How do you find the integral of #int 2secxtanx dx#?

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Answer 1 · 2015-09-26T20:01:57

2secx+C

#I=int 2secxtanxdx=2int (secx)'dx=2intd(secx)#

#I=2secx+C#

Explanation:

#(secx)' = secxtanx => 2secxtanx = 2(secx)'#

#I = int 2secxtanxdx = 2 int (secx)'dx#

Differential of a function is defined as:

#d(f(x)) = f'(x)dx => d(secx) = (secx)'dx#

#I = 2 int (secx)'dx = 2 int d(secx)#

Elementary formula from integration theory:

#int d(f(x)) = f(x) + C#

#I = 2 int d(secx) = 2secx + C#