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Q.
The integral value of the function $x \sec ^2 x$ is
Integrals
Solution:
Let $I=\int x \sec ^2 x d x$.
On taking $x$ as first function and $\sec ^2 x$ as second function and integrating by parts, we get
$I =x \int \sec ^2 x d x-\int\left[\frac{d}{d x}(x) \int \sec ^2 x d x\right] d x $
$=x \tan x-\int \tan x d x=x \tan x-\log |\sec x|+C $
$\Rightarrow I =x \tan x+\log |\cos x|+C $
$ {\left[\log |\sec x|=\log \left|\frac{1}{\cos x}\right|=\log 1-\log |\cos x|\right.} $
$ =-\log |\cos x|(\because \log =0)]$