**Class 12 Maths well explained NCERT solutions of exercise 7.2**

Here** Class 12 Maths well explained NCERT solutions of exercise 7.2 of chapter 7-Integrals** are created to clear your concept of methods solving** integration** of a function. These **NCERT solutions** are the** solutions of exercise 7.1 of chapter 7-Integral**s of **Class 12 maths NCERT** textbook prescribed by CBSE. All questions of **exercise 7.2 of chapter 7-Integrals** are solved by the expert by a step by step method. Although there are several methods used for solving **integration** but the method used in the** exercise 7.2** for solving the** integral** functions is substitution method

**What is the substitution method for calculation of integral function?-**Let two functions exist in the way ∫f(x).f'(x) dx, here second function is the derivative of the first function.

Let t = f(x)

Differentiating it both sides

dt/dx = f'(x)

dt = f'(x) dx

Substituting it in the integral

∫t dt = t²/2 +C , now substituting t = f(x) back in the integration

∫f(x).f'(x) dx = [f(x)]²/2

As an example Let’s calculate ∫2x cosx² dx =?

We have to find out

∫2x cosx² dx

here d/dx of x² is 2x, so ,

Let t = x²

dt/dx = 2x⇒ dt = 2x.dx

Substituting t = x² and dt = 2x dx in the integration, we have

∫cost dt = sint + C

Now substituting t = x² back to the integrals

∫cost dt = sin x² + C

Now, you can easily understand the method used here for calculating integration of the function used here in exercise 7.2 of chapter 7-Integrals

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** NCERT solutions of exercise 7.2 of chapter 7-Integrals**

Integrate the function in exercises 1 to 37.

Let 1 + x² = t

dt/dx = 2x

dt = 2x dx

We have

Ans.

Since 1/x is the d/dx of and

So, let t =

Substituting the value of dt

Substituting t = back in the integral

Hence the integration of the given function is

Ans. The given function is

Integrating it

Let 1 + logx =t

dt/dx = 1/x

dt =dx/x

Substituting the value of dt

Substituting the value of t back in the integral

Hence the required integration of the given function is

Q4. sinx.sin(cosx)

Ans. We are given sinx.sin(cosx)

Integrating it

Since d/dx of cosx =-sinx, so

Let t = cosx

dt/dx = -sinx

dt = -sinx dx

sinxdx= -dt

Substituting the value of sinx dx

= -(-cost) + C

= cost + C

Substituting back the value of t= cosx

= cos (cosx) + C

Q5. sin(ax +b) cos(ax +b)

Ans. We are given the function

sin(ax +b) cos(ax +b)

Divuding and multipling it by 2, we have

Now,integrating it

Let 2(ax+b) = t

dt/dx = 2a

dt/2a = dx

Substituting the value of dx

Subtituting back the value of t

Ans. We are given

Integrating it

Let t = ax +b

dt/dx = a

dx = dt/a

Substituting the value of dx and (ax + b)

Ans. We are given the function

Integrating it

Let t = x + 2⇒ x = t -2

dt/dx = 1⇒ dt = dx

Substituting value of x and dx

Substituting back the value of t

Therefore required integration of the given function is

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Ans. We are given the function

Integrating it

Let t = 1 + 2x²

dt/dx = 4x

xdx = dt/4

Substituting the value of xdx and (1 + 2x²), we have

Substituting the value of t back

Therefore the required value of integration is

Ans. We are given the function

Rewriting it

Integrating it

Let t = x² + x + 1

dt/dx = 2x + 1

We have dt = (2x +1)dx

Substituting back the value of t

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