A) \[\tan x+\frac{1}{3}{{\sec }^{3}}x+c\]
B) \[\frac{1}{6}\tan x+{{\sec }^{3}}x+c\]
C) \[\tan x-\frac{1}{3}{{\sec }^{3}}x+c\]
D) None of these
Correct Answer: A
Solution :
Let \[I=\int{{{\sec }^{4}}x}dx=\int{{{\sec }^{2}}x.{{\sec }^{2}}xdx}\] \[=\int{{{\sec }^{2}}x(1+{{\tan }^{2}}x)}dx\] \[=\int{{{\sec }^{2}}x}dx+\int{{{\sec }^{2}}x{{\tan }^{2}}xdx}\] Let \[sec\text{ }x=t\Rightarrow \text{ }ta{{n}^{2}}xdx=dt\] \[\therefore \] \[I=\tan x+\int{{{t}^{2}}}dt\] \[=\tan x+\frac{{{t}^{3}}}{3}+c\]
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