Evaluate \[ \int\sec^3x\,dx,\quad \int\sec^4x\,dx,\quad \int\tan^2x\,dx,\quad \int\tan^3x\,dx, \] \[ \int\tan^4x\,dx,\quad \int\sec^3x\tan^3x\,dx, \quad \int(\tan^4x+\tan^2x)\,dx, \] and \[ \int\frac{2\sin^2x}{\cos^3x}\,dx. \]
Solution
Standard reduction and \(u\)-substitutions give \[ \boxed{\int\sec^3x\,dx =\frac12\bigl(\sec x\tan x+\ln|\sec x+\tan x|\bigr)+C,} \] \[ \boxed{\int\sec^4x\,dx=\tan x+\frac13\tan^3x+C,} \] \[ \boxed{\int\tan^2x\,dx=\tan x-x+C,} \] \[ \boxed{\int\tan^3x\,dx=\frac12\tan^2x+\ln|\cos x|+C,} \] \[ \boxed{\int\tan^4x\,dx=\frac13\tan^3x-\tan x+x+C.} \] For the mixed powers, use \(u=\sec x\) in the first and \(u=\tan x\) in the second: \[ \boxed{\int\sec^3x\tan^3x\,dx=\frac15\sec^5x-\frac13\sec^3x+C,} \] \[ \boxed{\int(\tan^4x+\tan^2x)dx=\frac13\tan^3x+C.} \] Finally, \[ \frac{2\sin^2x}{\cos^3x}=2(\sec^3x-\sec x), \] so \[ \boxed{\int\frac{2\sin^2x}{\cos^3x}dx =\sec x\tan x-\ln|\sec x+\tan x|+C.} \] Using \((\sec x-\tan x)(\sec x+\tan x)=1\), this is equivalently \[ \sec x\tan x+\ln|\sec x-\tan x|+C. \]