\(\text{To evaluate:}\)

\(\int\tan^3 4x\ dx\)

\(\tan^2 x=\sec^2 x-1\)

\(\frac{d}{dx}(\sec x)=\sec x\tan x\)

\(\text{Consider}\int\tan^3 4xdx\)

\(=\int\tan^2 4x\tan 4x dx\)

\(=\int(\sec^2 4x-1)\tan 4xdx\)

\(\text{Now, let }\sec 4x=u\)

\(\Rightarrow4\sec 4x\tan4xdx=du\)

\(\Rightarrow\tan4xdx=\frac{du}{4\sec4x}\)

\(\text{Substituting all the values, we get,}\)

\(\int (u^2-1)\frac{du}{\sec4x}\)

\(=\int(u^2-1)\frac{du}{4u}\)

\(=\frac{1}{4}\int\frac{u^2-1}{u}du\)

\(=\frac{1}{4}\int u-\frac{1}{u}du\)

\(=\frac{1}{4}[\frac{u^2}{2}-\ln(u)]+C\)

\(=\frac{1}{4}[\frac{\sec^2 4x}{2}-\ln(\sec4x)]+C\)

\(\text{Hence, }\int\tan^3 4x\ dx=\frac{1}{4}[\frac{\sec^2 4x}{2}-\ln(\sec4x)]+C\)

\(\int\tan^3 4x\ dx\)

\(\tan^2 x=\sec^2 x-1\)

\(\frac{d}{dx}(\sec x)=\sec x\tan x\)

\(\text{Consider}\int\tan^3 4xdx\)

\(=\int\tan^2 4x\tan 4x dx\)

\(=\int(\sec^2 4x-1)\tan 4xdx\)

\(\text{Now, let }\sec 4x=u\)

\(\Rightarrow4\sec 4x\tan4xdx=du\)

\(\Rightarrow\tan4xdx=\frac{du}{4\sec4x}\)

\(\text{Substituting all the values, we get,}\)

\(\int (u^2-1)\frac{du}{\sec4x}\)

\(=\int(u^2-1)\frac{du}{4u}\)

\(=\frac{1}{4}\int\frac{u^2-1}{u}du\)

\(=\frac{1}{4}\int u-\frac{1}{u}du\)

\(=\frac{1}{4}[\frac{u^2}{2}-\ln(u)]+C\)

\(=\frac{1}{4}[\frac{\sec^2 4x}{2}-\ln(\sec4x)]+C\)

\(\text{Hence, }\int\tan^3 4x\ dx=\frac{1}{4}[\frac{\sec^2 4x}{2}-\ln(\sec4x)]+C\)