Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.1 | Set 2

Evaluate the following definite integrals:

Question 23. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = [(a² + b²)/2][π/2]

I = π(a² + b²)/4

Therefore, the value of is π(a² + b²)/4.

Question 24. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 2[sinπ/4 - cosπ/4 - 0 + 1]

I = 2[1/√2 - 1/√2 - 0 + 1]

I = 2 (1)

I = 2

Therefore, the value of  is 2.

Question 25. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 2√2[sinπ/4 - sin0]

I = 2√2[1/√2- sin0]

I = 2√2[1/√2] 

I = 2

Therefore, the value of is 2.

Question 26. 

Solution:

We have,

I = 

By using integration by parts, we get,

I = x ∫sinxdx - ∫(∫sin x (1)dx)dx

I = -xcosx - ∫(∫sin xdx)dx

I = -xcosx + ∫cosxdx

I = -xcosx + sinx

So we get,

I = 

I = [-π/2cosπ/2 + sinπ/2 + 0 - 0]

I = 0 + 1 + 0 – 0 

I = 1

Therefore, the value of is 1.

Question 27. 

Solution:

We have,

I = 

By using integration by parts, we get,

I = x∫cosxdx - ∫(∫cos x (1)dx)dx

I = xsinx - ∫(∫cosxdx)dx

I = xsinx - ∫sinxdx

I = x sin x + cos x

So we get,

I = 

I = [π/2sinπ/2 + cosπ/2 - 0 - cos0]

I = π/2 + 0 - 0 - 1 

I = π/2 - 1  

Therefore, the value of is π/2 - 1.

Question 28. 

Solution:

We have,

I = 

By using integration by parts, we get,

I = x²sinx - ∫(2x∫(cosx)dx)dx

I = x²sinx - ∫(2xsinx)dx

I = x²sinx - 2[-xcosx - ∫(1∫sinxdx)dx]

I = x²sinx - 2[-xcosx + ∫sinxdx]

I = x²sinx - 2[-xcosx + sinx]

I = x²sinx + 2xcosx - 2sinx

So we get,

I = 

I = [(π/2)²sinπ/2 + 2(π/2)cosπ/2 - 2sinπ/2 - 0 - 0 + sin0]

I = [π²/4 + 0 - 2 - 0 - 0 + 0]

I = π²/4 - 2 

Therefore, the value of  is π²/4 - 2.

Question 29. 

Solution:

We have,

I = 

By using integration by parts, we get,

I = -x²cosx - ∫(2x∫sinxdx)dx

I = -x²cosx + ∫(2xcosx)dx

I = -x²cosx + 2[xsinx - ∫(∫cosxdx)dx]

I = -x²cosx + 2[xsinx - ∫sinxdx]

I = -x²cosx + 2[xsinx + cosx]

I = -x²cosx + 2xsinx + 2cosx

So we get,

I = 

I = -(π/4)²cosπ/4 + 2π/4sinπ/4 + 2cosπ/4 + 0 - 0 - 2

I = -π²/16(1/ √2) + π/2(1/√2) + 2(1/√2) + 0 - 0 - 2

I = -π²/16√2 + π/2√2 + √2 -  2

Therefore, the value of is -π²/16√2 + π/2√2 + √2 -  2.

Question 30. 

Solution:

We have,

I = 

By using integration by parts, we get,

I = 1/2x²sin2x - ∫(2x∫cos2xdx)dx

I = 1/2x²sin2x - ∫(xsin2x)dx

I = 1/2x²sin2x - [-1/2xcos2x - ∫(∫sin2xdx)dx]

I = 1/2x²sin2x - [-1/2xcos2x + ∫1/2 cos2xdx]

I = 1/2x²sin2x - [-1/2xcos2x + 1/4sin2xdx]

I = 1/2x²sin2x + 1/2xcos2x - 1/4sin2xdx

So we get,

I = 

I = [1/2(π²/4)sinπ + 1/2(π/2)cosπ - 0 - 0 - 0 + 0]

I = -π/4

Therefore, the value of is -π/4.

Question 31. 

Solution:

We have,

I = 

I = 

I = 

I = 

By using integration by parts, we get,

I = 1/2[x³/3] + x²sin2x/2 - [x ∫sin2x - ∫(∫sin2xdx)dx]

I = 1/2[x³/3] + x²sin2x/2 + xcosx/2 - sin2x/4

So we get,

I = 

I = [1/6[π³/8] + 0 + 0 - π/8] 

I = π³/48 - π/8

Therefore, the value of  is π³/48 - π/8.

Question 32. 

Solution:

We have,

I = 

By using integration by parts, we get,

I = 

I = xlogx - ∫1dx

I = xlogx - x

So we get,

I = 

I = 2log2 - 2 - log1 + 1

I = 2 log 2 – 1

Therefore, the value of is 2 log 2 – 1.

Question 33. 

Solution:

We have,

I = 

By using integration by parts, we get,

I = 

I = 

I = 

I = 

So we get,

I = 

I = -log3/4 + log3 - log4 + log1/2 - log1 + log2

I = log3(1 - 1/4) - 2log2 + 0 - 0 + log2

I = 3/4log3 - log2

Therefore, the value of is 3/4log3 - log2.

Question 34. 

Solution:

We have,

I = 

I = 

I = 

By using integration by parts, we get,

I = 

I = e^xlogx

So we get,

I = 

I = e^eloge - e¹log1

I = e^e (1) – 0

I = e^e

Therefore, the value of is e^e.

Question 35. 

Solution:

We have,

I = 

Let log x = t, so we have,

=> (1/x) dx = dt

Now, the lower limit is, x = 1

=> t = log x

=> t = log 1

=> t = 0

Also, the upper limit is, x = e

=> t = log x

=> t = log e

=> t = 1

So, the equation becomes,

I = 

I = 

I = 

I = 1/2 - 0/2

I = 1/2

Therefore, the value of is 1/2.

Question 36. 

Solution:

We have,

I = 

I = 

By using integration by parts, we get,

I = 

I = 

I = 

I = x/logx

So we get,

I = 

I = 

I = 

I = e²/2 - e

Therefore, the value of is e²/2 - e.

Question 37. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 1/2[3log2 - log4 + log3]

I = 1/2[3log2 - 2log2 + log3]

I = 1/2[log 2 - log 3]

I = 1/2[log6]

I = log6/2

Therefore, the value of is log6/2.

Question 38. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = [1/5log6 + 3/√5tan^-1(√5) - 1/5log1 - 3/√5tan^-1(0)]

I = [1/5 log6 + 3√5 tan^-1(√5) - 0 - 0]

I = 1/5 log6 + 3√5 tan^-1(√5)

Therefore, the value of  is 1/5 log6 + 3√5 tan^-1(√5).

Question 39. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

Let x – 1/2 = t, so we have,

=> dx = dt

Now, the lower limit is, x = 0

=> t = x – 1/2

=> t = 0 – 1/2

=> t = 1/2

Also, the upper limit is, x = 2

=> t = x – 1/2

=> t = 2 – 1/2

=> t = 3/2

So, the equation becomes,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

Therefore, the value of  is .

Question 40. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 4/2√7[tan^-1(5/√7) - tan^-1(1/√7)]

I = 2/√7[tan^-1(5/√7) - tan^-1(1/√7)]

Therefore, the value of is 2/√7[tan^-1(5/√7) - tan^-1(1/√7)].

Question 41. 

Solution:

We have,

I = 

Let x = sin² t, so we have,

=> dx = 2 sin t cos t dt

Now, the lower limit is, x = 0

=> sin² t = 0

=> sin t = 0

=> t = 0

Also, the upper limit is, x = 1

=> sin² t = 1

=> sin t = 1

=> t = π/2

So, the equation becomes,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 

I = 1/4[π/2 - 0] - 1/16[sin2π - 0]

I = 1/4[π/2] - 1/16[0 - 0 ]

I = π/8

Therefore, the value of is π/8.

Question 42. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = [sin^-1(1/2) - sin^-1(-1/2)]

I = π/6 -(-π/6)

I = π/6 + π/6

I = π/3

Therefore, the value of  is π/3.

Question 43. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

I = 

I = [sin^-1(2/2) - sin^-1(-2/2)]

I = sin^-11 - sin^-1(-1)

I = π/2 - (-π/2) 

I = π/2 + π/2 

I = π

Therefore, the value of  is π.

Question 44. 

Solution:

We have,

I = 

I = 

I = 

Let x + 1 = t, so we have,

=> dx = dt

Now, the lower limit is, x = –1

=> t = x + 1

=> t = – 1 + 1

=> t = 0

Also, the upper limit is, x = 1

=> t = x + 1

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I = 

I = 

I = 1/2tan^-12/2 - 1/2tan^-10/2

I = 1/2tan^-11 - 1/2tan^-10

I = 1/2(π/4) - 0

I = π/8

Therefore, the value of is π/8.

Summary

This exercise typically covers:

• Further applications of the fundamental theorem of calculus
• Evaluation of definite integrals with algebraic and trigonometric functions
• Use of substitution method in definite integrals
• Properties of definite integrals
• Integrals involving composite functions