Formulas And Solved Examples Of Integrals Of Special Functions
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 Particular Functions Explained Clearly
1. What are integrals of some particular functions?
Integrals of some particular functions are standard antiderivatives of commonly used functions like trigonometric, exponential, and rational forms. These integrals are memorized because they appear frequently in calculus problems. Examples include:
- ∫ ex dx = ex + C
- ∫ sin x dx = −cos x + C
- ∫ cos x dx = sin x + C
- ∫ 1/x dx = ln|x| + C
Knowing these standard integrals helps in solving more complex integration problems efficiently.
2. What is the integral of e^x?
The integral of ex is ∫ ex dx = ex + C. The exponential function ex is unique because its derivative and integral are the same. For example, differentiating ex + C gives back ex, confirming the result.
3. What is the integral of sin x and cos x?
The integrals are ∫ sin x dx = −cos x + C and ∫ cos x dx = sin x + C. These results follow from basic differentiation rules:
- The derivative of −cos x is sin x.
- The derivative of sin x is cos x.
These are fundamental trigonometric integrals used in definite and indefinite integration.
4. What is the integral of 1/x?
The integral of 1/x is ∫ 1/x dx = ln|x| + C. The absolute value is necessary because the natural logarithm is defined only for positive arguments, but 1/x exists for both positive and negative x (except zero). This is a key standard result in integration formulas.
5. What is the integral of sec²x and cosec²x?
The integrals are ∫ sec²x dx = tan x + C and ∫ cosec²x dx = −cot x + C. These follow from differentiation identities:
- The derivative of tan x is sec²x.
- The derivative of cot x is −cosec²x.
These particular trigonometric integrals are frequently used in substitution problems.
6. What is the integral of 1/(1 + x²)?
The integral of 1/(1 + x²) is ∫ 1/(1 + x²) dx = tan−1x + C. This is a standard result related to inverse trigonometric functions because the derivative of tan−1x equals 1/(1 + x²).
7. What is the integral of 1/√(1 − x²)?
The integral of 1/√(1 − x²) is ∫ 1/√(1 − x²) dx = sin−1x + C. This result comes from the derivative of the inverse sine function. It is commonly used in problems involving inverse trigonometric integrals.
8. How do you integrate ax where a is a constant?
The integral of ax is ∫ ax dx = ax/ln a + C, where a > 0 and a ≠ 1. This formula is derived because the derivative of ax is ax ln a. For example, ∫ 2x dx = 2x/ln 2 + C.
9. What is the integral of tan x and cot x?
The integrals are ∫ tan x dx = −ln|cos x| + C and ∫ cot x dx = ln|sin x| + C. These results are obtained by rewriting:
- tan x = sin x / cos x
- cot x = cos x / sin x
Then substitution leads to logarithmic integrals.
10. Why are standard integrals of particular functions important in calculus?
Standard integrals of particular functions are important because they form the foundation for solving complex integration problems. They:
- Help in applying substitution and integration by parts.
- Are used in solving differential equations.
- Appear frequently in definite integrals and applications of integration.
Memorizing key formulas like ∫ ex dx, ∫ sin x dx, and ∫ 1/x dx makes integration faster and more accurate.
