Mathematics Integrals of some more types and Definite Integral FOR CBSE-NCERT

Please Wait... While Loading Full Video

Class 12 Chapter 7 - Integrals

Integrals of some more types and Definite Integral FOR CBSE-NCERT

Topic covered

`star` Integrals of some more types

`star` Integrals of some more types

Integrals of some more types

Some special types of standard integrals:

(i) `color{green}{ ∫ sqrt (x^2 -a^2 ) dx = x/2 sqrt (x^2 -a^2) - a^2/2 log | x + sqrt (x^2 -a^2) | +C}`

`"Prove :"` Let `I = ∫ sqrt (x^2 - a^2 ) .1 dx`

Taking constant function 1 as the second function and integrating by parts, we have

`I = x sqrt (x^2 -a^2) - ∫ 1/2 (2x)/(sqrt (x^2- a^2) ) x dx`

`= x sqrt (x^2 - a^2) - ∫ x^2/(sqrt (x^2-a^2) ) dx = x sqrt (x^2-a^2) - ∫ (x^2 -a^2 +a^2 )/( sqrt (x^2 -a^2 ) ) dx`

`= x sqrt (x^2 -a^2) - ∫ sqrt (x^2 -a^2) dx- a^2 ∫ (dx)/( sqrt (x^2-a^2) )`

`= x sqrt (x^2 -a^2) -I - a^2 ∫ (dx)/(sqrt (x^2 -a^2 ) )`

or `2I = x sqrt (x^2 -a^2) - a^2 ∫ (dx)/(sqrt (x^2 -a^2 ) )`

or `I = ∫ sqrt (x^2 -a^2) dx = x/2 sqrt (x^2 -a^2) - a^2/2 log | x + sqrt (x^2 -a^2) | +C`

Similarly, integrating other two integrals by parts, taking constant function 1 as the second function, we get

(ii) `color{green}{ ∫ sqrt (x^2 + a^2 ) dx =1/2 x sqrt (x^2 +a^2 ) + a^2/2 log | x+ sqrt (x^2 +a^2) | +C}`

(ii) `color{green}{∫ sqrt (a^2 - x^2) dx = 1/2 x sqrt (a^2 -x^2) + a^2/2 sin^(-1) x/a +C}`

Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric substitution `color{orange}{(i) x= sectheta (ii), x = a tanθ (iii) and x = a sinθ}.`

Some special types of standard integrals:

(i) `color{green}{ ∫ sqrt (x^2 -a^2 ) dx = x/2 sqrt (x^2 -a^2) - a^2/2 log | x + sqrt (x^2 -a^2) | +C}`

`"Prove :"` Let `I = ∫ sqrt (x^2 - a^2 ) .1 dx`

Taking constant function 1 as the second function and integrating by parts, we have

`I = x sqrt (x^2 -a^2) - ∫ 1/2 (2x)/(sqrt (x^2- a^2) ) x dx`

`= x sqrt (x^2 - a^2) - ∫ x^2/(sqrt (x^2-a^2) ) dx = x sqrt (x^2-a^2) - ∫ (x^2 -a^2 +a^2 )/( sqrt (x^2 -a^2 ) ) dx`

`= x sqrt (x^2 -a^2) - ∫ sqrt (x^2 -a^2) dx- a^2 ∫ (dx)/( sqrt (x^2-a^2) )`

`= x sqrt (x^2 -a^2) -I - a^2 ∫ (dx)/(sqrt (x^2 -a^2 ) )`

or `2I = x sqrt (x^2 -a^2) - a^2 ∫ (dx)/(sqrt (x^2 -a^2 ) )`

or `I = ∫ sqrt (x^2 -a^2) dx = x/2 sqrt (x^2 -a^2) - a^2/2 log | x + sqrt (x^2 -a^2) | +C`

Similarly, integrating other two integrals by parts, taking constant function 1 as the second function, we get

(ii) `color{green}{ ∫ sqrt (x^2 + a^2 ) dx =1/2 x sqrt (x^2 +a^2 ) + a^2/2 log | x+ sqrt (x^2 +a^2) | +C}`

(ii) `color{green}{∫ sqrt (a^2 - x^2) dx = 1/2 x sqrt (a^2 -x^2) + a^2/2 sin^(-1) x/a +C}`

Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric substitution `color{orange}{(i) x= sectheta (ii), x = a tanθ (iii) and x = a sinθ}.`

Q 3115178960

Find `∫ sqrt (x^2 +2x + 5 ) dx`
Class 12 Chapter 7 Example 23

Solution:

Note that

` ∫ sqrt (x^2 +2x + 5 ) dx = int sqrt ( (x+1)^2 +4) dx`

Put x + 1 = y, so that dx = dy. Then

` ∫ sqrt (x^2 +2x +5) dx = ∫ sqrt (y^2 + 2^2)` dy

` =1/2 y sqrt (y^2 +4) +4/2 log | y+ sqrt (y^2 +4) | +C` [using 7.6.2 (ii)]

` =1/2 (x+1) sqrt (x^2 +2x+5) +2 log | x+1 + sqrt ( x^2 +2x +5) | +C`

Q 3135180062

Find ` ∫ sqrt ( 3-2x - x^2) dx`
Class 12 Chapter 7 Example 24

Solution:

Note that ` ∫ sqrt (3- 2x- x^2) dx = ∫ sqrt (4- (x+1)^2 ) dx`

Put x + 1 = y so that dx = dy.

Thus ` ∫ sqrt (3- 2x - x^2) dx = ∫ sqrt (4- y^2) dy`

` =1/2 y sqrt (4- y^2) +4/2 sin^(-1) y/2 +C` [using 7.6.2 (iii)]

` =1/2 (x+1) sqrt (3- 2x - x^2) +2 sin^(-1) ( (x+1)/2) +C`

SiteLock