A) \[\tan \,x-\sec x+C\]
B) \[\log (1+\sec x)+C\]
C) \[\sec x+\tan \,x+C\]
D) \[\log \sin x+\log \cos x+C\]
Correct Answer: A
Solution :
Let \[I=\int{\frac{\sec \,x}{\sec x+\tan x}}dx\] \[=\int{\frac{\sec x(\sec x-\tan x)}{{{\sec }^{2}}\,x-{{\tan }^{2}}x}}dx\] \[\int{({{\sec }^{2}}x-\sec x\tan x)dx}\] \[=\tan x-\sec x+C\]
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