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

Answer 1

Samantha Adkison

2secx+C

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

#I=2secx+C#

Justification

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

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

The definition of a function's differential is:

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

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

Simple integration theory formula:

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

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

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Answer 2

Aurora Alvarado

To find the integral of ( \int 2\sec(x)\tan(x) , dx ), you can use substitution. Let ( u = \sec(x) + \tan(x) ). Then, ( du = (\sec(x)\tan(x) + \sec^2(x)) , dx ).

Notice that ( du = (\sec(x)\tan(x) + \sec^2(x)) , dx ) matches the integrand ( 2\sec(x)\tan(x) , dx ).

Thus, the integral becomes:

[ \int 2\sec(x)\tan(x) , dx = \int 2 , du = 2u + C ]

Where ( C ) is the constant of integration.

Finally, substitute back ( u = \sec(x) + \tan(x) ) to get the final answer:

[ \int 2\sec(x)\tan(x) , dx = 2(\sec(x) + \tan(x)) + C ]

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Answer from HIX Tutor

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