Antiderivatives: An Introduction to Indefinite Integrals

The rate of change of a function at a particular value x is known as the Derivative of that function. Anti-derivatives as the name suggests is the opposite of derivatives. An Anti derivative is commonly referred to as an Indefinite Integral. We can define an indefinite integral of F as follows: Any given function G is an indefinite integral of F or an indefinite integral of a function g if the derivative of that function G’ equals g. The notation of an indefinite integral of F or the indefinite integral of a function is, G(x) = Integral g(x) dx. From this notation, we can conclude that G(x) equals integral[g(x)]dx  if and only if G’(x) = g(x)

From the above we can understand that an Antiderivative is basically an Integral of a given function, which is set into a formula which helps us to take the indefinite integral of F. So, when we say the Integral it means indefinite integral of F. Here, we have to remember to add a constant “c” as every integral has an unknown constant which is added to the equation. Let us consider an example for a better understanding, given function y = x^2 + 3x + 5. The derivative y’ would be 2x +3. Let us now find the antiderivative of y’, that gives us integral[2x +3] dx = 2. X^(1+1)/(1+1) + 3. x^(0+1)/(0+1) + c = 2x^/2 + 3x/1 + c = x^2+3x +c, this function is same as the original function except that the constant 5 is missing, this is the reason why we need to add the constant “c” to the Integral of a function.  From this we can conclude that Anti Derivative is the reverse derivative or the indefinite integral.

When we solve an Integral, we eliminate the integral sign and dx to arrive to a function G(x), this function is the antiderivative.  For instance, indefinite integral of F of the function x^3 is given by x raised to the power (3+1) whole divided by (3+1), that is, integral(x^3)dx = x^(3+1)/(3+1)= x^4/4. In general we can write the indefinite integral of F of x^n  as x^(n+1)/(n+1)
Antiderivative of Sec X
indefinite integral of F is the Integral of Sec X
Multiplying sec(x) with 1 which is [sec(x)+tan(x)]/[tan(x)+sec(x)]
Integral[sec(x)] dx = Integral[Sec x][sec(x) +tan(x)]/[tan(x)+sec(x)]      

Let u= sec(x) + tan(x)
Differentiating on both sides,
du = [sec(x)tan(x) + sec^2(x)]dx
Substituting u= [sec(x) + tan(x)]du = [sec(x)tan(x) + sec^2(x)]dx ,
Integral[sec(x)][sec(x)+tan(x)]/[tan(x)+sec(x)] dx
 = Integral[sec^2(x)+sec(x)tan(x)]dx/[sec(x)+tan(x)]
= Integral[du/u]
Solving the integral we get,
 = ln|u|+ c
Again substituting u= sec(x) + tan(x)
= ln|sec(x)+tan(x)| + c
So, the Antiderivative of Sec X is the Integral[sec x]dx = ln|sec(x)+tan(x)| + c

A short note on derivative of Cot function

Cot is the short for the trigonometric function cotangent. It is the complementary function of the tan also called the tangent function. Cot is defined as the ratio of the adjacent side to the opposite side of the acute angle in a right triangle. It is also the reciprocal of the tan function.

Therefore symbolically it can be written like this:
Cot(x) = adjacent side/opposite = 1/tan(x)
We can graph the cot function using a table of values as follows:
X -pi -3pi/4 -pi/2 -pi/4 0 Pi/4 Pi/2 3pi/4 pi
Cot(x) Inf 1 0 -1 Inf 1 0 -1 inf

The graph would look as follows:

 From the graph we note that the cot function is not defined at the points –pi, 0 and pi. That is because at these points, the tan function has the values of 0,0 and 0 respectively. Since the cot function is the reciprocal of tan function, the cot is not defined at these points.

Derivative Cot X  function:

The above graph is of the function y = Cot(x). Now the derivative of Cot X would be the slope of tangent to the above curve at any point x. Therefore for example if we were to find the derivative of Cotx at the point x = pi/4, then the graph of the tangent line to the curve at that point would like this:

The blue line in the above picture is the tangent to the cot function at the point x= pi/4. The slope of this tangent is the derivative of cot x. In other words we can also say that the derivative of the cot function is the rate of change of cot function at a given point. This is also called the instantaneous rate of change of cot function at a point x=a. (In this case it is x = pi/4). Another way of stating the same thing is like this: the gradient of the curve of the function y = cot x at a point x=a, is called the derivative of the function at that point.

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The derivative of the function y = cot x at the point x = 0 can be found by making a tangent to the curve at the point x=0. Again from the above graph we see that, at x=0, the tangent would be vertical. So slope of tangent = not defined. Therefore derivative at this point is not defined.

Homogeneous and non homogeneous differential equation

Non Homogeneous Differential Equation: A differential equation is a linear differential equation if it is expressible in the form P0 d^n y/dx^n + P1 d^n-1 y/dx^n-1 + P2 d^n-2 y/dx^n-2 + … + Pn-1 dy/dx + Pn y = Q, Where P0, P1, P2, …., Pn-1, Pn and Q are either constants or functions of independent variable x.Thus, if a differential equation when expressed in the form of a polynomial involves the derivatives and dependent variable in the first power and there are no product of these, and also the coefficient  of the various terms are either constants or functions of the independent variable, then it is said to be linear differential equation. Otherwise, it is a non-differential equation.

Homogeneous differential equation: A function f(x, y) is called a homogeneous function of degree n if f(lambda x, lambda y) = lambda^n f(x, y).

For example, f(x, y) = x^2 – y^2 + 3xy is a homogeneous function degree 2, because f(lambda x, lambda y) = lambda^2 x^2 – lambda^2 y^2 + 3. Lambda x. lambda y = lambda^2 f(x, y).Let us understand Homogeneous Differential Equation Examples. Suppose we have to Solve x^2 y dx – (x^3 + y^3) dy = 0 . The given differential equation is x^2 y dx – (x^3 + y^3) dy = 0, dy/dx = (x^2 y)/(x^3 + y^3)……(i), Since each of the functions x^2 y and x^3 + y^3 is a homogeneous function of degree 3, so the given differential equation is homogeneous. Putting y = vx and dy/dx = v + x dv/dx in (i), we get  v + x dv/dx = vx^3 /(x^3 + v^3 x^3), v + x dv/dx = v/1 + v^3, x dv/dx = [v/(1 + v^3)] – v, x dv/dx = (v – v – v^4)/(1 + v^3), x dv/dx = -v^4/(1 + v^3), x (1 + v^3) dv = -v^4 dx, (1 + v^3)/v^4 dv = -dx/x, (1/v^4 + 1/v) dv = -dx/x, Integrating both sides we get v^-3/-3 + log v = -log x + C=> -1/3v^3 + log v + log x = C=>-1/3 x^3/y^3 + log(y/x . x) = C=-1/3 . x^3/y^3 + log y = C, which is the required Homogeneous Solution Differential Equation.

Linear Homogeneous Differential Equation:  A differential equation is linear if the dependent variable (y) and its derivative appear only in first degree. The general form of a linear differential equation of first order is  dy/dx + Py = Q, where P and Q are functions of x (or constants).For example, dy/dx + xy = x^3, x dy/dx + 2y = x^3, dy/dx + 2y = sin x etc. are linear differential equations. Let us Find the General Solution of the Homogeneous Differential Equation  (1 + y^2) dx + (1 + x^2) dy = 0. Now   (1 + y^2) dx + (1 + x^2) dy = 0 => dx/(1 + x^2) + dy/(1 + y^2) = 0.On integration, we get  tan^-1 x + tan^-1 y = tan^-1 C, (x + y)/(1 – xy) = C, X + y = C (1 – xy).