Question

Evaluate $\int xsecx.tanxdx=$

• A. $x\sec x+\log |\tan (\pi /2+x/2)|+c$
• B. $x\sec x-\log$$|\tan (\pi /4+x/2)|+c$
• C. $x\sec x-\log |\tan (\pi /4+x)|+c$
• D. $x\sec x+\log |\tan x/2|+c$

Answer 1

Answer: (B)

$x\sec x-\log$$|\tan (\pi /4+x/2)|+c$

$\int x\sec x\tan xdx=\int x\left(\sec x\right(\tan x\left)\right)dx$

$=x\int \left(\sec x\tan x\right)dx$ $-\int 1\int \left(\sec x\tan x\right)dx$ ....... [Using integration parts]

$=x\sec x-\int \sec xdx$

$=x\sec x-\log |\sec x+\tan x|+c$

$=x\sec x-\log \left|\tan \right(\pi /4+x/2\left)\right|+c$ ..... $[\because \sec x+\tan x=\tan (\pi /4+x/2\left)\right]$

$\therefore \int x\sec x\tan xdx=x\sec x-\log \left|\tan \right(\pi /4+x/2\left)\right|+c$