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• RD Sharma Class 12 Solutions for Maths

# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.2 | Set 2

### Question 25. Evaluate ∫(tan⁡x + cot⁡x)2 dx

Solution:

We have, ∫(tan⁡x + cot⁡x)2 dx

By using formula (x + y)2 = x2 + y2 + 2xy

We get, ∫(tan2x + cot2⁡x + 2tan⁡x cot⁡x)dx

= ∫ (sec2⁡x – 1 + cosec2x – 1 + ((2 × 1)/cot⁡x) × cot⁡x)dx

= ∫ (sec2⁡x + cosec2⁡x)dx

= ∫sec2xdx + ∫cosec2⁡xdx

= tan⁡x – cot⁡x + c

### Question 26. Evaluate ∫(1 – cos⁡2x)/(1 + cos⁡2x) dx

Solution:

We have, ∫(1 – cos⁡2x)/(1 + cos⁡2x) dx

= ∫(2sin2⁡x)/(2cos2⁡x) dx

= ∫tan2xdx

= ∫(sec2x – 1)dx

= ∫sec2⁡xdx – 1∫dx

= tan⁡x – x + c

### Question 27. Evaluate ∫(cos⁡x)/(1 – cos⁡x) dx

Solution:

We have, ∫(cos⁡x)/(1 – cos⁡x) dx

= ∫(cos⁡x(1 + cos⁡x))/((1 – cos⁡x)(1 + cos⁡x)) dx

= ∫(cos⁡x + cos2⁡x)/(1 – cos2x) dx

= ∫(cos⁡x + cos2⁡x)/(sin2⁡x) dx

= ∫(cos⁡x)/(sin2⁡x) dx + ∫(cos2x)/(sin2⁡x) dx            [Since, cosx/sinx = cotx]

= ∫cot⁡x × cosec⁡xdx + ∫(cosec2⁡x – 1)dx                 [Since, cot2x = cosec2x – 1]

= -cosec⁡x – cot⁡x – x + c

### Question 28. Evaluate ∫cos2x – sin2⁡x/√(1 + cos⁡4x) dx

Solution:

We have, ∫cos2x – sin2⁡x/√(1 + cos⁡4x) dx

= ∫(cos2⁡x – sin2x)/√(2cos2⁡2x) dx

= 1/√2 ∫(cos2x – sin2⁡x)/(cos⁡2x) dx

= 1/√2∣(cos2x – sin2⁡x)/(cos2⁡x – sin2⁡x) dx

= 1/√2∫1 × dx

= x/√2 + c

### Question 29. Evaluate ∫ 1/(1 – cos⁡x) dx

Solution:

We have, ∫ 1/(1 – cos⁡x) dx

= ∫1/(1 – cos⁡x) × (1 + cos⁡x)/(1 + cos⁡x) × dx

= ∫(1 + cos⁡x)/(1 – cos2x) × dx

= ∫(1 + cos⁡x)/(sin2x) × dx

= ∫1/(sin2x) dx + ∫(cos⁡x)/(sin22⁡x) dx

= ∫cosec2xdx + ∫cot⁡x × cosec⁡x dx

= -cot⁡x – cosec⁡x + c

### Question 30. Evaluate ∫1/(1 – sin⁡x) dx

Solution:

We have, ∫1/(1 – sin⁡x) dx

= ∫1/(1 – sin⁡x) × (1 + sin⁡x)/(1 + sin⁡x) × dx

= ∫(1 + sin⁡x)/(1 – sin2⁡x) × dx

= ∫(1 + sin⁡x)/(cos2⁡x) × dx

= ∫(1/(cos2x) + (sin⁡x)/(cos2⁡x)) × dx

= ∫1/(cos2⁡x) dx + ∫(sin⁡x)/(cos2⁡x) × dx

= ∫sec2⁡xdx + ∫tan⁡x sec⁡x dx

= tan⁡x + sec⁡x + c

### Question 31. Evaluate ∫(tan⁡x)/(sec⁡x + tan⁡x) dx

Solution:

We have, ∫(tan⁡x)/(sec⁡x + tan⁡x) dx

= ∫(tan⁡x)/(sec⁡x + tan⁡x) × (sec⁡x – tan⁡x)/(sec⁡x – tan⁡x) × dx

= ∫(tan⁡x(sec⁡x – tan⁡x))/(sec2⁡x – tan2⁡x) × dx

= ∫(tan⁡xsec⁡x – tan2⁡x)dx

= ∫sec⁡tan⁡xdx – ∫(sec2x – 1)dx

= ∫sec⁡xtan⁡xdx – ∫sec2⁡xdx + 1∫dx

= sec⁡x – tan⁡x + x + c

### Question 32. Evaluate ∫(cosec⁡x)/(cosec⁡x – cot⁡x)dx

Solution:

We have, ∫(cosec⁡x)/(cosec⁡x – cot⁡x)dx

= ∫(cosec⁡x)/(cosec⁡x – cot⁡x) × (cosec⁡x + cot⁡x)/(cosec⁡x + cot⁡x) × dx

= ∫(cosec⁡x(cosec⁡x + cot⁡x))/(cosec2⁡x – cot2x) × dx

= ∫(cosec2⁡x + cosec⁡x cot⁡x)dx

= ∫cos⁡ec2xdx + ∫cosec⁡x cotx dx

= -cot⁡x – cosec⁡x + c

### Question 33. Evaluate ∫1/(1 + cos⁡2x) dx

Solution:

We have, ∫1/(1 + cos⁡2x) dx

= ∫ 1/(2cos2⁡x) × dx

= 1/2 ∫sec2⁡x × dx

= 1/2 × tan⁡x + c

= (tan⁡x)/2 + c

### Question 34. Evaluate∫1/(1 – cos⁡2x) dx

Solution:

We have, ∫1/(1 – cos⁡2x) dx

= ∫1/(2sin2⁡x)dx

= 1/2 ∫cosec2⁡x dx

= (-1)/2 × cot⁡x + c

= (-cot⁡x)/2 + c

### Question 35. Evaluate ∫tan-1⁡[(sin⁡2x)/(1 + cos⁡2x)]dx

Solution:

We have, ∫tan-1⁡[(sin⁡2x)/(1 + cos⁡2x)]dx

= ∫tan-1[(2sin⁡xcos⁡x)/(2cos2⁡x)]dx

= ∫tan-1⁡[(sin⁡x)/(cos⁡x)]dx

= ∫tan-1(tan⁡x)dx

= ∫xdx

= x2/2 + c

### Question 36. Evaluate ∫cos-1(sin⁡x)dx

Solution:

We have, ∫cos-1(sin⁡x)dx

= ∫cos-1⁡[cos⁡(π/2 – x)]dx

= ∫(π/2 – x)dx

= π/2 ∫dx – ∫xdx

= π/2 × x – x2/2 + c

### Question 37. Evaluate ∫ cot-1⁡(sin⁡x)dx

Solution:

We have, ∫ cot-1⁡(sin⁡x)dx

= ∫cot-1⁡[(sin⁡2x)/(1 – cos⁡2x)]dx

= ∫cot-1⁡((cos⁡x)/(sin⁡x))dx

= ∫cot-1⁡(cotx)dx

= ∫xdx

= x2/2 + c

### Question 38. Evaluate ∫ sin-1⁡((2tan⁡x)/(1 + tan2⁡x))dx

Solution:

We have, ∫ sin-1⁡((2tan⁡x)/(1 + tan2⁡x))dx

= ∫ sin-1⁡(sin⁡2x)dx

= ∫2xdx

= 2∫xdx

= (2x2)/2 + c

= x2 + c

### Question 39. Evaluate ∫((x3 + 8)(x – 1))/(x2 – 2x + 4) dx

Solution:

We have, ∫((x3 + 8)(x – 1))/(x2 – 2x + 4) dx

= ∫((x + 2)(x2 – 2x + 4)(x – 1))/(x2 – 2x + 4) dx

= ∫(x + 2)(x – 1)dx

= ∫(x2 – x+2x – 2)dx

= ∫(x2 + x – 2)dx

= x3/3 + x2/2 – 2x + c

### Question 40. Evaluate ∫(atan⁡x + bcot⁡x)2 dx

Solution:

We have, ∫(atan⁡x + bcot⁡x)2 dx

By using formula (x + y)2 = x2 + y2 + 2xy , we get

= ∫(a2 tan2⁡x + b2cot2x + 2ab tan⁡x cot⁡x)dx

= ∫[a2 (sec2⁡x – 1) + b2(cosec2x – 1) + 2ab]dx

= ∫[a2 sec2x – a2 + b2cosec2⁡x – b2 + 2ab]dx

= a2tan⁡x – a2x – b2 cot⁡x – b2x + 2abx + c

= a2tan⁡x – b2 cot⁡x – (a2 + b2 – 2ab)x + c

### Question 41. Evaluate ∫(x3 – 3x2 + 5x – 7 + x2 ax)/(2x2) dx

Solution:

We have, ∫(x3 – 3x2 + 5x – 7 + x2 ax)/(2x2) dx

= 1/2 ∫x3/x2dx – 3/2∫x2/x2dx + 5/2∫x/x2dx – 7/2∫x-2dx + 1/2∫(x2ax)/x2dx

= 1/2 × x2/2 – 3/2x + 5/2 log⁡x – 7/2 x-1 + 1/2ax/(log⁡a) + c

= 1/2 [x2/2 – 3x + 5log⁡x + 7/x + ax/(log⁡a)] + c

### Question 42. Evaluate ∫cos⁡x/(1 + cos⁡x) dx

Solution:

We have, ∫cos⁡x/(1 + cos⁡x) dx  …..(1)

Now solve

Since, cos⁡x = cos2x/2 – sin2⁡x/2 and cos⁡x + 1 = 2cos2⁡x/2

So, we get cos⁡x/(1 + cos⁡x) = 1/2[1 – tan2x/2]

Now put this value in eq(1), we get

= 1/2 ∫(1 – tan2x/2)dx

= 1/2 ∫(1 – sec2⁡x/2 + 1)dx

= 1/2 ∫(2 – sec2⁡x/2)dx

= 1/2 [2x – (tan⁡x/2)/(1/2)] + c

= x – tan⁡x/2 + c

### Question 43. Evaluate∫(1 – cos⁡x)/(1 + cos⁡x) dx

Solution:

We have, ∫(1 – cos⁡x)/(1 + cos⁡x) dx  ….(1)

Now solve

(1 – cos⁡x)/(1 + cos⁡x) = (2sin2⁡x)/(2cos2⁡x)

= tan2⁡x/2

= (sec2x/2 – 1)               [Since, 2sin2⁡x/2 = 1 – cos⁡x and 2cos2⁡x/2 = 1 + cos⁡x]

Now put this value in eq(1), we get

= ∫(sec2⁡x/2 – 1)dx

= tan(x/2)/(1/2) – x + c

= 2tan⁡x/2 – x + c

### Question 44. Evaluate ∫{3sin⁡x – 4cos⁡x + 5/(cos2x) – 6/(sin2⁡x) + tan2⁡x – cot2⁡x}dx

Solution:

We have, ∫{3sin⁡x – 4cos⁡x + 5/(cos2x) – 6/(sin2⁡x) + tan2⁡x – cot2⁡x}dx

= 3∫sin⁡xdx – 4∫cos⁡xdx + 5∫sec2⁡dx – 6∫cosec2⁡x + ∫tan2⁡xdx – ∫cot2⁡xdx

= 3∫sin⁡xdx – 4∫cos⁡xdx + 5∫sec2⁡xdx – 6∫cosec2⁡x + ∫(sec2⁡x – 1)dx – ∫(cosec2⁡x – 1)dx

= 3∫sin⁡xdx – 4∫cos⁡xdx + 6∫sec2xdx – 7∫cosec2xdx

= -3cos⁡x – 4sin⁡x + 6tan⁡x + 7cot⁡x + c

### Question 45. If f'(x) = x – 1/x2 and f(1) = 1/2, find f(x)?

Solution:

Given that ∫f'(x) = x – 1/x2

and f(1) = 1/2

We have to find f(x)

So, ∫f'(x) = ∫xdx – ∫1/x2dx

f(x) = x2/2 + x-1 + c

f(x) = x2/2 + 1/x + c

f(x) = x2/2 + 1/x + c     …..(i)

As we know that

f(1) = 1/2

12/2 + 1/1 + c = 1/2

1/2 + 1 + c = 1/2

c = -1

On putting c = -1 in (i), we get

f(x) = x2/2 + 1/x – 1

### Question 46. If f'(x) = x + b, f(1) = 5, f(2) = 13, find f(x)?

Solution:

Given that f'(x) = x + b

and f(1) = 5, f(2) = 13

We have to find f(x)

So, ∫f'(x) = ∫(x + b)dx

f(x) = x2/2 + bx + c   …….(i)

As we know that

f(1) = 5

12/2 + b × 1 + c = 5

1/2 + b + c = 5

b + c = 9/2    …….(ii)

Also, f(2) = 13

22/2 + b × 2 + c = 13

2 + 2b + c = 13

2b + c = 11     …….(iii)

Now, subtract eq(ii) from eq(iii), we get

b = 11 – 9/2

b = 13/2

Now, put b = 13/2 in eq(ii), we get

13/2 + c = 9/2

c = 9/2 – 13/2

c = (9 – 13)/2

= (-4)/2

= -2

Now, on putting b = 13/2 and c = -2 in equation (i), we get

f(x) = x2/x + 13/2x – 2

f(x) = x2/2 + 13/2x – 2

### Question 47. If f'(x) = 8x3 – 2x, f(2) = 8, find f(x)?

Solution:

Given that f'(x) = 8x3 – 2x

and f(2) = 8

We have to find f(x)

So, ∫f'(x)dx = ∫(8x3 – 2x)dx

f(x) = ∫(8x3 – 2x)dx

= ∫8x3dx – ∫2xdx

= (8x4)/4 – (2x2)/2 + c

= 2x4 – x2 + c

f(x) = 2x4 – x2 + c  ……….(i)

As we know that f(2) = 8

So, f(2) = 2(2)4 – (2)2 + c = 8

32 – 4 + c = 8

28 + c = 8

c = -20

Now, Put c = -20 in eq(i), we get

f(x) = 2x4 – x2 – 20

### Question 48. If f'(x) = asin⁡x + bcos⁡x and f'(0) = 4, f(0) = 3, f(π/2) = 5, find f(x)?

Solution:

Given that, f'(x) = asin⁡x + bcos⁡x

and f'(0) = 4, f(0) = 3, f(π/2) = 5

We have to find f(x)

So,

∫f'(x) = ∫(asin⁡x + bcos⁡x)dx

f(x) = -acos⁡x + bsin⁡x + c

f(x) = -acos⁡x + bsin⁡x + c ………(i)

As we know that f'(0) = 4

So, f'(0) = asin⁡0 + bcos⁡0 = 4

a × 0 + b × 1 = 4

b = 4

Also, f(0) = 3

f(0) = -acos⁡0 + bsin⁡0 + c = 3

-a + 0 + c = 3

c – a = 3                  ……..(ii)

Also, f(π/2) = 5

f(π/2) = -acos⁡(π/2) + bsin⁡(π/2) + c = 5

-a × 0 + b × 1 + c = 5

b + c = 5

4 + c = 5                    [Since, b = 4]

c = 5 – 4

c = 1

Now, put c = 1 in eq(ii), we get 1 – a = 3

-a = 3 – 1

-a = 2

a = -2

Now, put a = -2, b = 4, and c = 1 in eq(i), we get

f(x) = -(-2)cos⁡x + 4sin⁡x + 1

f(x) = 2cos⁡x + 4sin⁡x + 1

### Question 49. Write the primitive or anti-derivative of f(x) = √x + 1/√x.

Solution:

We have, f(x) = √x + 1/√x

∫f(x) = ∫(√x + 1/√x)dx

= ∫x1/2dx + ∫ x-1/2 dx

= 2/3 x3/2 + 2x1/2 + c

Hence, the primitive or anti-derivative of f(x) is 2/3 x3/2 + 2x1/2 + c.