Question

Let $f(a)=\underset{0}{\overset{a}{\int }}\ln (1+\tan a\tan x)dx$, then $f^{\prime }\left(\frac{\pi }{4}\right)$ equals

• A. $\frac{\pi }{4}+\frac{\ln 2}{2}$
• B. $\frac{\pi }{2}+\frac{\ln 2}{2}$
• C. $\frac{\pi }{4}+\ln 2$
• D. $\frac{\pi }{2}+\ln 2$

Answer 1

Answer: (A)

$I=\underset{0}{\overset{a}{\int }}\ln (1+\tan a\tan x)dx$ ...(1)
$I=\underset{0}{\overset{a}{\int }}(\ln (1+\tan a\tan (a-x)))dx\text{ (Applyking) }$
$=\underset{0}{\overset{a}{\int }}\ln \left(1+\frac{(\tan a(\tan a-\tan x))}{1+\tan a\tan x}\right)dx=\underset{0}{\overset{a}{\int }}\ln \left(\frac{1+{\tan }^{2}a}{1+\tan atanx}\right)dx$....(2)
$2I=\underset{0}{\overset{a}{\int }}\ln \left({\sec }^{2}a\right)dx=a\cdot \ln {\sec }^{2}a$
$I=a\ln (\sec a)$
$\frac{d}{da}(a\ln \sec a)=\ln (\sec a)+a\left(\frac{\sec a}{\sec a}\right)\tan a=a\tan a+\ln \sec a$