# Integral of secx – the circular function of an angle

Secx or secant of x can be written in various forms such as, sec, cosec, and or cot can be defined as a trigonometric function like, sin, cos and tan. Forms of secx include the following:

• sec x = 1cos x
• cosec x 1sin x
• cot x = 1tan x = cos x sin x

Secant can as well be obtained from complex number: secx = 2/[e^(ix) + e^(-ix)], where i is the imaginary unit that equals to sqrt (-1). In contrast to cos(x), finding the antiderivative of secant of x needs few calculus tricks. Moreover, the antiderivative of secx has many identical forms that can puzzle students who want to see how good they are at computing. Below are few and most commonly used functional forms:

• ∫ sec(x) dx = ln[sec(x) + tan(x)] = ln [1 + sin(x)] – ln[cos(x)] = -ln[secant of x – tan(x)] = ln[cos(x)] – ln [1 – sin(x)] = 0.5*ln [1 + sin(x)] – 0.5*ln [1 – sin(x)] = ln[cos(x2) + sin(x2)] – ln[cos(x2) – sin(x2)]

You can also check this is similar via trigonometric identities and properties of logarithms. Described below are the 2 methods used to obtain the antiderivative of secant of x:

• Method 1: this method is by far the standard way of checking the secx as you will find in most textbooks of calculus. It uses an intuitive multiplication trick that changes the function into something that will easily lead itself to u substitution. To begin with, multiply the function sec(x) by the unit fraction in the following way:
• 1 = [ (secant of x + tan (x)sec(x) + tan(x)) ], this gives a new integral
• ∫ sec(x) dx = ∫ (secant of x + tan (x)sec(x) + tan(x))] dx = ∫ [sec(x)^2 + (secant of x + tan (x)sec(x) + tan(x))] dx = ∫ [(sec(x)tan(x) + sec(x)^2sec(x) + tan(x))]dx
• Now let u be equal to: sec (x) + tan (x), the function in the denominator of this integral. Taking the derivative of both sides of this substituted equation produces the differentials: du = [ sec (x) tan (x) + sec(x)^2 ] dx. Since secx derivative is sec(x)tan(x) while the derivative of tan(x) is sec(x)^2). Our integral changed into
• ∫ [sec(x)tan(x) + sec(x)^2sec(x) + tan(x)] dx= ∫ duu = ∫ (1u)du
• The integral of 1u is ln(u) + c. This makes the reverse substitution to give us

ln(u) + c= ln[sec(x) + tan(x)] + c

• Method 2: this method is not as intuitive as previously explained method, yet it can still change the integral of secx into something with an identifiable antiderivative. The first step is to rewrite the integral as: ∫ secant of x dx = ∫ 1/cos(x) dx. Next, make the substitution: u = sin(x) which in turn produces a new equation; cos(x) = sqrt [1 – sin(x)^2] = sqrt (1 – u^2)
• Another new equation is: x = arcsin(u) and dx = 1sqrt(1 – u^2) du
• Now our integral will transform to the following:

∫ 1/cos(x) dx = ∫ 1sqrt(1 – u^2) *[1sqrt(1 – u^2) ] du = ∫ 1(1 – u^2)du = ∫ [1(1 – u) [1(1 – u) ]du

The technique of partial fractions allows us to express the product of two fractions as a sum. The use of this strategy to the integrand gives us:

• [ 1/(1 – u) ]*[ 1/(1 + u) ] = 0.5/(1 – u) + 0.5/(1 + u)
• Now the u integral is; 0.5*∫ 1/(1-u) du + 0.5*∫ 1(1+ u) du= -0.5*Ln(1 – u) + 0.5*Ln(1 + u) + c
• Making the reverse substitution, we will get: -0.5*Ln[1 – sin(x)] + 0.5*Ln[1 + sin(x)] + c:
• This function is equivalent to that which is obtained via Method 1 as it will be shown in the next section

## Equating Various Forms of integral of secx

To see that the functions ln[sec(x) + tan (x)] and -0.5*ln [1 – sin(x)] + 0.5*ln [1 + sin(x)] are equal to each other, we re-write the 2nd equation in its identical form 0.5*ln [ (1 + sin(x)(1- sin⁡(x) ]

Now we the product of the below equation: 1 =(1 – sin(x)(1 – sin(x) ) which gives: 0.5*ln [(1 – sin(x)^2)(1 – sin(x) ^2/] = 0.5*ln [cos(x)^2 (1 – sin(x) ^2] = 0.5*ln {[ [ cos(x)(1 – sin(x)]^2} = 2 * 0.5 * ln [ cos(x)(1 – sin(x)] = ln [ cos(x)(1 – sin(x)]. With the help of logarithmic property, we have: ln [ cos(x)(1 – sin(x) = -ln [ (1 – sin(x)cos(x))] = -ln [1cos(x) – sin(x)cos(x)] = -ln [ secant of x – tan(x)]

Next we use the trigonometric identity:

• 1 + tan(x)^2 = sec(x)^2
• >sec(x)^2 – tan(x)^2 = 1
• [sec(x) + tan(x)] *[sec(x) – tan(x)] = 1
• sec(x) + tan(x) = 1sec(x) – tan(x)]

This shows that the expressions secant of x + tan(x) and sec(x) – tan(x) are reciprocals of each another. At last, we can use this to obtain:

• -ln [ sec(x) – tan(x)] = -ln[ 1sec(x) + tan(x)) ] = ln[ sec(x) + tan(x)]

### Derivative of secant and secant to curve

Derivative of secant:

dx (sec x) = sec x. tan x

ddx (sec x) = ddx 1cos x

= (cos x )(1)-(cos x )(cos x)∧2)

=0+sin x cos∧2 x

=1cos x. sin x cos x

=sec x. tan x

To find the antiderivative of any given function, we will have to determine the indefinite integral of the function i.e.:

• Int f(x)dx = Int sec x dx
• Next is to substitute the expression of function: f(x)= secx by f(x) = 1/cos x
• Write out the integral: Int dx/cos x = Int cos xdx/(cos x)^2
• Using the trigonometry fundamental formula: (cos x)^2 = 1 – (sin x)^2
• Int cos xdx/(cos x)^2 = Int cos xdx/[1 – (sin x)^2]
• You notice that: sin x = t
• cos x*dx = dt
• Rewriting the integral of secx in t gives: Int cos xdx/[1 – (sin x)^2] = Int dt/(1 – t^2)
• Analyze the integral: 1/(1 – t^2) = 1/(1-t)(1+t)
• Then separate the integral of secx into a partial fraction like below
• <li

• 1 = A(1+t) + B(1-t)
• 1 = A + At + B – Bt
• Factorizing the above equation; 1 = t(A-B) + A+B
• The coefficient of t from the left side has to be equal to the coefficient of t on the right side of the equation:

A-B = 0

A = B

A+B = 1

2B = 1

B = A = 1/2

1(1-t) (1+t)= 12(1-t) + 12(1+t)

In t dt(1-t) (1+t) = In t dt2(1+t) + In t dt2(1+t)

Intdt/(1-t) (1+t) = (12) ln |1-t| + (12) ln|1+t| + C

A secant to a curve A is a straight line that cuts a curve. Hence, look at the secant line that cuts the curve at points P and Q. Then the slope of that secant is given below:

• ΔyΔx= y2-y1/x2-x2

Using a triangle, it is possible to explain how secant of x can be derived. The secant of the angles in a right angle triangle for instance is the length of its hypotenuse divided by its adjacent length on both sides. Of all the available trigonometric functions known in calculus such as: secant, cosecant, etc. sin, cos, and tan are the most and widely used trigonometric functions. Secant, cosecant and all others are seldom used

For every trig function, there is the inverse function which works in the reverse form. These inverse or reverse functions have similar name but with an arc written in front of them. It is therefore known that the inverse function of secant of x is arcsec x. few example to give a clear understanding of what we have been talking about:

• Sec 60 = 2.000. This also mean secant of 600 gives 2.000
• Also, Arcsec 2.0 = 60. This can be expressed as the reverse of the first example. So, this means that angle with 2.0 as its secant is 600

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