# Integrate the function$x\;\sec^2x$

Toolbox:
• (i)When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts$\int udv=uv-\int vdu$
• (ii)$\int\sec^2xdx=\tan x.$
• (iii)$\int \tan xdx=-log|\cos x|+c.$
Given I=$\int x\sec^2xdx.$

Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts

Let u=x,on differentiating with respect to x,

We get du=dx.

and let $dv=\sec^2xdx.$

On integrating we get,

v=tan x.

Now substituting for u,v,du and dv we get,

$\sec^2xdx=(x\tan x)-\int \tan x.dx.$

On integrating we get

$I=x\tan x-(-log|\cos x|+c.$

$\;\;\;=x\tan x+log |\cos x|+c.$