Find $\int \tan ^{2}x\sec ^{4}x\,dx.$

Solution Here, $\sec x$ is raised to a positive even power. We factor $\sec ^{2}x$ from $\sec ^{4}x$ and use the identity $\sec ^{2}x=1+\tan ^{2}x.$ Then $$\begin{eqnarray*} \begin{array}{rcl@l} \int \tan ^{2}x\sec ^{4}x\,dx &=&\int \tan ^{2}x\sec ^{2}x\cdot \sec ^{2}x\,dx & {\color{#0066A7}{\hbox{Factor \(\sec^2 x\) from \(\sec^4 x\).}}} \\[5pt] &=&\int \tan ^{2}x(1+\tan ^{2}x)\sec ^{2}x\,dx & {\color{#0066A7}{\hbox{\(\sec ^{2}{x=1+}\tan ^{2}{x}\)}}} \nonumber \\[5pt] &=&\int u^{2}(1+u^{2})\,du & {\color{#0066A7}{\hbox{Substitute \(u=\tan x\);}}} \nonumber \\[0pt] &&&{\color{#0066A7}{\hbox{\(du=\sec ^{2} x\,dx\).}}}\nonumber \\[5pt] &=&\int (u^{2}+u^{4})\,du = \dfrac{u^{3}}{3}+\frac{u^{5}}{5}+C \underset{\underset{\color{#0066A7}{\hbox{\(u=\tan x\)}}}{\color{#0066A7}{\uparrow}}}{=}\dfrac{\tan ^{3}x}{3}+\dfrac{\tan ^{5}x}{5}+C\\[-8pt] \end{array} \end{eqnarray*}$$