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