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Integration by Reduction Formulae - Indefinite Integration, Class 12, Maths PDF Download

F. INTEGRATION BY REDUCTION FORMULAE


Ex.47  CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration


Sol.

CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration

 

Ex.48 Integration of 1/(x2 + k)n.


Sol.

CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration
Above is the reduction formula for CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration [1/(x2 + k)n]dx . By repeated application of this formula the integral shall reduce to that of  CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration

 


Ex.49 Integrate (x + 2) / (2x2 + 4x + 3)2.


Sol. Here (d/dx) (2x2 + 4x + 3) = 4x + 4.
CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration
Now put x + 1 = t and then applying the reduction formula, we get
CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration


Ex.50 Integrate (2x + 3)/(x2 + 2x + 3)2.


Sol. Here (d/dx) (x2 + 2x + 3) = 2x + 2
CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration

CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration

Ex.51 If Im=

CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration


Sol.

CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration

Ex.52 If Im,n = CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration


Sol. We have, Im,n = CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integrationcosm x.cosnx dx
CBSE, Class 12, IIT JEE, Syllabus, Preparation, NCERT, Important, Indefinite Integration
As we have cos (n – 1) x = cos nx cos x + sin nx . sinx

Integration by Reduction Formulae - Indefinite Integration, Class 12, Maths

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FAQs on Integration by Reduction Formulae - Indefinite Integration, Class 12, Maths

1. What is the reduction formula for indefinite integration?
Ans. The reduction formula for indefinite integration is a formula that allows us to reduce the degree of a given integral by using a substitution or integration by parts. It is derived by repeatedly applying the reduction formula until the integral becomes solvable.
2. How do we use reduction formulae to find indefinite integrals?
Ans. To use reduction formulae to find indefinite integrals, we follow these steps: 1. Identify the integral that needs to be solved. 2. Apply the reduction formula by substituting a suitable value for the variable or using integration by parts. 3. Simplify the resulting expression and solve for the value of the integral. 4. If the integral still cannot be solved, apply the reduction formula again until the degree of the integral is reduced enough to be solvable.
3. What is the purpose of using reduction formulae in integration?
Ans. The purpose of using reduction formulae in integration is to simplify complex integrals by reducing their degree. This makes it easier to solve the integral and find the antiderivative. Reduction formulae provide a systematic approach to solving integrals and can be particularly useful when dealing with integrals involving trigonometric functions or powers of a variable.
4. Can reduction formulae be used to solve all types of integrals?
Ans. Reduction formulae can be used to solve many types of integrals, particularly those involving powers of a variable or trigonometric functions. However, there may be cases where reduction formulae are not applicable or do not lead to a solvable integral. In such cases, alternative methods like substitution or partial fractions may be required.
5. Are reduction formulae applicable only for indefinite integration?
Ans. Reduction formulae can be used for both indefinite and definite integration. While the focus of reduction formulae is often on finding antiderivatives and solving indefinite integrals, they can also be applied to definite integrals by using the limits of integration in the final step of the solution. The reduction formula is still used to simplify the integral, but the definite limits are considered to obtain the final result.
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