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Q.
Let $f(a)=\int\limits_0^a \ln (1+\tan a \tan x) d x$, then $f^{\prime}\left(\frac{\pi}{4}\right)$ equals
Integrals
Solution:
$I=\int\limits_0^a \ln (1+\tan a \tan x) d x$ ...(1)
$I =\int\limits_0^{ a }(\ln (1+\tan a \tan ( a - x ))) dx \text { (Applyking) } $
$=\int\limits_0^{ a } \ln \left(1+\frac{(\tan a (\tan a -\tan x ))}{1+\tan a \tan x }\right) dx =\int\limits_0^{ a } \ln \left(\frac{1+\tan ^2 a }{1+\tan atan x }\right) dx$....(2)
$2 I =\int\limits_0^{ a } \ln \left(\sec ^2 a \right) dx = a \cdot \ln \sec ^2 a $
$I = a \ln (\sec a ) $
$\frac{ d }{ da }( a \ln \sec a )=\ln (\sec a )+ a \left(\frac{\sec a }{\sec a }\right) \tan a = a \tan a +\ln \sec a$