Here find the proof of integral of tan x is integral of sec x + C where C is a constant with alternative solutions.  

Solution 1 - Integral of tan x is:

 

I=∫tan(x)dx

 

=∫sin(x)cos(x)dx

 

Assume, u=cos(x)

 

⟹du=d dx cos(x)dx

 

⟹du=−sin(x)dx

 

⟹−du=sin(x)dx

 

Substitution of x in I gives us,

 

I=∫−1u du

 

=−ln(|u|)

 

Substitution of u, gives us,

 

I=−ln (|cos(x)|)+C

       

Solution 2

   

(integral) tan x dx = - ln |COs x| + C

 

= ln | (COs x)-1 | + C

 

= ln |sec x| + C

 

Hence,

 

(integral) tan x dx = - ln |COs x| + C

   

 

= ln |sec x| + C

       

Solution 3

   

(integral) tan x dx = (integral) sin x/cos x dx

 

 

set, u = cos x.

 

then we find

 

du = - sin x dx

 

substitute du=-sin x, u=cos x

 

 

(integral) sin x/cos x dx = - (integral) (-1) sin x dx/cos x

 

= - (integral) du/u

 

 

Solve the integral

 

 

= - ln |u| + C

 

substitute back u=cos x

 

Hence,

 

= - ln |cos x| + C

       

Solution 4

 

Integral of tan x tan⁡ x = −log(cos x)−log⁡(cos⁡ x)

 

or log(sec x) log ⁡(sec⁡ x)

 

 

We can integrate tan x using the substitution rule.

 

Here I=∫tan x dx

 

I=∫tan⁡ x dx

 

⇒I=∫sin x cos x dx

 

∵tan x=sin x cos x

 

⇒I=∫sin⁡ xcos⁡ x dx

 

 

∵tan⁡ x=sin⁡ x cos⁡ x

 

Put cos x=t cos⁡ x=t

 

⇒−sin x dx=dt or sin x dx=−dt

 

⇒−sin⁡ x dx = dt or sin⁡ x dx=−dt

 

so, I=−∫dt t I=−∫dt t

 

⇒I=−logt+C⇒I=−logt+C

 

put the value of ‘t’

 

I=−log(cos x)+C

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