A) \[\frac{\pi }{4}+\frac{ln2}{2}\]
B) \[\frac{\pi }{2}+\frac{ln2}{2}\]
C) \[\frac{\pi }{4}ln2\]
D) \[\frac{\pi }{2}+ln2\]
Correct Answer: A
Solution :
\[I\,=\int\limits_{0}^{a}{\ln \,(1+\tan \,a.\tan x)\,dx}\] ? \[I=\int\limits_{0}^{a}{\,(\ln (1+\tan a\,\tan (a-x)))dx}\](Apply king) \[=\int\limits_{0}^{x}{\ln \,\left( 1+\frac{(\tan a(\tan a-\tan x))}{1+\tan a\,\tan x} \right)dx}\]\[=\int\limits_{0}^{a}{\ln \left( \frac{1+{{\tan }^{2}}}{1+\tan \,a\,\tan } \right)dx}\] ? \[2I=\int\limits_{0}^{a}{\,\ln \,({{\sec }^{2}}a)\,dx=a.\,\ln \,{{\sec }^{2}}a}\] \[I=a\,\ln \,(\sec a)\] \[\frac{d}{da}(a\,\ln \,\sec a)\,=\ln \,(\sec a)+a\left( \frac{\sec a}{\sec a} \right)\,\tan a\]\[=a\tan a+\ln \,\sec a\]
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