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What is the integral of #(secxtanx)(1+secx)dx#?

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Mar 24, 2015

#1/2 (1+secx)^2+C#

#int (secxtanx)(1+secx)dx=int (1+secx)(secxtanx)dx #

Let #u=1+secx#. Then #du=secxtanxdx#

#int udu=1/2u^2+C# So,

#int (secxtanx)(1+secx)dx=1/2(1+secx)^2+C#

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