Question

Sol. [Key Concept: Use the king property: adding $I$ for $x$ and $- x$ removes the $1 + e^{sin\ x}$ factor, giving $2I = \int\ 32\cos^{4}x\ dx$; reduce $\cos^{4}$ with power-reduction.]

Apply (P-5)

$$I = \int_{- \pi/4}^{\pi/4}\mspace{2mu}\frac{32\cos^{4}\theta}{1 + e^{- sin\theta}}d\theta $$Add $2I = \int_{- \pi/4}^{\pi/4}\mspace{2mu} 32\cos^{4}\theta\text{ }d\theta = 2\int_{0}^{\pi/4}\mspace{2mu} 32\cos^{4}\theta\text{ }d\theta$

$${I = 32\int_{0}^{\pi/4}\mspace{2mu}\cos^{4}\theta d\theta }{= 8\int_{0}^{\pi/4}\mspace{2mu}\left( 2\cos^{2}\theta \right)^{2}\text{ }d\theta = 8\int_{0}^{\pi/4}\mspace{2mu}(1 + cos2\theta)^{2}\text{ }d\theta }{= 8\int_{0}^{\pi/4}\mspace{2mu} 1 + 2cos2\theta + \cos^{2}2\theta\text{ }d\theta }{= 8\int_{0}^{\pi/4}\mspace{2mu} 1 + 2cos2\theta + \frac{1 + cos4\theta}{2}\text{ }d\theta }{= 8\left\lbrack \frac{3\theta}{2} + \frac{2sin2\theta}{2} + \frac{sin4\theta}{8} \right\rbrack_{0}^{\pi/4} }{= 8\left\lbrack \left( \frac{3\pi}{8} + 1 + 0 \right) \right\rbrack = 3\pi + 8 }$$