User:Xenocidius/Specialist/Q4

i.
$$ \begin{align} \int_0^\frac{3\pi}{2} \! e^{\sin x} \,dx & \approx \frac{1}{2} (e^{\sin 0} + e^{\sin \frac{\pi}{2}}) \frac{\pi}{2} + \frac{1}{2} (e^{\sin \frac{\pi}{2}} + e^{\sin \pi}) \frac{\pi}{2} + \frac{1}{2} (e^{\sin \pi} + e^{\sin \frac{3\pi}{2}}) \frac{\pi}{2} \\ & = \frac{\pi}{4} ((e^0 + e^1) + (e^1 + e^0) + (e^0 + e^{-1})) \\ & = \frac{\pi}{4} (2e + \frac{1}{e} + 3) \\ & \approx 6.91 \end{align} $$

ii.
$$ \begin{align} \int_0^\frac{3\pi}{2} \! e^{\sin x} \,dx & \approx \frac{\pi}{2} e^{\sin \frac{\pi}{4}} + \frac{\pi}{2} e^{\sin \frac{3\pi}{4}} + \frac{\pi}{2} e^{\sin \frac{5\pi}{4}} \\ & = \frac{\pi}{2} (e^{\frac{\sqrt 2}{2}} + e^{\frac{\sqrt 2}{2}} + e^{-\frac{\sqrt 2}{2}}) \\ & \approx 7.15 \end{align} $$

iii.
$$ \begin{align} y & = e^{\sin x} \\ \therefore \sin x & = \ln y \\ \therefore x & = \sin^{-1} (\ln y) \end{align} $$

$$ \begin{align} \int_1^e \! \sin^{-1} (\ln y) \,dy & \approx \frac{\pi}{2} e - \frac{1}{2} (e^{\sin 0} + e^{\sin \frac{\pi}{2}}) \frac{\pi}{2} \\ & \approx 1.35 \end{align} $$