How To Integrate sec^4(x)


 

\[\begin{array}{l} \smallint se{c^{\bf{4}}}x{\rm{d}}x\\\\ \smallint se{c^2}xse{c^2}x{\rm{d}}x\\ \smallint \left( {1 + {{\tan }^2}x} \right){\sec ^2}x{\rm{d}}x\\ By\;Substitution\\ Suppose \Rightarrow z = tanx\\ dz = se{c^2}xdx\\ \smallint \left( {1 + {z^2}} \right){\rm{d}}z\;\\ z + \frac{{{z^3}}}{3}\\ \tan x + \frac{{{{\tan }^3}x}}{3} + c \end{array}\]

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