Prove that the integral of tan x is integral of sec x + C where C is a constant?
By Tutorsvalley December 18th, 2019
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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