 # Reduction Formula

## Reduction Formula in Integration

A reduction formula is considered as an important method of integration. Integration by reduction formulas enable us to solve complex integration problems. It can be used for trigonometric functions, power of elementary functions, product of two or more complex functions etc. These functions cannot be integrated easily. Hence, to simplify the process of integration, we can use some reduction formulas for determining the solution of integral problems These reduction formulas help us to minimize the degree of integrals and formulate the integrals in a finite number of steps.

Some of the commonly used reduction formulas for integrals including the common functions are discussed below.

### Reduction Formulas For Trigonometric Function

Here, we will discuss some important reduction formulas for trigonometric functions.

• ∫ Sinn(y) dy = -sinn-1 (y) cos (y)/n + n-1/n Sinn-2 (y) dy

• ∫ yn Sinn(y) dy = -yncos (y) + n ∫ yn-1 cos (y) dy

• ∫ yn Cos(y) dy = ynSin (y) - n ∫ yn-1 sin (y) dy

• ∫ tann (y) dy = - tann-1 (y)/n-1 – ∫ tann-2 (y) dy

• ∫ Sinn(y) dy Cosm(y) dy = sinn+1 (y) cosm-1(y)/n+m + m-1/n+m ∫ Sinn(y) Cosm-2(y)dy

### Reduction Formula for Exponential Function

• ∫ yn emy dy = 1/mynemy –n/m yn-1 emydy

• ∫ emy / yn dy = emy (n-1)yn-1 + m/n-1 ∫ emx/xn-1 y

• ∫ dy / sinhny = -1/n sinhn-1 y cosh y – (n-1/n) ∫ sinh n-2 y dy

• ∫ Sinhn y dy = -1/n Sin hn-1 cosh y – n-1/n ∫ Sinhn-2 y dy

### Reduction Formula for Algebraic Functions

• ∫ yn / ayn + b dy = y/a-b/a ∫ dy/ayn + b

• ∫ dy (ay2 + by + C)n = - 2ay- b/ (n-1)(b2 – 4ac) (ay2 + by + C)n-1 -2(2n-3)a/(n-1)(b2 – 4ac) ∫ dy/(ay2 + by + C)n-1 , n ≠ 1

• ∫ dy/(y2- a2)n = x/2(n-1)a2(y2 – a2)n-1 – 2n-3/2(n-1)a2 ∫ dy(y2 – a2)n-1 , n ≠ 1

• ∫ dy/(y2 + a2)n = x/2(n-1)a2(y2 – a2)n-1 + 2n-3/2(n-1)a2 ∫ dy(y2 – a2)n-1 , n ≠ 1

### Reduction Formulas For Trigonometric Functions

• ∫ Inn y dy = yn+1 Inm y/ n +1 xn - m/n+1 ∫ xnInm-1 y dy

• ∫ Inm y/yn = Inmy + (n-1)yn +1 + m/n-1  ∫ Inm-1y/yn dy

### Reduction Formula For Inverse Trigonometric Function

• ∫ yn arcsin y dy = yn+1/ n+1 arcsin y -1/n+1 ∫ yn-1/ $\sqrt{1-y^2dy}$

• ∫ yn arc cos y dy = yn+1/ n+1 arc cos y -1/n+1 ∫ yn-1/ $\sqrt{1-y^2dy}$

• ∫ yn arctan y dy = yn+1/ n+1 arctan y -1/n+1 ∫ yn-1/ $\sqrt{1-y^2dy}$

### Quiz Time

1. Find  $\int_{0}^{π/2}$ sin⁶ (x) dx

1. 0

2. π/8

3. π/4

4. 15π/96

2. Find $\int_{0}^{π/2}$ sin10 (x) cos (x) dx

1. 1

2. 0

3. 13π/1098

4. 21π /2048

### Reduction Formula Example with Solutions

1. Given the Reduction Formula

In = ∫ sin y dy = 1/n cos y sinn-1 y + n-1/n In-2

Calculate

∫ Sin4 y dy;

Solution:

Using the reduction formula with n= 4 gives

∫ Sin4 y dy = -1/4 cos y sin3 y + ¾ I2

We need to calculate I2 = sin2 y dy with corresponds to n= 2

Using the integrals equations we get,

∫ Sin2 y dy = y/2-1/2 sin y cos y + K

No combining all together gives

∫ Sin4 y dy

= -¼ Cos y Sin3 y + ¾ [y/2 – 1/ sin y cos y + K]

= -1/4 -¼ Cos y Sin3 y + 3/8y – cos y sin y + K’

= 3/8 y  -1/4 sin (2y) + 1/32 sin 4(y) + K’

Note: We used the value of K and K’ as the value of constants are actually different.

2. Calculate the Integral

y⁷(8 + 3y⁴)⁸ dy

Solution:

u = 8 + 3y⁴ ￫ du = 12y³dy  ￫ y³dy = 1/12 du

Let's rewrite the integer

Calculate the integral

∫ y7 (8 + 3y4)⁸ dy =∫ y4y3 (8 + 3y4)8 dy = ∫ y4(8 + 3y4)y3dy

Now, consider that we can transform the entire y’s in the integrand excluding for the y4 that is in the front. We can consider from the substitution that we can solve it for y4 to get,

y4 = 1/3(u-8)

Now, we will do substitution and calculate the integral

y7 (8 + 3y4) =1/12 1/3 (u-8)u8 du = 1/36 u9 – 8u8 du = 1/36 (1/10 u10 -8/9u9 + C

= 1/36(1/10 (8 + 3Y4)10 - 8/9 (8 + 3y4)9) + C

1. How to calculate the Reduction Formulas?

integration such as Integration by substitution, integration by parts, integration by trigonometric functions, integration by partial fractions etc. The key concept of reduction formulas is to express an integral parameter (eg. power) of a function, denoted by In in terms of an integral that includes a lower value of the parameter ( lower power) of the function ,For example,In-1, or In-2.. This makes the reduction formulas a type of recurrence relation. In other words, the reduction formulas denotes the integrals

In = f (y,n)dy,

In terms of

Ik  = F(y,k) dy,

k < n

2. What are the different types of Reduction Formulas in integral calculus?

Ans. The reduction formulas are used to determine some integrals. These reduction formulas enable us to minimise the degree of the integer and calculate the integrals in a  finite number of steps. The reduction can be applied to different functions including the trigonometric functions such as trigonometric functions for sin, trigonometric functions cos, trigonometric functions, exponential functions, logarithmic functions etc. The formulas for reduction is divided into 4 types such as

• Reduction Formula For Trigonometric Functions

• Reduction FormulaFor Exponential Functions

• Reduction Formula For Inverse Trigonometric functions

• Reduction Formula For Hyperbolic Trigonometric Functions

• Reduction Formula For Algebraic Functions