Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.1 | Set 3

Evaluate the following definite integrals:

Question 45. 

Solution:

We have,

I = 

Let 2x + 1 = t², so we have,

=> 2 dx = 2t dt

=> dx = t dt

Now, the lower limit is, x = 1

=> t² = 2x + 1

=> t² = 2(1) + 1

=> t² = 3

=> t = √3 

Also, the upper limit is, x = 4

=> t² = 2x + 1

=> t² = 2(4) + 1

=> t² = 9

=> t = 3 

So, the equation becomes,

I = 

I = 

I = 

I = 

I = 

I = 1/4 [3⁵/5 – 3 – (√3)⁵/5 + √3]

I = 1/4[243/5 – 3 – 9√3/5 + √3]

I = 1/4((243 – 15 – 9√3 + 5√3)/5)

I = 1/4[(228 – 4√3)/5]

I = 1/4[4(57 – √3)/5]

I = (57 – √3)/5 

Therefore, the value of is (57 – √3)/5.

Question 46. 

Solution:

We have,

I = 

By using binomial theorem in the expansion of (1 – x)⁵, we get,

I = 

I = 

I = 

I = 

I = 1/2 – 5/3 + 10/4 – 10/5 + 5/6 – 1/7

I = 1/2 – 5/3 + 5/3 – 2 + 5/6 – 1/7

I = 1/2 – 2 + 5/6 – 1/7 

I = 1/42

Therefore, the value of  is 1/42.

Question 47. 

Solution:

We have,

I = 

I = 

I = 

I = 

By using integration by parts, we get,

I = 

I = 

I = e^x/x

So we get,

I = 

I = e²/2 – e¹/1

I = e²/2 – e

Therefore, the value of is e²/2 – e.

Question 48. 

Solution:

We have,

I = 

By using integration by parts in first integral, we get,

I = 

I = xe^2x/2 – (1/2)(e^2x/2) + 2/π[1 – 0]

I = xe^2x/2 – e^2x/4 + 2/π

So we get,

I = 

I = [e²/2 + e²/4 – 0 + 1/4] + 2/π

I = e²/4 + 1/4 + 2/π

Therefore, the value of  is e²/4 + 1/4 + 2/π.

Question 49. 

Solution:

We have,

I = 

By using integration by parts in first integral, we get,

I = 

I = 

I = 

I = 

So we get,

I = 

I = 

I = [e¹(1 – 1) – e⁰(0 – 1)] + 2√2/π

I = [0 – (-1)] + 2√2/π

I = 1 + 2√2/π

Therefore, the value of  is 1 + 2√2/π.

Question 50. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = -e^π cotπ/2 + e^π/2 cotπ/4

I = 0 + e^π/2(1)

I = e^π/2

Therefore, the value of is e^π/2.

Question 51. 

Solution:

We have,

I = 

I = 

I = 

I = 

By using integration by parts in first integral, we get,

I = 

I = 

I = 

I = 1/√2[sinπ(2e^π) – 0]

I = 1/√2[0 – 0]

I = 0

Therefore, the value of is 0.

Question 52. 

Solution:

We have,

I = 

By using integration by parts, we get,

I = e^xcos(x/2 + π/4) + 1/2∫e^xsin(x/2 + π/4)

I = e^x cos(x/2 + π/4) + 1/2[ e^xsin(x/2 + π/4) – 1/2 ∫e^xcos(x/2 + π/4)dx]

I = e^xcos(x/2 + π/4) + 1/2e^xsin(x/2 + π/4) – 1/4I

5I/4 = -3/ 2√2(e^2π + 1)

I = -3√2/5(e^2π + 1)

Therefore, the value of is -3√2/5(e^2π + 1).

Question 53. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 2/3[2^3/2 – 1] + 2/3[1 – 0]

I = 

I = 2^5/2/3 

Therefore, the value of is 2^5/2/3.

Question 54. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = -log3 + log2 + 2[log4 – log3]

I = -log3 + log2 + 2[2log2 – log3]

I = -log3 + log2 + 4log2 – 2log3

I = 5log2 – 3log3

I = log2⁵ – log3³

I = log32 – log27

I = log32/27 

Therefore, the value of is log32/27.

Question 55. 

Solution:

We have,

I = 

I = 

I = 

Let cos x = t, so we have,

=> – sin x dx = dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x =  π/2

=> t = cos x

=> t = cos π/2

=> t = 0

So, the equation becomes,

I = 

I = 

I = 

I = [0 – 1/3] – [0 – 1]

I = [-1/3] – [-1]

I = -1/3 + 1

I = 2/3

Therefore, the value of  is 2/3.

Question 56. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = -sinπ + sin0 

I = 0

Therefore, the value of  is 0.

Question 57. 

Solution:

 We have,

I = 

Let 2x = t, so we have,

=> 2x dx = dt

Now, the lower limit is, x = 1

=> t = 2x

=> t = 2(1)

=> t = 2

Also, the upper limit is, x =  2

=> t = 2x

=> t = 2(2)

=> t = 4

So, the equation becomes,

I = 

I = 

I = 

By using integration by parts in first integral, we get,

I = 

I = 

I = 

I = e⁴/4 – e²/2

Therefore, the value of  is e⁴/4 – e²/2.

Question 58. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = [sin^-1(1) – sin^-1(-1)]

I = π/2 – (-π/2)

I = π/2 + π/2

I =  π

Therefore, the value of  is π.

Question 59. If , find the value of k.

Solution:

We have,

=> 

=> 

=> 

=> 

=> tan^-12k/4 – tan^-10 = π/16

=> tan^-12k/4 – 0 = π/16

=> tan^-12k/4 = π/16

=> tan^-12k = π/4

=> 2k = tanπ/4

=> 2k = 1

=> k = 1/2

Therefore, the value of k is 1/2.

Question 60. If , find the value of k.

Solution:

We have,

=> 

=> 

=> 

=> 

=> a³ – 0 = 8

=> a³ = 8

=> a = 2

Therefore, the value of a is 2.

Question 61. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = -[√2cos3π/2 – √2cosπ]

I = -(-√2 – 0)

I = √2

Therefore, the value of  is √2.

Question 62. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = [-4cosπ/2 + 4cos0] + [4sinπ/2 – 4sin0]

I = 0 + 4 + 4 – 0

I = 8

Therefore, the value of  is 8.

Question 63. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = (π/8 – 1/4) – (3/4(π/8 – 1/4) – 1/16)

I = π/8 – 1/4 – (3π/32 – 3/16 – 1/16)

I =  π/8 – 1/4 – (3π/32 – 1/4)

I = π/8 – 1/4 – 3π/32 + 1/4

I = π/8 – 3π/32

I = (4π – 3π)/32

I = π/32 

Therefore, the value of  is π/32.

Question 64. 

Solution:

We have,

I = 

By using integration by parts we get,

I = 

I = 

I = 

I = 

I = 

So we get,

I = 

I = log3/2 – 1/8log3

I = 3/8log3

Therefore, the value of  is 3/8log3.

Question 65. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = [tanπ/3 – tanπ/6] + [-cotπ/3 + cotπ/6]

I = [√3 – 1/√3] + [- 1/√3 – √3]

I = 2[√3 – 1/√3]

I = 4/√3

Therefore, the value of  is 4/√3.

Question 66. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

Therefore, the value of  is .

Question 67. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = -log2/4 + log2/2 – 1/4 + 1/2

I = log2/4 + 1/4

Therefore, the value of  is log2/4 + 1/4.