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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 = exlogx

So we get,

I = 

I = eeloge – e1log1

I = ee (1) – 0

I = ee

Therefore, the value of is ee.

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.


Article Tags :
Class 12
Mathematics
RD Sharma Solutions
School Learning
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