Reduction Formula

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