Answer
$- \frac{1}{4} +\frac{1}{2} \ln 2$
Work Step by Step
$$\eqalign{
& \text{Let }I=\int_0^{\pi /4} {\frac{{{{\sin }^3}x}}{{\cos x}}} dx \cr
& {\text{Split }}{\sin ^3}x{\text{ as }}{\sin ^2}x\sin x \cr
& I = \int_0^{\pi /4} {\frac{{{{\sin }^2}x\sin x}}{{\cos x}}} dx \cr
& {\text{Rewrite }}{\sin ^2}x{\text{ using }}{\sin ^2}x + {\cos ^2}x = 1 \cr
& I= \int_0^{\pi /4} {\frac{{\left( {1 - {{\cos }^2}x} \right)}}{{\cos x}}\left( {\sin x} \right)} dx \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = \cos x,{\text{ }}du = - \sin xdx \cr
& {\text{The new limits of integration are:}} \cr
& x = \frac{\pi }{4} \to u = \cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2} \cr
& x = \frac{\pi }{4} \to u = \cos \left( 0 \right) = 1 \cr
& {\text{Substituting}} \cr
& I=\int_0^{\pi /4} {\frac{{\left( {1 - {{\cos }^2}x} \right)}}{{\cos x}}\left( {\sin x} \right)} dx = \int_1^{\sqrt 2 /2} {\frac{{\left( {1 - {u^2}} \right)}}{u}\left( { - 1} \right)du} \cr
& = \int_1^{\sqrt 2 /2} {\frac{{{u^2} - 1}}{u}du} \cr
& = \int_1^{\sqrt 2 /2} {\left( {u - \frac{1}{u}} \right)du} \cr
& {\text{Integrating}} \cr
& I = \left[ {\frac{{{u^2}}}{2} - \ln \left| u \right|} \right]_1^{\sqrt 2 /2} \cr
& = \left[ {\frac{{{{\left( {\sqrt 2 /2} \right)}^2}}}{2} - \ln \left( {\frac{{\sqrt 2 }}{2}} \right)} \right] - \left[ {\frac{{{{\left( 1 \right)}^2}}}{2} - \ln \left( 1 \right)} \right] \cr
& = \frac{1}{4} - \ln \left( {\frac{{\sqrt 2 }}{2}} \right) - \frac{1}{2} \cr
& = - \frac{1}{4} - \ln \left( {\frac{{\sqrt 2 }}{2}} \right) \cr
& = - \frac{1}{4} - \ln \left( 2^{-1/2} \right) \cr
& = - \frac{1}{4} +\frac{1}{2} \ln 2 \cr} $$
