`color{red} star` Comparison between differentiation and integration

`color{red} {star` Integration by substitution

`color{red} {star` Integration by substitution

1. Both are operations on functions.

2. Both satisfy the property of linearity, i.e.,

(i) `color{blue} {d/(dx) [ k_1 f_1 (x) + k_2 f_2 (x) ] = k_1 d/(dx) f_1 (x) + k_2 d/(dx) f_2 (x)}`

(ii) ` color{blue} {∫ [ k_1 f_1 (x) + k_2 f_2 (x) ] d_x = k_1 ∫ f_1 (x) dx + k_2 ∫ f_2 (x) dx}`

Here `k_1 `and `k_2 `are constants.

3. We have already seen that all functions are not differentiable. Similarly, all functions are not integrable. We will learn more about nondifferentiable functions and nonintegrable functions in higher classes.

4. The derivative of a function, when it exists, is a unique function. The integral of a function is not so. However, they are unique upto an additive constant, i.e., any two integrals of a function differ by a constant.

5. When a polynomial function P is differentiated, the result is a polynomial whose degree is 1 less than the degree of P. When a polynomial function P is integrated, the result is a polynomial whose degree is 1 more than that of P.

6. We can speak of the derivative at a point. We never speak of the integral at a point, we speak of the integral of a function over an interval on which the integral is defined.

7. The derivative of a function has a geometrical meaning, namely, the slope of the tangent to the corresponding curve at a point. Similarly, the indefinite integral of a function represents geometrically, a family of curves placed parallel to each other having parallel tangents at the points of intersection of the curves of the family with the lines orthogonal (perpendicular) to the axis representing the variable of integration.

8. The derivative is used for finding some physical quantities like the velocity of a moving particle, when the distance traversed at any time t is known. Similarly, the integral is used in calculating the distance traversed when the velocity at time t is known.

9. Differentiation is a process involving limits. So is integration.

10. The process of differentiation and integration are inverses of each other as discussed in Section.

2. Both satisfy the property of linearity, i.e.,

(i) `color{blue} {d/(dx) [ k_1 f_1 (x) + k_2 f_2 (x) ] = k_1 d/(dx) f_1 (x) + k_2 d/(dx) f_2 (x)}`

(ii) ` color{blue} {∫ [ k_1 f_1 (x) + k_2 f_2 (x) ] d_x = k_1 ∫ f_1 (x) dx + k_2 ∫ f_2 (x) dx}`

Here `k_1 `and `k_2 `are constants.

3. We have already seen that all functions are not differentiable. Similarly, all functions are not integrable. We will learn more about nondifferentiable functions and nonintegrable functions in higher classes.

4. The derivative of a function, when it exists, is a unique function. The integral of a function is not so. However, they are unique upto an additive constant, i.e., any two integrals of a function differ by a constant.

5. When a polynomial function P is differentiated, the result is a polynomial whose degree is 1 less than the degree of P. When a polynomial function P is integrated, the result is a polynomial whose degree is 1 more than that of P.

6. We can speak of the derivative at a point. We never speak of the integral at a point, we speak of the integral of a function over an interval on which the integral is defined.

7. The derivative of a function has a geometrical meaning, namely, the slope of the tangent to the corresponding curve at a point. Similarly, the indefinite integral of a function represents geometrically, a family of curves placed parallel to each other having parallel tangents at the points of intersection of the curves of the family with the lines orthogonal (perpendicular) to the axis representing the variable of integration.

8. The derivative is used for finding some physical quantities like the velocity of a moving particle, when the distance traversed at any time t is known. Similarly, the integral is used in calculating the distance traversed when the velocity at time t is known.

9. Differentiation is a process involving limits. So is integration.

10. The process of differentiation and integration are inverses of each other as discussed in Section.

`\color{green} ✍️` Methods of Integration

1. Integration by Substitution

2. Integration using Partial Fractions

3. Integration by Parts

`color{red} {ul ("Integration by Substitution")`

● In this section, we consider the method of integration by substitution.

● The given integral ` color{red} {∫ f (x) dx}` can be transformed into another form by changing the independent variable `x` to `t` by substituting `color{red}"x = g (t)."`

Consider `color{red} { I= ∫ f(x) dx}` Now Put `x = g(t)` so that ` color{green} {(dx)/(dt) = g' (t)}`

We write `color{red} {dx = g′(t) dt}`

Thus `color{red} {I = ∫ f(x) dx = ∫ f (g(t)) g'(t) dt}`

● This change of variable formula is one of the important tools available to us in the name of `color{green}{" integration by substitution."`

● Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the examples.

1. Integration by Substitution

2. Integration using Partial Fractions

3. Integration by Parts

`color{red} {ul ("Integration by Substitution")`

● In this section, we consider the method of integration by substitution.

● The given integral ` color{red} {∫ f (x) dx}` can be transformed into another form by changing the independent variable `x` to `t` by substituting `color{red}"x = g (t)."`

Consider `color{red} { I= ∫ f(x) dx}` Now Put `x = g(t)` so that ` color{green} {(dx)/(dt) = g' (t)}`

We write `color{red} {dx = g′(t) dt}`

Thus `color{red} {I = ∫ f(x) dx = ∫ f (g(t)) g'(t) dt}`

● This change of variable formula is one of the important tools available to us in the name of `color{green}{" integration by substitution."`

● Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the examples.

Q 3135378262

Integrate the following functions w.r.t. x:

(i) `sin mx `

(ii) `2x sin (x^2 + 1)`

(iii) ` (tan^4 sqrt x sec^2 sqrt x)/( sqrt x)`

(iv) ` (sin (tan^(-1) x) )/(1+x^2)`

Class 12 Chapter 7 Example 5

(i) `sin mx `

(ii) `2x sin (x^2 + 1)`

(iii) ` (tan^4 sqrt x sec^2 sqrt x)/( sqrt x)`

(iv) ` (sin (tan^(-1) x) )/(1+x^2)`

Class 12 Chapter 7 Example 5

Solution

(i) We know that derivative of mx is m. Thus, we make the substitution

mx = t so that mdx = dt.

Therefore, ` int sin mx dx = 1/m int sin t dt = - 1/m cos t +C = -1/m cos mx + C`

(ii) Derivative of `x^2 + 1` is 2x. Thus, we use the substitution `x^2 + 1 = t` so that

`2x dx = dt`.

Therefore, ` ∫2x sin (x^2 +1) dx = ∫sin t dt = – cos t + C = – cos (x^2 + 1) + C`

(iii) Derivative of `sqrt x` is `1/2 x^(-1/2) = 1/(2 sqrt x)` Thus, we use the substitution

`sqrt x = t` so that `1/(2 sqrt x) dx =dt ` giving `dx =2t dt`

Thus ` int ( tan^4 sqrt x sec^2 sqrt x)/( sqrt x) dx = int (2t tan^4 t sec^2 t dt)/t = 2 int tan^4 t sec^2 t dt`

Again, we make another substitution` tan t = u` so that `sec^2 t dt = du`

Therefore, ` 2 int tan^4 sec^2 t dt = 2 int u^4 du = 2 u^5/5 +C`

`= 2/5 tan^5 t +C` (since u = tan t)

`= 2/5 tan^5 sqrt x + C ` (since `t = sqrt x` )

Hence, ` int (tan^4 sqrt x sec^2 sqrt x)/(sqrt x) dx = 2/5 tan^5 sqrt x +C`

Alternatively, make the substitution `tan sqrt x = t`

(iv) Derivative of `tan^(-1) x = 1/(1+x^2)` . Thus, we use the substitution

`tan^(-1) x = t` so that ` (dx)/( 1+x^2) = dt`

Therefore , ` int (sin (tan^(-1) x) )/(1+x^2) dx = int sin t dt = - cos t +C = - cos (tan^(-1 )x) +C`

Now, we discuss some important integrals involving trigonometric functions and

their standard integrals using substitution technique. These will be used later without

reference.

Q 3115478360

Find the following integrals:

(i) ` int sin^3 x cos^2 x dx`

(ii) `int (sin x)/( sin (x+a) ) dx`

(iii) ` int 1/(1+ tan x) dx`

Class 12 Chapter 7 Example 6

(i) ` int sin^3 x cos^2 x dx`

(ii) `int (sin x)/( sin (x+a) ) dx`

(iii) ` int 1/(1+ tan x) dx`

Class 12 Chapter 7 Example 6

(i) We have

` int sin^3 x cos^2 x dx = int sin^2 x cos^2 x (sin x ) dx`

`= int (1- cos^2 x) cos^2 x (sin x ) dx`

Put t = cos x so that dt = – sin x dx

Therefore, ` ∫ sin^2 x cos^2 x (sin x) dx = − ∫ (1 – t^2 ) t^2 dt`

`= - ∫ (t^2 - t^4 ) dt = - ( t^3/3- t^5/5) +C`

`= -1/3 cos^3 x + 1/5 cos^5 x +C`

(ii) Put x + a = t. Then dx = dt. Therefore

`int (sin x )/( sin (x+a) ) dx = int (sin (t-a) )/(sin t) dt`

`= int ( sin t cos a- cos t sin a )/( sin t)dt `

`= cos a ∫dt – sin a ∫cot t dt`

`= (cos a) t – (sin a) [ log | sin t | + C_1 ]`

`= (cos a) (x + a) – (sin a) [ log | sin (x + a) | + C_1]`

`= x cos a + a cos a – (sin a) log | sin (x + a) | – C_1 sin a`

Hence, `int (sin x)/( sin (x+a) ) dx = x cos a - sin a log | sin (x+a) | +C` ,

where, `C = – C_1 sin a + a cos a`, is another arbitrary constant.

(iii) `int (dx)/( 1+ tan x) = int (cos x dx)/(cos x + sin x )`

`= 1/2 int ( ( cos x + sin x + cos x - sin x ) dx)/( cos x + sin x)`

`=1/2 int dx + 1/2 int ( cos x - sin x )/(cos x + sin x)dx`

`= x/2 + C_1/2 + 1/2 int (cos x - sin x)/(cos x + sin x) dx` ..........(1)

Now, consider `I = int (cos x - sin x)/(cos x + sin x) dx`

Put cos x + sin x = t so that (cos x – sin x) dx = dt

Therefore `I= int (dt)/t = log | t| +C_2 = log | cos x + sin x | +C_2`

Putting it in (1), we get

` int (dx)/(1+ tan x) = x/2 + C_1/2 +1/2 log | cos x + sin x | + C_2/2`

`= x/2 + 1/2 log | cos x + sin x | + C_1/2 +C_2/2`

` =x/2 +1/2 log | cos x + sin x | +C , (C =C_1/2 + C_2/2)`