Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 531: 60

$$\int_{0}^{\pi / 3} \sec ^{3 / 2} x \tan x d x =\frac{4\sqrt{2}}{3}-\frac{2}{3}$$

$$ \int_{0}^{\pi / 3} \sec ^{3 / 2} x \tan x d x $$ Since \begin{align*} \int_{0}^{\pi / 3} \sec ^{3 / 2} x \tan x d x&=\int_{0}^{\pi / 3} \sec ^{1 / 2}x(\sec x \tan x )d x\\ &=\frac{2}{3}\sec^{3/2}x\bigg|\int_{0}^{\pi / 3} \\ &=\frac{4\sqrt{2}}{3}-\frac{2}{3} \end{align*}