Reduction Formula in Integration: Types and Solved Example

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Shwetha S

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Reduction formula important method of integration. It is effective in solving difficult and complex integration questions. It is used for various functions including trigonometric, logarithmic, exponential, and many others. To simplify the integration of these functions, the reduction formula comes into existence. These reduction formulas tend to calculate the solution of integral complex problems and help us to reduce the integral degree and evaluate the integrals in a limited number of steps. The reduction formula follows the rules of integration and is derived from the formulas of integration.

Read More: Applications of the Integrals

Key Terms: Integrals, Integration, Trigonometric, Logarithmic, Exponential, Algebraic Functions, Inverse Trigonometric Function


Formulas for Reduction in Integration

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There are several types of reduction formulas that are required to solve complex problems related to different functions. The different types of reduction formulas include: Reduction Formula for Exponential Functions, Reduction Formula for Inverse Trigonometric Functions, Reduction Formulas For Trigonometric Functions, Reduction Formula for Algebraic Functions, Reduction Formula For Trigonometric Functions, Reduction Formula for Logarithmic Functions, and Reduction Formula for Hyperbolic Trigonometric Functions

Integrals Detailed Video Explanation

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Reduction Formula for Exponential Functions

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  • ∫ yn emy dy = 1/mynemy –n/m yn-1 emydy
  • ∫ dy / sinhny = -1/n sinhn-1 y cosh y – (n-1/n) ∫ sinh n-2 y dy
  • ∫ emy / yn dy = emy (n-1)yn-1 + m/n-1 ∫ emx/xn-1 y
  • ∫ Sinhn ydy = -1/n Sin hn-1 cosh y – n-1/n ∫ Sinhn-2 y dy

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Reduction Formula for Inverse Trigonometric Functions

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  • ∫ yn arcsin y dy = yn+1/ n+1 arcsin y -1/n+1 ∫ yn-1/\( \sqrt{} \) 1-y2dy
  • ∫ yn arccos y dy = yn+1/ n+1 arc cos y -1/n+1 ∫ yn-1\( \sqrt{} \) 1-y2dy
  • ∫ yn arctan y dy = yn+1/ n+1 arctan y -1/n+1 ∫ yn-1\( \sqrt{} \) 1-y2dy

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Reduction Formula For Trigonometric Functions

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  • ∫ 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
  •  \(\frac{dx}{sin^n x} \) \(\frac{cos x}{(n-1) sin^{n-1} x} + \frac{(n-2)}{(n-1)}\)∫ \(\frac{dx}{sin^{n-2} x'} \) n ≠ 1
  •  \(\frac{dx}{cos^n x} \) \(\frac{sin x}{(n-1) cos^{n-1} x} + \frac{(n-2)}{(n-1)}\)∫ \(\frac{dx}{cos^{n-2} x'} \) n ≠ 1

Read More: Integration by Parts


Reduction Formula for Algebraic Functions

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  • ∫ 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

Read More: Difference Between Variance and Standard Deviation


Reduction Formula For Trigonometric Functions

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  • ∫ 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

Read More: Continuous Integration


Reduction Formula for Logarithmic Functions

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  • ∫ xn Inm dx = \(\frac{x^{n+1}ln^mx}{n+1} - \frac{m}{n+1}\)∫ xn Inm – 1xdx
  • \(\frac{ln^mx}{x^n} = - \frac{ln^mx}{(n-1)x^{n+1}} + \frac{m}{n-1} \)∫ \(\frac{ln^{m-1}x}{x^n}\), n ≠ 1

Read More: u substitution formula


Reduction Formula for Hyperbolic Trigonometric Functions

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  • ∫sinhnx dx = −(1/n) sinhn−1 x cosh x − (n−1/n) ∫sinhn−2x dx
  • ∫coshnx dx = (1/n) sinh x coshn−1 x − (n−1/n) ∫coshn−2x dx
  • ∫tanhnx dx = −(1/n−1) tanhn−1x + ∫tanhn−2x dx, n≠1
  • ∫cothnx dx = −(1/n−1)cothn−1x + ∫cothn−2x dx, n≠1
  • ∫ \(\frac{dx}{sinh^nx} = - \frac{cosh x}{(n-1) sinh^{n-1}x} + \frac{(n-2)}{(n-1)}\)∫ \(\frac{dx}{sinh^{n-2}x'}\), n≠1
  • ∫ \(\frac{dx}{cosh^nx} = - \frac{sinh x}{(n-1) cosh^{n-1}x} + \frac{(n-2)}{(n-1)}\)∫ \(\frac{dx}{cosh^{n-2}x'}\), n≠1
  • ∫sinhnxcoshmxdx = \(\frac{sinh^{n+1}xcosh^{m-1}x}{n+m} + \frac{(m-1)}{(n+m)}\)∫ sinhnxcoshm-2xdx
  • ∫sechnxdx = \(\frac{sech^{n-2}xtanhx}{n-1} + \frac{n-2}{n-1}\)∫sechn-2xdx,  n≠1

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Things to Remember

  • Reduction formulas tend to calculate the solution of integral complex problems and help us to reduce the integral degree and evaluate the integrals in a limited number of steps.
  • Integration is a method to calculate the antiderivative of a given function.
  • The different types of categories of reduction formula include the reduction formula for trigonometric functions, inverse trigonometric functions, exponential functions, logarithmic functions, algebraic functions, and hyperbolic trigonometric functions. 
  • ∫ yn emy dy = 1/mynemy –n/m yn-1 emydy
  • ∫sinhnx dx = −(1/n) sinhn−1 x cosh x − (n−1/n) ∫sinhn−2x dx
  • ∫xn lnmx dx = \(\frac{x^{n+1}ln^{m}x}{n+1}\) - \(\frac{m}{n+1}\) ∫xnlnm−1xdx
  • ∫ Sinn(y) dy = -sinn-1 (y) cos (y)/n + n-1/n Sinn-2 (y) dy

Read More: Three-Dimensional Geometry Introduction


Previous Year Questions

  1. The area enclosed between this curve and the coordinate axes… [JKCET – 2017]
  2. The area (in sq units) of the region described by… [JEE Main – 2014]
  3. The area (in s units) of the quadrilateral formed by the tangents… [JEE Main – 2015]
  4. The area (in s units) bounded by the curve y… [COMEDK UGET – 2013]
  5. If the line x = b bisects the area bounded by the curves, C1​ and C2 [JEE Main – 2020]
  6. The greatest integer less than or equal to “t”…  [JEE Main – 2019]
  7. [x] denotes the greatest integer less than 20Cr​ … [JEE Main – 2019]
  8. The value of the integral… [UPSEE – 2018]
  9.  \(\int {x^3 - 1}\over {x^3 - x}\)dx… [COMEDK UGET – 2015]
  10. The only integral root of the equation … [AMUEEE – 2016]

Sample Questions

Ques 1. Calculate the integral of ∫tan5(2x) dx (5 Marks)

Ans. With the help of reduction formula:

∫tann(u) du = (1/ n−1) tann−1(u) − ∫tann−2(u) du

Substituting the values:

a = 2x

∴ ½ da = dx

Hence,

∫tan5 (2x) dx = ½ [∫tan5(a) da]

= ½ [¼ tan4(a) − ∫tan3(a) da]

= ½ [¼ tan4(a) – [½ tan2(a) – ∫tan(a) da]]

= 1/8 tan4(a) − ¼ tan2(a) + ½ ln sec(a) + C

= 1/8 tan4(2x) − ¼ tan2(2x) + ½ ln sec(2x) + C

Ques 2. Evaluate: x3Log2x. (3 Marks)

Ans: Applying the reduction formula we can conveniently find the integral of the given expression.

xnlogmx.dx = \(\frac{x^{n+1}log^mx}{n+1} - \frac{m}{n+1}\)xnlogm-1x.dx

x3log2x.dx = \(\frac{x^{4}log^2x}{4} - \frac{2}{4}\)x3logx.dx

\(\frac{x^{4}log^2x}{4} - \frac{1}{2}\)(\(\frac{x^{4}logx}{4} - \frac{1}{4}\)x3.dx)

\(\frac{x^{4}log^2x}{4} - \frac{x^4logx}{8} + \frac{x^4}{32} = C\)

Ans.  x3log2x.dx =\(\frac{x^{4}log^2x}{4} - \frac{x^4logx}{8} + \frac{x^4}{32} = C\)

Ques 3. Explain the different types of reduction formulas. (3 Marks)

Ans. An Integration is a method to calculate the antiderivative of a given function. Reduction formulas tend to calculate the solution of integral complex problems and help us to reduce the integral degree and evaluate the integrals in a limited number of steps. The different types of reduction formulas are: 

  • Reduction Formula for Exponential Functions
  • Reduction Formula for Inverse Trigonometric Functions
  • Reduction Formulas For Trigonometric Functions
  • Reduction Formula for Algebraic Functions
  • Reduction Formula For Trigonometric Functions
  • Reduction Formula for Logarithmic Functions
  • Reduction Formula for Hyperbolic Trigonometric Functions

Ques 4. Calculate y7(8 + 3y4)8 dy. (4 Marks)

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

Rewriting these values:

Calculating the integral:

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

Now, transform the entire y’s in the integrand excluding for the first y4. Solving the same for y4 using substitution, we get

y4 = 1/3(u-8)

Now, substituting and evaluating the final result:

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

Ques 5. Evaluate In = ∫ sin y dy = 1/n cos y sinn-1 y + n-1/n In-2(5 Marks)

Ans.Using the reduction formula with n= 4, we have,

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

Calculating 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/siny cosy+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’

Both values of K’ and K are different.

Therefore, 3/8 y -1/4 sin (2y) + 1/32 sin 4(y) + K’ is the required answer.

Ques 6. Calculate the integral of Sin5x. (5 Marks)

Ans. Example, sin5 xdx = sin4 x sinxdx

sin4 x + sinx

4sin3 xcos x – -cos x

sin4 xcos x – -4 sin3 xcos2 xdx

sin4 xcos x + 4sin3x(1 – sin2x)dx

sin4 xcos x + 4sin3xdx – 4sin3xdx

5sin5xdx = – sin4xcosx + 4sin3xdx

sin5xdx = – \(\frac{1}{5}\)sin4xcosx + \(\frac{4}{5}\)sin5xdx

This is the required answer.

Ques 7. Calculate the integral of Cos2(x)Sin(x)dx. (3 Marks)

Ans. cos2(2x)sin(x)dx =  \(\frac{1}{2}\)[1 + cos(4x)]sin (x)dx

\(\frac{1}{2}\)cos(4x) + \(\frac{1}{2}\)sin (x)cos(4x)dx

\(\frac{1}{2}\) cos(4x)dx + \(\frac{1}{4}\)sin (- 3x) + sin(5x)dx

\(\frac{1}{8}\)sin(4x) – \(\frac{1}{4}\)∫ sin(3x)dx + \(\frac{1}{4}\) sin(5x)dx

\(\frac{1}{8}\)sin(4x) + \(\frac{1}{12}\)cos(3x) – \(\frac{1}{20}\)cos(5x) + C, x ∈ IR.

Ques 8. According to the Given Reduction Formula In = ∫ sin y dy = 1/n cos y sinn-1 y + n-1/n In-2.  Find out: ∫ Sin4 y dy. (5 Marks)

Ans. According to the reduction formula of integrals with n= 4, the equation would be

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

Now calculating I2 = sin2 y dy with corresponds to n= 2

On applying the integrals equations, we get the equation as:

∫ Sin2 y dy = y/2-1/2 sin y cos y + K

After combining all these, it gives the result:

∫ Sin4 y dy

= -¼ Cos y Sin3 y + ¾ y/2–1/sinycosy+Ky/2–1/sinycosy+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’

This is the required answer. 

Note: The values of both the constants K and K’ are different.

Ques 9. Evaluate the Integral y7(8 + 3y4)8 dy. (5 Marks)

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

Rewriting the equation

Evaluating the integral

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

All the y’s in the integrand except the y4 in the front can be substituted.

Solving for y4, we will apply the substitution and calculate the result as:

y4 = 1/3(u-8)

Now, substituting the values and then calculating the result

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

This is the required answer.

Ques 10. What is the reduction formula? (2 Marks)

Ans. Higher-order integrals are frequently calculated in integration using a reduction formula. Higher degree expressions need a lot of time and effort to solve, hence in this case, the reduction formulas are provided as simple expressions with a degree n. A collection of four formulas representing the reduction formulas has been provided here.

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                        CBSE CLASS XII Previous Year Papers

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