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 are the main types of standard integrals introduced in the chapter 'Integrals of Some Particular Functions' for Class 12 Maths?
The chapter covers key standard integrals including:
- ∫dy/(y² – a²) and ∫dy/(a² – y²)
- ∫dy/(y² + a²)
- ∫dy/√(y² – a²) and ∫dy/√(a² – y²)
- Forms like ∫dy/(ay² + by + c)
- Integrals with linear numerators: ∫(py + q)/(ay² + by + c)
These forms are essential for solving complex integration problems as per the CBSE 2025-26 syllabus.
2. How do you solve integrals involving quadratic expressions, such as ∫dy/(ay² + by + c), as per the CBSE Class 12 approach?
For integrals like ∫dy/(ay² + by + c):
- First, complete the square in the denominator to convert it to the form a(t² ± k²).
- Set y + (b/2a) = t
- This reduces the problem to a standard form: ∫dt/(t² ± k²), which is solved using known formulas.
- This approach matches the stepwise methodology specified in class 12 maths for CBSE board exams.
3. What are the most common mistakes students make when applying standard integral formulas in board exams?
Students often face marks deduction due to:
- Mistaking the form of the integrand (e.g., mixing up y² – a² with a² – y²)
- Incorrect substitutions or missing steps during transformation
- Omitting the constant of integration (C)
- Missing out on absolute value signs in logarithmic integrals
Careful rewriting and checking each step is important for full marks.
4. Why is substitution used for integrating functions with roots, such as ∫dy/√(y² – a²), and how do you choose an effective substitution?
Substitution transforms complicated irrational integrals into standard, easily solvable forms. For ∫dy/√(y² – a²), set y = a sec t or y = a cosh t so that the root simplifies using trigonometric identities. Choose substitutions that match the integrand's structure to reduce complexity and ensure alignment with the NCERT/CBSE exam methods.
5. How can you distinguish between the integrals ∫dy/(y² – a²) and ∫dy/(a² – y²) and remember their correct formula during board exams?
The difference lies in the position of y² and a²:
- For ∫dy/(y² – a²), use: 1/2a log |(y – a)/(y + a)| + C
- For ∫dy/(a² – y²), use: 1/2a log |(a + y)/(a – y)| + C
Remember their order closely as it's a frequent exam trap. Write out each step to avoid confusion.
6. What are the main conceptual differences between integrating rational and irrational functions in this chapter?
For rational functions (polynomials in numerator and denominator), apply partial fraction decomposition. For irrational functions (those involving roots like √(y² ± a²)), use appropriate substitutions such as trigonometric or hyperbolic identities. Identifying the type of function guides your choice of method and is critical for solving CBSE-style questions accurately.
7. How do you approach a problem where the numerator is linear and the denominator is quadratic, for example, ∫(py + q)/(ay² + by + c) dy?
Express the numerator as a combination of the derivative of the denominator and a constant: (py + q) = A × d/dy(ay² + by + c) + B. This lets you split the integral into two simpler parts, each matching a standard integral form. Matching coefficients carefully is crucial according to CBSE exam pattern.
8. In what ways do standard integrals apply to real-world scenarios relevant to Class 12 physics and engineering topics?
Standard integrals are used to:
- Calculate areas under curves and between functions
- Determine volumes of solids of revolution
- Find centroids and moments of inertia of complex shapes
- Compute work done by variable forces
- Evaluate electric potential in physics problems
These applications are routinely part of integrated CBSE assessments.
9. How can students verify the accuracy of their indefinite integral solutions during CBSE exams?
After integration, differentiate your solution. If the derivative matches the original integrand, your solution is correct. This reverse-check process is a strong board exam strategy and ensures accuracy.
10. What is the significance of including the constant of integration (C) in all indefinite integral answers, especially for board evaluations?
The constant C represents all possible antiderivatives. Neglecting it results in incomplete answers and marks loss. As per CBSE 2025–26 marking guidelines, always write '+ C' to receive full credit for indefinite integrals.
11. How do standard integrals typically feature in CBSE Class 12 board exam questions?
Questions may ask for:
- Direct evaluation using standard integrals
- Problems requiring appropriate substitution before using a standard result
- Mixed forms involving partial fractions and substitution
- Application-based questions in real-life or physics contexts
Both short and long answer formats are common, so mastering all types is essential for scoring well.
12. What conceptual mistakes should be avoided when handling logarithmic forms in integration, such as ∫dy/(y² – a²)?
Common errors include:
- Omitting the absolute value bars around expressions inside the logarithm
- Ignoring the constant of integration C
- Flipping the order of numerator and denominator in log arguments
Paying attention to each detail matches the CBSE/NCERT marking scheme and prevents loss of marks.
