Integral is a method, to sum up, the functions on a larger scale. In this article, let us discuss the integrals of some particular functions that are generally used for calculations. These integrals have a variety of applications in the real-life as well, such as to find the area between the curves, finding the volume, finding the average value of a function, centre of mass, kinetic energy, amount of work done, and many more.
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Particular Functions
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Proof of The Integral Functions
Now that you know about these integral functions and their values, let us take a look at the proof of each of these functions.
Integral of Function 1
∫ dy / (y2 – a2) = 1/2a log |(y – a) / (y + a)| + C
As you know,
1 / (y2 – a2) = 1 / (y – a) (y + a)
Solving this,
= 1/2a [(y + a) – (y – a) / (y – a) (y + a)]
Reducing it further,
= 1/2a [1/(y – a) – 1/(y + a)]
Therefore, ∫ dy / (y2 – a2) = 1/2a [∫ dy / (y – a) – ∫ dy / (y + a)]
Solving this,
= 1/2a [log |(y – a) – log |(y + a)] + C
Hence,
= 1/2a log |(y – a) / (y + a)| + C
Integral of function 2
∫ dy / (a2 – y2) = 1/2a log |(a + y) / (a – y)| + C
As you,
1 / (a2 – y2) = 1 / (a – y) (a + y)
Solving,
= 1/2a [(a + y) + (a – y) / (a – y) (a + y)]
Hence,
= 1/2a [1/(a – y) + 1/(a + y)]
Therefore, ∫ dy / (a2 – y2) = 1/2a [∫ dy / (a – y) + ∫ dy / (a + y)]
When you solve,
= 1/2a [– log |(a – y) + log |(a + y)] + C
Hence,
= 1/2a log |(a + y) / (a – y)| + C
Integral of Function 3
∫ dy / (y2 + a2) = 1/a tan–1 (y/a) + C
Substitute y = a tan t, so you have dy = a sec2 t dt.
Therefore,
∫ dy / (y2 + a2) = ∫ [(a sec2 t dt) / (a2 tan2 t + a2)]
Solving,
∫ dy / (y2 + a2) = 1/a ∫ dt = t/a + C
Re-substitute the value of t,
∫ dy / (y2 + a2) = 1/a tan–1 (y/a) + C
Integral of Function 4
∫ dy / √ (y2 – a2) = log |y + √ (y2 – a2)| + C
Substitute y = a sec t
So, dy = a sec t tan t dt.
Therefore,
∫ dy / √ (y2 – a2) = ∫ a sec t tan t dt / √ (a2 sec2 t – a2)
Solving,
∫ dy / √ (y2 – a2) = ∫ sec t dt = log |sec t + tan t| + C1
Substituting the value of t again,
∫ dy / √ (y2 – a2) = log |(y/a) + √ [(y2 – a2) / a2]| + C1
Solving,
= log |y + √(y2 – a2)| – log |a| + C1
Hence,
= log |y + √(y2 – a2)| + C
where, C = C1 – log |a|
Integral of Function 5
∫ dy / √ (a2 – y2) = sin–1 (y/a) + C
Substitute y = a sin t
dy = a cos t dt.
Therefore,
∫ dy / √ (a2 – y2) = ∫ a cos t dt / √ (a2 – a2 sin2 t)
Solving,
∫ dy / √ (a2 – y2) = ∫ t dt = t + C
Substituting the value of t,
∫ dy / √ (a2 – y2) = sin–1 (y/a) + C
Integral of Function 6
∫ dy / √ (y2 + a2) = log |y + √ (y2 + a2)| + C
Substitute y = a tan t,
dy = a sec2 t dt
Therefore,
∫ dy / √ (y2 + a2) = ∫ a sec2 t dt / √ (a2 tan2 t + a2)
Solving,
∫ dy / √ (y2 – a2) = ∫ sec t dt = log |sec t + tan t| + C1
Re-substituting the value of t,
∫ dy / √ (y2 – a2) = log |(y/a) + √ [(y2 + a2) / a2]| + C1
Solving,
= log |y + √(y2 + a2)| – log |a| + C1
Hence,
= log |y + √(y2 + a2)| + C
where, C = C1 – log |a|
Integral of Function 7
∫ dy / (ay2 + by + c)
You can write this as
ay2 + by + c = a [y2 + (b/a)y + (c/a)]
Solving,
a [(y + b/2a)2 + (c/a – b2/4a2)]
Substitute (y + b/2a) = t and you would get dy = dt.
Substitute (c/a – b2/4a2) = ±k2.
Therefore,
ay2 + by + c = a (t2 ± k2)
where the signs + or – depend on the sign of the equation (c/a – b2/4a2).
Therefore,
∫ dy / (ay2 + by + c) = 1/a ∫ dt / (t2 ± k2)
You can evaluate this equation by using one or more of the above siy integration formulas shown. Remember that you can also solve for the equation ∫ dy / √ (ay2 + by + c) in a similar manner.
Integral of Function 8
∫ [(py + q) / (ay2 + by + c)] dy,
where p, q, a, b, c are known to be constants.
To solve this, you must find the constants A and B such that,
(py + q) = A d/dy (ay2 + by + c) + B, which is equal to = A (2ay + b) + B
To determine ‘A’ and ‘B’, first, equate from both the sides of the coefficients of y and the constant terms. ‘A’ and ‘B’ can then be obtained and therefore, the integral is reduced to any one of the known forms.
Solved Example
Find the integral of (y + 3) / √ (5 – 4y + y2) with respect to y.
Solution
You can express
y + 3 = A d/dy (5 – 4y + y2) + B = A (– 4 – 2y) + B
Equating the coefficients, you get
A = – ½ and B = 1
Therefore,
∫ [(y + 3) / √ (5 – 4y + y2)] dy = – ½ ∫ [(– 4 – 2y) / √ (5 – 4y + y2)] dy + ∫ dy / √ (5 – 4y + y2)
= – ½ I1 + I2 … (a)
Solving I1
Substitute (5 – 4y + y2) = t,
(– 4 – 2y) dy = dt
Therefore,
I1 = ∫ [(– 4 – 2y) / √ (5 – 4y + y2)] dy = ∫ dt / √ t = 2 √ t + C1
= 2 √ (5 – 4y + y2) + C1 … (b)
Solving I2
I2 = ∫ dy / √ (5 – 4y + y2) =
∫ dy / √ [9 – (y + 2)2]
Substitute (y + 2) = t,
dy = dt
Therefore,
I2 = ∫ dt / √ (32 – t2) = sin–1 (t/3) + C2
Solving,
= sin–1 [(y + 2) / 3] + C2 … (c)
Substitute (b) and (c) in (a),
∫ [(y + 3) / √ (5 – 4y + y2)] dy = – ½ I1 + I2
= – √ (5 – 4y + y2) + sin–1 [(y + 2) / 3] + C
where C = C2 = C1/2.
FAQs on Integrals of Some Particular Functions
1. What is integration?
Integration refers to the act of bringing together smaller components into one single system. This single system functions as one. In mathematics, an integral assigns the numbers to the functions in such a way that it can describe displacement, area, volume, and other concepts which arise by combining infinitesimal data. Integration is amongst the two main operations of calculus; its inverse operation, differentiation, is the other one.
Given that a function f of a real variable y and an interval [a, b] of the real line, the definite integral is given by the equation as shown below:
∫ab f(y)dy.
2. Explain the applications of integration in real life.
Integration has several uses in real life. It plays a role in the fields of physics, engineering, medical science, graphic designing and research analysis.
Applications of the indefinite integral lie in how to find the displacement from velocity and the velocity from acceleration by using the indefinite integral. The area under the curve and the area between two different curves are also found out with the help of integration.
The volume of the solid of revolution explains you about how to use integration for finding the volume of an object having curved sides, for example, wine barrels.
The centroid of an area refers to the centre of mass. You can use integration for finding the centroid of an area having curved sides.
Moments of inertia explains finding the resistance of a rotating body. You use integration if the shape has curved sides.
Work done by a variable force shows finding the work done on an object if the force is not constant. This includes Hooke's Law for springs which uses integration.
Electric charges have a force between them which varies depending upon the amount of charge and the distance between these charges. You use integration for calculating the work done when the charges are separated.
The average value of a curve is calculated using integration. Head Injury Criterion is an application of the average value and used in the road safety research.
The force by the liquid pressure varies and depends on the shape of the object and its depth. You use integration to find this force.
