Integrals of Some Particular 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.

1. 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

2. 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

3. 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

4. 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|

5. 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

6. 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|

7. 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.

8. 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.