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.

Summary

Exercise 19.2 Set 2 in Chapter 19 on Indefinite Integrals focuses on integrating rational functions where the numerator is linear (ax + b) and the denominator is quadratic (ax^2 + bx + c). This set of problems requires students to use the method of completing the square in the denominator, followed by partial fraction decomposition. The resulting fractions typically lead to a combination of logarithmic and inverse tangent (arctan) functions. This exercise aims to enhance students' skills in handling rational functions with irreducible quadratic denominators, reinforcing their understanding of completing the square, partial fraction decomposition, and integration techniques for different types of functions, including logarithmic and inverse trigonometric integrals.