Answer
$$\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {\frac{{\sin \theta }}{{{{\cos }^3}\theta }}} d\theta \cr
& {\text{set }}u = \cos \theta {\text{ then }}du = - \sin \theta d\theta ,{\text{ }} - du = \sin \theta d\theta \cr
& {\text{use the substitution in the integral }}\int {\frac{{\sin \theta }}{{{{\cos }^3}\theta }}} d\theta \cr
& \int {\frac{{\sin \theta }}{{{{\cos }^3}\theta }}} d\theta = \int {\frac{{ - du}}{{{u^3}}}} \cr
& {\text{property of the exponents}} \cr
& = - \int {{u^{ - 3}}} du \cr
& {\text{use }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \cr
& = - \frac{{{u^{ - 2}}}}{{ - 2}} + C \cr
& = \frac{1}{{2{u^2}}} + C \cr
& {\text{replace }}u{\text{ with }}\cos \theta \cr
& = \frac{1}{{2{{\cos }^2}\theta }} + C \cr
& = \frac{1}{2}{\sec ^2}\theta + C \cr
& {\text{then}} \cr
& \int_0^{\pi /4} {\frac{{\sin \theta }}{{{{\cos }^3}\theta }}} d\theta = \frac{1}{2}\left[ {{{\sec }^2}\theta } \right]_0^{\pi /4} \cr
& {\text{evaluate the limits }} \cr
& = \frac{1}{2}\left[ {{{\sec }^2}\left( {\frac{\pi }{4}} \right) - {{\sec }^2}\left( 0 \right)} \right] \cr
& {\text{simplifying}} \cr
& = \frac{1}{2}\left( {2 - 1} \right) \cr
& = \frac{1}{2} \cr} $$
