Question

Evaluate: $\int \sec x\log (\sec x+\tan x)dx$

• A. ${\left[\log (\sec x+\tan x)\right]}^2$
• B. $\frac{1}{2}\left[\log (\sec x+\tan x)\right]$
• C. $\frac{1}{2}{\left[\log (\sec x-\tan x)\right]}^2$
• D. $\frac{1}{2}{\left[\log (\sec x+\tan x)\right]}^2$

Answer 1

Answer: (D)

$\frac{1}{2}{\left[\log (\sec x+\tan x)\right]}^2$

Let $I=\int \sec x\log (\sec x+\tan x)dx$

Put $\log (\sec x+\tan x)=t$
$\Rightarrow =\frac{secxtanx+se{c}^{2}x}{secx+tanx}dx=\sec xdx=dt$

Therefore
$I=\int tdt=\frac{1}{2}{t}^2=\frac{1}{2}{\left[\log (\sec x+\tan x)\right]}^2$

Hence, option 'D' is correct.