Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.4

Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.4

Exercise 7.4 of NCERT Class 12 Mathematics Part II focuses on evaluating definite integrals. It involves applying standard integration techniques, including the substitution method, to solve integrals within given limits. The exercise helps reinforce the fundamental theorem of calculus and the concept of finding the area under a curve between two points.

Question 1: 3x²/(x⁶ + 1)

Solution:

Let x³ = t

So, 3x²dx = dt

=> ∫3² ∕ (x⁶ + 1) dx

= ∫dt ∕ (t² + 1)

= tan^-1 t + C

= tan^-1 (x³) + C

Question 2: 1 ∕ √(1+4x²)

Solution:

Let 2x = t

So, 2dx = dt

=> ∫ 1 ∕ √(1 + 4x² )dx

= 1 ∕ 2 [ log | t + √(t² + 1) | ] + C

= 1 ∕ 2 [ log | 2x + √(4x² + 1) | ] + C

Question 3: 1 / (√(2-x)² + 1)

Solution:

Let 2 - x = t

=> -dx = dt

=> ∫ 1 ∕ ( √(2 - x)² +1) dx

= -∫ 1 ∕ (√(t² + 1) dt

= -log | t + √(t² + 1) | + C

= -log |2 - x + √((2 - x)²+ 1) | + C

Question 4: 1 ∕ √(9 - 25x²)

Solution:

Let 5x = t

So, 5dx = dt

=> ∫ 1 ∕ √ (9 - 25x²) dx

= 1 ∕ 5 ∫ 1 ∕ √ (9 - t²) dt

= 1 ∕ 5 ∫ 1 ∕ √(3² - t²) dt

= 1 ∕ 5 sin^-1(t ∕ 3) + C

= 1 ∕ 5 sin^-1(5x ∕ 3) + C

Question 5: 3x ∕ (1 + 2x⁴)

Solution:

Let √2x² = t

So, 2√2x dx = dt

∫ 3x ∕ √ (1 + 2x⁴) dx

= 3∕ 2√2 ∫ dt ∕ (1+t²)

= 3∕ 2√2 [ tan^-1 t] + C

= 3∕ 2√2 [ tan^-1(√2 x²)] + C

Question 6: x² ∕ (1 - x⁶)

Solution:

Let x³ = t

So, 3x² dx = dt

∫ x² ∕ (1 - x⁶) dx

= 1 ∕ 3 ∫ dt ∕ (1 - t²)

= 1 ∕ 3 [1 ∕ 2 log |(1 + t) ∕ (1 - t)| ] + C

= 1 ∕ 6 log |(1 + x³) ∕ (1 - x³) | + C

Question 7: (x - 1) ∕ √ (x² - 1)

Solution:

∫ (x - 1) ∕ √ (x² - 1) dx

= ∫ (x ∕ √ (x² - 1) dx - ∫ 1 ∕ √ (x² - 1) dx------------- (i)

For ∫ x ∕ √ (x² - 1) dx , Let x²-1 = t => 2x dx = dt

∫ x ∕ √ (x² - 1) dx

= 1 ∕ 2 ∫ dt ∕ √t

= 1 ∕ 2 ∫ t^-1/2 dt

= 1 ∕ 2 [ 2t^1/2]

= √t

= √(x² - 1)

From (1), we obtain

∫ (x - 1) ∕ √ (x² - 1) dx

∫ (x ∕ √ (x² - 1) dx - ∫ 1 ∕ √ (x² - 1) dx

= √ (x²- 1) - log | x+ √(x²- 1)| + C

Question 8: x ∕ √ (x⁶+a⁶)

Solution:

Let x³ = t

So, 3x² dx = dt

∫ x² ∕ √(x⁶ + a⁶) dx

= ∫dt ∕ √ (t² + (a³)² )

= 1 ∕ 3 log |t + √(t² + a⁶) | + C

= 1 ∕ 3 log |x³ + √(x⁶ + a⁶) | + C

Question 9: sec²x ∕ √ (tan²x + 4)

Solution:

Let tan x = t

So, sec²x dx = dt

∫ sec²x dx ∕ √(tan²x + 4)

= ∫ dt ∕ √(t² + 2²)

= log |t + √(t² + 4) | + C

= log | tan x + √(tan²x + 4) | + C

Question 10: 1 ∕ √ (x²+2x+2)

Solution:

1 ∕ √ (x² + 2x + 2) = 1 ∕ √ ((x + 1)² + (1)²) dx

Let x + 1 = t

So, dx = dt

=> ∫ 1 ∕ √ (x² + 2x + 2) dx = ∫ 1 ∕ √(t² + 1) dt

= log | t + √(t² + 1) | + C

= log | (x + 1) + √((x + 1)² + 1) | + C

= log | (x + 1) + √(x² + 2x + 1) | + C

Question 11: 1 ∕ (9x² + 6x + 5)

Solution:

∫ 1 ∕ (9x²+6x+5) dx = ∫1 ∕ ((3x+1)² + (2)²) dx

Let (3x+1) = t

So, 3dx = dt

∫1 ∕ ((3x+1)² + (2)²) dx

= 1 ∕ 3 ∫ 1 ∕ (t²+2²) dt

= 1 ∕ 3 [ 1 ∕ 2 tan^-1❲t ∕ 2❳] + C

= 1 ∕ 6 tan^-1((3x+1) ∕ 2) + C

Question 12: 1 ∕ (√(7-6x-x²))

Solution:

7-6x-x² can be written as 7- (x²+6x+9-9)

Therefore

7- (x²+6x+9-9)

= 16 - (x²+6x+9)

= 16 - (x+3)²

= (4)² - (x+3)²

So, ∫ 1 ∕ (√(7-6x-x²) ) dx

= ∫ 1 ∕ √ ( (4)² - (x+3)² ) dx

Let x+3 = t

=> dx = dt

=> ∫ 1 ∕ √ ( (4)² - (x+3)² ) dx

=> ∫ 1 ∕ √ ( (4)² - (t)² ) dt

=> sin^-1(t ∕ 4) + C

=> sin^-1((x+3) ∕ 4) + C

Question 13: 1 ∕ √((x-1)(x-2))

Solution:

(x-1)(x-2) can be written as x²-3x+2

Therefore,

x²-3x+2

= x-3x + 9 ∕ 4 - 9 ∕ 4 + 2

= (x - 3 ∕ 2)² - 1 ∕ 4

= (x- 3 ∕ 2 )² - (1 ∕ 2)²

So, ∫ 1 ∕ √((x-3∕2)²-(1 ∕ 2)²) dx

Let x - 3 ∕ 2 = t

So, dx = dt

∫1 ∕ √((x-3∕2)²-(1 ∕ 2)²) dx

= ∫ 1 ∕ √ (t² - (1 ∕ 2)²) dt

= log | t + √(t² - (1 ∕ 2)²) + C

= log | (x- 3 ∕ 2) + √(x²-3x+2) + C

Question 14: 1 ∕ √ (8+3x-x²)

Solution:

8+3x-x² can be written as 8 - (x - 3x + 9 ∕ 4 - 9 ∕ 4)

Therefore,

8 - (x²-3x + 9 ∕ 4 - 9 ∕ 4)

= 41 ∕ 4 - (x-3 ∕ 2)²

=> ∫ 1 ∕ √ (8+3x-x²) dx = ∫ 1 ∕ √ (41 ∕ 4 - (x- 3 ∕ 2 )²) dx

Let x- 3 ∕ 2 = t

So, dx = dt

=> ∫ 1 ∕ √ (41 ∕ 4 - (x- 3 ∕ 2 )²) dx

= ∫ 1 ∕ √ ( (√41 ∕ 2)² - t²) dt

= sin^-1{t ∕ √41∕2} + C

= sin^-1{(x-3/2) ∕ √41∕2} + C

= sin^-1{(2x-3) ∕ √41} + C

Question 15: 1 ∕ √((x-a)(x-b))

Solution:

(x-a)(x-b) can be written as x² - (a+b)x +ab

Therefore,

x² - (a+b)x +ab

= x² - (a+b)x + (a+b) ∕ 4 - (a+b) ∕ 4 + ab

= [x - (a+b) ∕ 2 ]² - (a-b)² ∕ 4

= ∫ 1 ∕ √ (x-a)(x-b) dx

= ∫ 1 ∕ √{x - (a+b) ∕ 2}² - {(a-b)/2}² dx

= ∫ 1 ∕ √ [ t² - ({a-b}/2)²] dt

= log | t + √ [ t² -( {a-b}/2)²] | + C

= log | { x - (a+b)/2} + √(x-a)(x-b) | + C

Question 16: 1(4x+1) ∕ √(2x²+x-3)

Solution:

Let 4x + 1 = A d/dx (2x+x-3) + B

=> 4x +1 = A(4x+1) + B

=> 4x+1 = 4 Ax + A + B

Equating the coefficients of x and constant term on both sides, we obtain

4A = 4 => A = 1

A+B = 1 => B = 0

Let 2x² + x - 3 = t

So, (4x+1) dx = dt

=> ∫ (4x+1) ∕ √ (2x² +x-3) dx

= ∫ 1 ∕ √t dt

= 2√t + C

= 2 √2x+x-3 + C

Question 17: (x+2) ∕ √(x²-1)

Solution:

Let x+2 = A d/ dx (x²-1) + B ----------- (1)

=> x + 2 = A (2x) + B

Equating the coefficients of x and constant term on both sides, we obtain

2A = 1 => A = 1/2

B = 2

From (1), we obtain

(x+2) = 1/2 (2x) + 2

Then, ∫ (x+2) ∕ √ (x²-1) dx

= ∫ (1/2 (2x) + 2 ) / √ (x² - 1) dx

= 1/2 ∫ 2x ∕ √ (x² - 1) dx + ∫ 2 ∕ √(x-1) dx ----------(2)

In 1/2 ∫ 2x ∕ √ (x² - 1) dx ,

let x²-1 = t

So, 2x dx = dt

= 1/2 ∫ 2x ∕ √ (x² - 1) dx

= 1/2 ∫ dt ∕ √t

= 1/2 [2√t ]

= √t

= √(x²-1)

Then, ∫ 2 ∕ √(x²-1) dx = 2 ∫ 1 ∕ √(x²-1) dx

= 2 log | x + √ (x²-1) |

From equation (2), we obtain

∫ (x+2) ∕ √ (x²-1) dx = √ (x²-1)+ 2 log | x + √ (x²-1) | +C

Question 18: (5x-2) ∕ (1+2x+3x²)

Solution:

Let 5x-2 = A d ∕ dx (1+2x+3x²) + B

=> 5x-2 = A(2+6x) + B

Equating the coefficients of x and constant term on both sides, we obtain

5 = 6A => A = 5 ∕ 6

2A + = -2 => B = - 11 ∕ 3

So, 5x -2 = 5 ∕ 6(2+6x) + (-11 ∕ 3)

=> ∫ (5x-2) ∕ (1+2x+3x²) dx

=> ∫ {5∕6 (2+6x) - 11 ∕3} ∕ (1+2x+3x²) dx

=> 5 ∕ 6 ∫ (2+6x) ∕ (1+2x+3x²) dx - 11 ∕ 3 ∫ 1 ∕ (1+2x+3x²) dx

Let I₁ = ∫ (2+6x) ∕ (1+2x+3x²) dx and I₂ = ∫ 1 ∕ (1+2x+3x²) dx

∫ (5x-2) ∕ (1+2x+3x) dx = 5 ∕ 6 I₁ - 11 ∕ 3 I₂

I = ∫(2+6x) ∕ (1+2x+3x²) dx

Let 1+2x+3x² = t

=> (2+6x)dx = dt

I₁ = ∫ dt/ t

I₁ = log|t|

I₁ = log|1+2x+3x²|--------------(2)

I₂ = ∫ 1 ∕ (1+2x+3x²) dx

1+2x+3x² can by written as 1+3{x²+2 ∕ 3 x}

Therefore,

1+3{x²+2 ∕ 3 x}

= 1+3{x² +2∕3 x + 1 ∕ 9 - 1∕ 9 }

= 1 + 3{x+1 ∕3}² - 1 ∕ 3

= 2 ∕ 3 + 3 {x + 1 ∕ 3}²

= 3 [{x+1 ∕ 3}² + 2 ∕ 9 ]

= 3 [{x+1 ∕ 3}² + {√2 ∕ 3}² ]

I = 1 ∕ 3 ∫ 1 ∕ [(x+1 ∕ 3)² + (√2 ∕ 3)² ] dx

= 1 ∕ 3 [1 ∕ (√2 / 3) tan^-1{ (x+1/3) ∕ (√2/3)}]

= 1 ∕ 3 [ 3 ∕ √2 tan^-1{ (3x+1) ∕ √2} ]

= 1 ∕ √2 tan^-1{(3x+1) ∕ √2 } ---------------------(3)

Substituting equation (2) and (3) in equation (1), we obtain

∫ (5x-2) ∕ (1+2x+3x²) dx = 5 ∕ 6[ log|1+2x+3x2| ] - 11/3 [ 1 ∕ √2 tan-1{(3x+1) ∕ √2 } ] + C

= 5 ∕ 6[ log|1+2x+3x²| ] - 11/3√2 tan^-1{(3x+1) ∕ √2 } + C

Question 19: (6x+7) ∕ (√((x-5)(x-4))

Solution:

(6x+7) ∕ (√((x-5)(x-4)) = (6x+7) ∕ √ (x²-9x+20)

Let 6x+7 = A d∕ dx (x²-9x+20) + B

=> 6x+7 = A(2x-9)+B

Equating the coefficients of x and constant term on both sides, we obtain

2A = 6 => A = 3

-9A+B = 7 => B = 34

6x+7 = 3(2x-9)+34

∫ (6x+7) ∕ √(x²-9x+20) dx

= ∫ (3(2x-9)+34) ∕ √(x²-9x+20) dx

= 3 ∫ (2x-9) ∕ √(x²-9x+20) dx + 34 ∫ 1 ∕ √(x²-9x+20) dx

Let I₁ = ∫ (2x-9) ∕ √(x²-9x+20) dx and I₂ = ∫ 1 ∕ √(x²-9x+20) dx

∫ (6x+7) ∕ √(x²-9x+20) dx = 3I₁ + 34I_{2--------------- (1)}

I₁ = ∫ (2x-9) ∕ √(x²-9x+20) dx

Let x²-9x+20 = t

=> (2x-9)dx = dt

=> I₁ = dt / √t

I = 2√(x²-9x+20)-------------(2)

and I₂ = ∫ 1 ∕ √(x²-9x+20) dx

x²-9x+20 can be written s x²-9x+20 + 81/4 - 81/4

Therefore,

x²-9x+20 + 81/4 - 81/4

= { x-9/4}² - 1/4

= { x-9/4}2 - (1/2)²

^=> I₂ = ∫ 1 ∕ √((x-9/2)² - (1/2)²) dx

I = log |{ x-9/2}+ √(x²-9x+20) |---- (3)

Substituting equation (2) and (3) in equation (1), we obtain

∫ (6x+7) ∕ √(x²-9x+20) dx

= 3[ 2√(x²-9x+20) ] + 34 log |{ x-9/2}+ √(x²-9x+20) | + C

= 6√(x²-9x+20) + 34 log |{ x-9/2}+ √(x2-9x+20) | + C

Question 20: (x+2) ∕ √(4x-x²)

Solution:

Let x+2 = A d/dx(4x-x²) + B

=> x+2 = A(4-2x) + B

Equating the coefficients of x and constant term on both sides, we obtain

-2A = 1

=> A = -1/2

4A+B = 2 => B = 4

=> (x+2) = -1/2(4-2x)+4

∫ (x+2) ∕ √(4x-x²) dx = ∫ { -1/2 (4-2x)} ∕ √(4x-x²) dx

= -1/2 ∫ (4-2x) ∕ √(4x-x²) dx + 4 ∫ 1 ∕ √(4x-x²) dx

Let I₁ = ∫ (4-2x) ∕ √(4x-x²) dx and I₂= ∫1 ∕ √(4x-x²) dx

∫ (x+2)∕ √(4x-x²) dx = -1/2 I₁ + 4I_{2----------------------------------------(1)}

Then, I₁ = ∫(4-2x) ∕ √(4x-x²) dx

Let 4x-x² = t

=> (4-2x)dx = dt

=> I₁ = ∫ dt∕ √(t) = 2√t = 2√(4x-x²)------------------------------(2)

I₂ = ∫1 ∕ √(4x-x²) dx

=> 4x-x² = -(-4x+x²)

= (-4x+x²+4-4) = 4-(x-2²)

= (2)² - (x-2)²

So, I = I1 = ∫ 1 ∕ √{(2)² - (x-2)²} dx = sin^-1{ (x-2)/2}---------------------(3)

Using equation (2) and (3) in equation (1), we obtain

∫ (x+2) ∕ √(4x-x²) dx = -1/2 [2√(4x-x²) ] + 4sin^-1{ (x-2)/2} + C

= -√(4x-x²) + 4sin^-1{ (x-2)/2} + C

Question 21: (x+2) ∕ √(x²+2x+3)

Solution:

∫(x+2) ∕ √(x²+2x+3) dx

1 ∕ 2 ∫ 2(x+2) ∕ √(x²+2x+3) dx

1 ∕ 2 ∫ (2x+4) ∕ √(x²+2x+3) dx

1 ∕ 2 ∫ (2x+2) ∕ √(x²+2x+3) dx + 1 ∕ 2 ∫ 2 ∕ √(x²+2x+3) dx

1 ∕ 2 ∫ (2x+2) ∕ √(x²+2x+3) dx + ∫ 1 ∕ √(x²+2x+3) dx

Let I₁ = ∫ (2x+2) ∕ √(x²+2x+3) dx and I₂ = ∫ 1 ∕ √(x²+2x+3) dx

So, ∫(x+2) ∕ √(x²+2x+3) dx = 1 ∕ 2 I₁ + I_{2 -------------------(1)}

Then, I₁ = ∫ (2x+2) ∕ √(x²+2x+3) dx

Let x² +2x+3 = t

=> (2x+2) dx = dt

I₁ = ∫dt/ √t = 2√t = 2√(x²+2x+3)----------------(2)

I₂ = ∫ 1 ∕ √(x²+2x+3) dx

=> x²+2x+3 = x²+2x+1+2 = (x+1)² + (√2)²

I = ∫ 1 ∕ √{ (x+1)² + (√2)²} dx = log | (x+1) + √(x²+2x+3) |-------------- (3)

Using equation (2) and (3) in equation (1), we obtain

∫(x+2) ∕ √(x²+2x+3) dx = 1/2 [2√(x²+2x+3) ] + log | (x+1) + √(x²+2x+3) | + C

=> √(x²+2x+3)+ log | (x+1) + √(x²+2x+3) | + C

Question 22: (x+3) ∕ (x²-2x-5)

Solution:

Let (x+3) = A d/dx (x²-2x-5) + B

(x+3) = A(2x-2) + B

Equating the coefficients of x and constant term on both sides, we obtain

2A = 1 => A = 1/2

-2A+B = 3 => B = 4

(x+3) = 1/2 (2x-2)+4

=> ∫ (x+3) ∕ (x²-2x-5) dx = ∫ {/2 (2x-2) + 4 } ∕ (x²-2x-5) dx

=> 1/2∫(2x-2) ∕(x²-2x-5) dx +4 ∫ 1 ∕ (x²-2x-5) dx

=> I_{1 =} ∫(2x-2) ∕ (x²-2x-5) dx and I_{2 =} ∫ 1 ∕ (x²-2x-5) dx

∫ (x+3) ∕ (x²-2x-5) dx = 1/2 I₁ + 4I₂--------------- (1)

Then, I₁ = ∫(2x-2) ∕ (x²-2x-5) dx

Let x²-2x-5 = t

(2x-2)dx = dt

=> I₁ = ∫ dt ∕ t = log|t| = log | x²-2x-5 |----------------- (2)

I₂ = ∫ 1 ∕ (x²-2x-5) dx

= ∫ 1 ∕ {(x²-2x+1) -6} dx

= ∫ 1 ∕ {(x-1)² -(√6)²} dx

= 1 ∕ 2√6 log |{x-1-√6} ∕ {x-1+√6} |----------------- (3)

Substituting equation (2) and (3) in equation (1), we obtain

∫ (x+3) ∕ (x²-2x-5) dx = 1 ∕ 2 log | x²-2x-5 | + 4 ∕ 2√6 log |{x-1-√6} ∕ {x-1+√6} | + C

= 1 ∕ 2 log | x²-2x-5 | + 2 ∕ √6 log |{x-1-√6} ∕ {x-1+√6} | + C

Question 23: (5x+3) ∕√ (x²+4x+10)

Solution:

Let 5x+3 = A d/dx (x²+4x+10) + B

=> 5x+3 = A(2x+4) + B

Equating the coefficients of x and constant term on both sides, we obtain

2A = 5 => A = 5/2

4A+B = 3 => B = -7

5X+3 = 5/2 (2X+4) - 7

∫(5x+3) ∕√ (x²+4x+10) dx

=> ∫ {5/2 (2x+4-7)} ∕ √{x²+4x+10} dx

=> 5/2 ∫(2x+) ∕ √{x²+4x+10} dx - 7 ∫1/√{x²+4x+10} dx

Let I₁ = ∫(2x+) ∕ √{x²+4x+10} dx and I₂ = ∫1/ √{x²+4x+10} dx

∫(5x+3) ∕√ (x²+4x+10) dx = 5/2 I₁ - 7I₂-------------(1)

Then I₁ = ∫(2x+) ∕ √{x²+4x+10} dx

Let x²+4x+10 = t

So, (2x+4)dx = dt

=> I₁ = ∫ dt /√t = 2√(x²+4x+10) -----------------(2)

I₂ = ∫1/ √{x²+4x+10} dx

= ∫1/ √{(x²+4x+4)+6} dx

= ∫1/ √{(x+2)² +(√6)²} dx

= log | (x+2)√{x²+4x+10} |--------------------- (3)

Using equation (2) and (3) in equation (1), we obtain

∫ (5x+3) ∕√ (x²+4x+10) dx = 5/2 [ 2√(x2+4x+10) ] -7 log | (x+2)√{x²+4x+10} | + C

= 5 [ √(x2+4x+10) ] -7 log | (x+2)√{x²+4x+10} | + C

Question 24: ∫ dx ∕ (x²+2x+2) equals

(A) x tan^-1(x+1) + C

(B) tan^-1(x+1) + C

(C) (x+1) tan^-1x + C

(D) tan^-1x + C

Solution:

To solve the integral \int \frac{dx}{x^2+2x+2}, we can use the method of completing the square.

First, let's complete the square in the denominator:

x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1

Now, we can rewrite the integral as:

\int \frac{dx}{(x + 1)^2 + 1}

Next, we can make a substitution to simplify the integral. Let u = x + 1, then du = dx.

Substituting u = x + 1 and du = dx, we get:

\int \frac{du}{u^2 + 1}

Now, this is a standard integral:

\int \frac{du}{u^2 + 1} = \arctan(u) + C

Finally, substituting back u = x + 1 and adding the constant of integration C, we get the final result:

\int \frac{dx}{x^2+2x+2} = \arctan(x + 1) + C

So, the solution to the integral is arctan(x + 1) + C, where C is the constant of integration.

Correct Opition is B

Question 25: ∫ dx ∕ √(9x-4x²) equals

(A) 1/9 sin^-1{ (9x-8)/8} + C

(B) 1/2 sin^-1{ (8x-9)/9} + C

(C) 1/3 sin^-1{ (9x-8)/8} + C

(D) 1/2 sin^-1{ (9x-8)/8} + C

Solution:

Factor out -4 from the square root to simplify:

\int \frac{dx}{\sqrt{-4(x^2 - \frac{9}{4}x)}}

x^2 - \frac{9}{4}x = \left(x - \frac{9}{8}\right)^2 - \left(\frac{9}{8}\right)^2

= \frac{1}{2i} \int \frac{dx}{\sqrt{\left(x - \frac{9}{8}\right)^2 - \left(\frac{9}{8}\right)^2}}

Perform a trigonometric substitution: Let x - \frac{9}{8} = \frac{9}{8} \sin(\theta)

Substitute dx = \frac{9}{8} \cos(\theta) d\theta and simplify.

x - \frac{9}{8} = \frac{9}{8} \sin(\theta)

\frac{1}{2} \sin^{-1} \left( \frac{9x - 8}{8} \right) + C

Correct Solution: is B

• Class 12 NCERT Solutions- Mathematics Part (ii) – Chapter 7
• NCERT Solutions For Class 12 Maths Part ii – Chapter 7
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7