otherwise, show that
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { e } ^ { 4 x + 2 \sin 2 x } \cos ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 8 } \left( \mathrm { e } ^ { 2 \pi } - 1 \right)$$
Figure 3
The curve shown in Figure 3, has equation
$$y = 6 \mathrm { e } ^ { 2 x + \sin 2 x } \cos x$$
The region \(R\), shown shaded in Figure 3, is bounded by the positive \(x\)-axis, the positive \(y\)-axis and the curve.
The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid.
(b) Use the answer to part (a) to find the volume of the solid formed, giving the answer in simplest form.