$\displaystyle \int{{{\sec }^{3}}x\,dx}=\sec x\tan x-\int{\left( {{\sec }^{3}}x-\sec x \right)\,dx},$ so $\displaystyle 2\int{{{\sec }^{3}}x\,dx}=\sec x\tan x+\int{\sec x\,dx},$ and few post ago I told ya how to integrate the last integral.
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Originally Posted by Krizalid $\displaystyle \int{{{\sec }^{3}}x\,dx}=\sec x\tan x-\int{\left( {{\sec }^{3}}x-\sec x \right)\,dx},$ so $\displaystyle 2\int{{{\sec }^{3}}x\,dx}=\sec x\tan x+\int{\sec x\,dx},$ and few post ago I told ya how to integrate the last integral. I cannot thank you enough for your help. Thanks for the patience and tremendous help.
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