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

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Answer 1 · 2018-03-13T15:01:03

#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#

Answer 2 · 2018-03-13T15:03:05

The answer is #=secx +C#

Perform the substitution

#u=secx#

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

#=tanxsecxdx#

Therefore,

#inttanxsecxdx=intdu=u#

#=secx+C#

Answer 3 · 2018-07-21T11:33:14

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#