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Integrate the function\[x\;\sec^2x\]

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


answered Feb 11, 2013 by sreemathi.v