Question

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

Answer 1

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

Explanation:

`int secx*tanx*dx`

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

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

=`1/cosx +C`

=`secx+C`

Answer 2

The answer is `=secx +C`

Explanation:

Perform the substitution

`u=secx`

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

`=tanxsecxdx`

Therefore,

`inttanxsecxdx=intdu=u`

`=secx+C`

Answer 3

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`