Finding the Integral \(\int \tan ^{3}x\,\sec ^{4}x\,dx\)

Find \(\int \tan ^{3} x\sec ^{4} x\,dx\).

Solution Here, \(\tan x\) is raised to the odd power \(3.\) We factor \(\tan x\) from \(\tan ^{3}x\) and use the identity \(\tan ^{2} x=\sec^{2}x-1.\) \[ \begin{eqnarray*} \begin{array}{@rcl@l} \int \tan ^{3}x\sec ^{4}x\,dx &=& \int \tan ^{2}x\,\tan x\,\sec ^{4}x\,dx & {\color{#0066A7}{\hbox{Factor tan \(x\) from \(\tan^{3} x\).}}}\\[6pt] &=& \int (\sec ^{2}x-1) \tan x\sec ^{4}x\,dx & {\color{#0066A7}{\hbox{\(\tan^{2} x = \sec^{2} x - 1\)}}}\\[6pt] &=& \int ( \sec ^{2}x-1) \sec^{3}x\sec x\tan x\,dx & {\color{#0066A7}{\hbox{Factor sec \(x\) from \(\sec^{4} x\).}}}\\[6pt] &=& \int (u^{2}-1) u^{3}du & {\color{#0066A7}{\hbox{Substitute \(u=\sec x\);}}}\\[-4pt] &&& {\color{#0066A7}{\hbox{\(du = \sec x\) \(\tan x\,dx\).}}}\\[6pt] &=& \int (u^{5}-u^{3}) \,du =\dfrac{u^{6}}{6}-\dfrac{u^{4}}{4}+C \underset{\underset{\color{#0066A7}{\hbox{\(u=\sec x\)}}}{\color{#0066A7}{\uparrow}}}{=} \dfrac{\sec ^{6}x}{6}-\dfrac{\sec ^{4}x}{4}+C &\\[-13pt] \end{array} \end{eqnarray*} \]