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What's the integral of #int secx tanx dx#?

Answer 1

Samantha Adkison

The answer is #=secx +C#

Execute the replacement.

#u=secx#

#du=(1/cosx)'=-1/cos^2x*-sinxdx=sinx/cos^2xdx#

#=tanxsecxdx#

Consequently,

#inttanxsecxdx=intdu=u#

#=secx+C#

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Answer 2

Isabella Ackerman

#int secx*tanx*dx=secx+C#

#int secx*tanx*dx#

=#int 1/cosx*sinx/cosx*dx#

=#int sinx/(cosx)^2*dx#

=#1/cosx +C#

=#secx+C#

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Answer 3

Ezra Adams

Method-1:

#\int sec x\tan x\ dx#

#=\int d(sec x)#

#=\sec x+C#

Method-2:

#\int sec x\tan x\ dx#

#=\int \frac{1}{\cos x}\frac{\sin x}{\cos x}\ dx#

#=\int \frac{\sin x}{\cos^2 x}\ dx#

#=\int \frac{-d(\cos x)}{\cos^2 x}#

#=-\int(\cos x)^{-2}d(\cos x)#

#=-\frac{(\cosx)^{-2+1}}{-2+1}+C#

#=(\cos x)^{-1}+C#

#=\sec x+C#

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Answer 4

William Abdalla

The integral of (\int \sec(x) \tan(x) , dx) is (\ln|\sec(x) + \tan(x)| + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.