A) \[\sin x\cos x\]
B) \[{{\sec }^{2}}x\]
C) \[\tan x\]
D) \[\log \left| \tan x \right|\]
Correct Answer: C
Solution :
[c] Let \[I=\int{\sec \,\,x.\cos ec\,\,x\,\,dx}\] \[=\int{\frac{2}{2\sin \,\,x\,\,\cos \,\,x}dx}\] \[=\,\,\,\,2\int{\frac{2}{\sin 2x}dx-2\int{\frac{1}{\frac{2\tan x}{1+{{\tan }^{2}}x}}}}\] \[[\because \sin 2x=\frac{2\tan x}{1+{{\tan }^{2}}x}]\] \[=\int{\frac{{{\sec }^{2}}x}{\tan x}dx}\] Let \[\tan x=t\Rightarrow {{\sec }^{2}}dx=dt\] So, \[I=\int{\frac{dt}{t}=\log \left| t \right|+c=\log \left| \tan x \right|+c}\] But \[\int{\sec x\cos ec\,\,x\,\,dx=\log \left| g(x) \right|+c}\] \[\therefore g(x)=\tan \,\,x\]
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