Class 12 Maths NCERT Solutions of exercise 7.3 of the chapter 7-Integrals - Future Study Point

Class 12 Maths NCERT Solutions of exercise 7.3 of the chapter 7-Integrals

Ex.7.3-Integrals

Class 12 Maths NCERT Solutions of exercise 7.3 of the chapter 7-Integrals

Ex.7.3-Integrals

 

Class 12 Maths NCERT solutions of exercise 7.3 of the chapter 7-Integrals are the solutions of exercise 7.3 of the chapter 7-Integrals of class 12 NCERT maths textbook.  Class 12 Maths NCERT Solutions of exercise 7.3 of the chapter 7-Integrals are compulsory to be studied for every maths student of class 12 for clearing the concept of the methods used for solving the questions based on the integration of the functions. In this exercise 7.3, you will study the integration of different types of complex trigonometric functions. The solutions of exercise 7.3 are required the inputs of class 11 chapter 3- trigonometric functions because for simplification of complex trigonometric functions the trigonometric identities are required to use for the integration of the functions.

As an example, we can not integrate the function directly Cos²2x, so using the trigometric identity cosA = 2cos²A – 1⇒ cos²A = (1+cosA)/2, so cos²2x = (1+cos4x)/2

Now, integrating the simplified function ((1+cos4x)/2), we have

1/2∫(1+cos4x)dx = 1/2[∫1dx +∫cos4xdx] = x/2 +(1/2)(sin4x)/4 +C= x/4 + (1/8)sin4x +C

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Class 12 Maths NCERT solutions of Chapter 7 Integrals

Exercise 7.1- Integrals

Exercise 7.2-Integral

Exercise 7.3-Integrals

Exercise 7.4 -Integral

 

Class 12 Maths NCERT Solutions of exercise 7.3

Q1.Integrate sin²(2x +5)

Ans.We are given the function

sin²(2x +5)

Applying the trigonometric identity

cos2A=1 – 2sin²A

Now, integrating it

Integrating each terms individually

Q2.Integrate sin3x cos4x.

Using the trigonometric identity

2sinAcosB = sin(A+B)+ sin (A-B)

 

Now, integrating it

Integrating each term individually

Q3.Integrate cos2x cos4x cos6x

Ans. We are given the function

cos2x cos4x cos6x

Using the trigonometric identity

Now, integrating the given function

 

Again,applying the  identities cos²2x= (1 + cos4x)/2 and

 

Q4.Integrate sin³(2x +1)

Ans. We are given the function

sin³(2x +1)

Integrating the function

Rewriting the given function

Applying the trigonometric identity

 

sin²(2x+1) = 1 – cos²(2x+1)

Using the substitution method

Let t = cos(2x+1)

Substituting cos(2x+1) =t and sin(2x+1)dx=-dt/2

 

Now,substituting back the value of t= cos(2x+1)

Q5.cos³xsin³x

Ans. We are given

cos³xsin³x

Integrating it

Using the identy sin²x = 1-cos²x

Let t= cosx

-dt = sinx dx

Substituting cosx and sinxdx by t and -dt respectively

Substituting back the value of t= cosx

Q6.sinx sin 2x sin3x

Ans.We are given the function

sinx sin 2x sin3x

Integrating it

Using the trigonometric identity

Using cos(-x) = cosx

Substituting the value of sin2x sin3x

 

Using the identity 2sinx cosx = sin2x, and sinx cos5x = 1/2[sin(x+5x) +sin(x-5x)], we get

 

Using sin(x-5x) = sin(-4x) = -sin4x

 

 

Class 12 Maths NCERT Solutions of exercise 7.3 of the chapter 7-Integrals

Q7.Integrate sin4x sin8x

Ans. We are given

sin4x sin8x

Integrating it

Using the trigonometric identity

sinA.sinB = 1/2(cos(A-B) -cos(A+B)

 

 

Q8. Integrate

Ans. We are given

Applying the identity 1-cosx = 2sin²x/2 and 1+cox = 2cos²x/2

Also applying  tan²x/2 =sec²x/2 – 1

Now, integrating it

 

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The money collected by us will be used for the education of poor students who leaves their study because of a lack of money.

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