A) \[\frac{1}{4}{{\sec }^{4}}x+c\]
B) \[4{{\sec }^{4}}x+c\]
C) \[\frac{{{\sec }^{3}}x}{3}+c\]
D) \[3{{\sec }^{3}}x+c\]
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{{{\sec }^{4}}tnx\,dx}=\int_{{}}^{{}}{{{\sec }^{3}}x\sec x\tan x\,dx}\] Put \[t=\sec x\Rightarrow dt=\sec x\tan x\,dx,\] then it reduces to \[\int_{{}}^{{}}{{{t}^{3}}dt}=\frac{{{t}^{4}}}{4}+c=\frac{1}{4}{{\sec }^{4}}x+c.\]
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