## Thomas' Calculus 13th Edition

$$\frac{3}{7}$$
\eqalign{ & \int_0^{\pi /4} {\frac{{{{\sec }^2}x}}{{{{\left( {1 + 7\tan x} \right)}^{2/3}}}}} dx \cr & {\text{Use the substitution method}}{\text{: }} \cr & u = 1 + 7\tan x,\,\,\,\,du = 7{\sec ^2}xdx,\,\,\,\,\,dx = \frac{{du}}{{7{{\sec }^2}x}} \cr & {\text{Then}}{\text{,}} \cr & \int {\frac{{{{\sec }^2}x}}{{{{\left( {1 + 7\tan x} \right)}^{2/3}}}}} dx = \int {\frac{{{{\sec }^2}x}}{{{u^{2/3}}}}} \left( {\frac{{du}}{{7{{\sec }^2}x}}} \right) \cr & = \frac{1}{7}\int {\frac{{du}}{{{u^{2/3}}}}} \cr & = \frac{1}{7}\int {{u^{ - 2/3}}du} \cr & {\text{Integrate by the power rule}} \cr & = \frac{1}{7}\left( {\frac{{{u^{1/3}}}}{{1/3}}} \right) + C \cr & = \frac{3}{7}\root 3 \of u + C \cr & {\text{Write in terms of }}x;{\text{ replace }}u = 1 + 7\tan x \cr & = \frac{3}{7}\root 3 \of {1 + 7\tan x} + C \cr & \cr & \int_0^{\pi /4} {\frac{{{{\sec }^2}x}}{{{{\left( {1 + 7\tan x} \right)}^{2/3}}}}} dx = \left( {\frac{3}{7}\root 3 \of {1 + 7\tan x} } \right)_0^{\pi /4} \cr & {\text{Evaluate}} \cr & = \frac{3}{7}\left( {\root 3 \of {1 + 7\tan \left( {\frac{\pi }{4}} \right)} - \root 3 \of {1 + 7\tan \left( 0 \right)} } \right) \cr & = \frac{3}{7}\left( {\root 3 \of {1 + 7} - \root 3 \of {1 + 0} } \right) \cr & = \frac{3}{7}\left( {2 - 1} \right) \cr & = \frac{3}{7} \cr}