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Q. $ \int{{{\sec }^{3}}\,\,x\,\,dx} $

J & K CETJ & K CET 2014Integrals

Solution:

Let $ I=\int{{{\sec }^{3}}\,x\,dx} $
$=\int{\underset{I}{\mathop{\sec }}\,x \,{{\underset{II}{\mathop{\sec }}\,}^{2}}}x\,\,dx $ $ \sec x\tan x-\int{\sec x\,ta{{n}^{2}}x\,dx} $ (using integration by parts)
$=\sec x\tan x-\int{\sec x\,({{\sec }^{2}}x-1)\,dx} $
$ \Rightarrow $ $ I=\sec x\tan x-\int{{{\sec }^{3}}\,x\,dx+\int{sec\,x\,\,dx}} $
$ \Rightarrow $ $ I=\sec x\tan x-I+\log |sec \,x+tanx|+2C $
$ \Rightarrow $ $ 2I=\sec x\tan x+\log |sec \,x+\tan x|+2C $
$ \Rightarrow $ $ I=\frac{1}{2}[\sec x\,\tan \,x+\log |\sec x+\tan x|]+C $