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 + tan^2⁡x))dx

Solution:

We have, ∫ sin^-1⁡((2tan⁡x)/(1 + tan^2⁡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 a^x)/(2x²) dx

Solution:

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

= 1/2 ∫x³/x²dx – 3/2∫x²/x²dx + 5/2∫x/x²dx – 7/2∫x^-2dx + 1/2∫(x²a^x)/x²dx

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

= 1/2 [x²/2 – 3x + 5log⁡x + 7/x + a^x/(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

= ∫x^1/2dx + ∫ x^-1/2 dx

= 2/3 x^3/2 + 2x^1/2 + c

Hence, the primitive or anti-derivative of f(x) is 2/3 x^3/2 + 2x^1/2 + c.