Class 12 RD Sharma Mathematics Solutions Chapter 19 Indefinite Integrals Exercise 19.13 | Set 1 (original) (raw)
Last Updated : 23 Jul, 2025
Chapter 19 of RD Sharma's Class 12 Mathematics textbook focuses on Indefinite Integrals, a fundamental concept in calculus. Exercise 19.13 | Set 1 specifically deals with integration techniques for functions involving √(ax + b). This set of problems helps students develop their skills in manipulating and integrating expressions with square roots, which is crucial for more advanced calculus applications.
Important Formulas and Concepts
**Basic Indefinite Integral: ∫ f(x) dx = F(x) + C, where F'(x) = f(x)
**Integration of √(ax + b):
**∫ √(ax + b) dx = (2/3a) * (ax + b)^(3/2) + C
**Substitution Method:
**Let u = ax + b, then du = a dx
**∫ √(ax + b) dx = (1/a) ∫ √u du
**Integration by Parts:
**∫ u dv = uv - ∫ v du
**Trigonometric Substitution: For √(ax + b), use x = (1/a)(tan²θ - b)
**Partial Fractions : Used when integrating rational functions
**Rationalization: Multiply numerator and denominator by √(ax + b) to simplify the integrand
Class 12 RD Sharma Mathematics Solutions -Exercise 19.13 | Set 1
Question 1. Evaluate ∫ x/ √x4+a4 dx
**Solution:
Let us assume I = ∫ x/ √x4+a4 dx
= ∫ x/ √(x2)2+(a2)2 dx (i)
Put x2 = t
2x dx = dt
x dx = dt/2
Put the above value in eq. (i)
= 1/2 ∫ dt/√t2 +(a2)2
Integrate the above eq. then, we get
= 1/2 log |t+ √t2+(a2)2| + c [since ∫ 1/√x2+a2 dx =log|x +√x2+a2| + c]
= 1/2 log |x2+ √(x2)2+(a2)2| + c
Hence, I = 1/2 log |x2+ √x4+a4| + c
Question 2. Evaluate ∫ sec2x/ √tan2x+4 dx
**Solution:
Let us assume I =∫ sec2x/ √tan2x+4 dx (i)
Put tanx = t
sec2x dx = dt
Put the above value in eq. (i)
= ∫ dt/ √t2+(2)2
Integrate the above eq. then, we get
= log|t +√t2+(2)2| + c [since ∫ 1/√x2+a2 dx =log|x +√x2+a2| + c]
= log|tanx +√tan2x+(2)2| + c
Hence, I = log|tanx +√tan2x+4| + c
Question 3. Evaluate ∫ ex/ √16-e2x dx
**Solution:
Let us assume I =∫ ex/ √16-e2x dx (i)
Put ex = t
ex dx = dt
Put the above value in eq. (i)
= ∫ dt/ √(4)2-(t)2
Integrate the above eq. then, we get
= sin-1(t/4) + c [ since ∫1/ √a2 - x2 dx = sin-1(x/a) + c]
= sin-1(ex/4) + c
Hence, I = sin-1(ex/4) + c
Question 4. Evaluate ∫ cosx/ √4+sin2x dx
**Solution:
Let us assume I =∫ cosx/ √4+sin2x dx (i)
Put sinx = t
cosx dx = dt
Put the above value in eq. (i)
= ∫ dt/ √(2)2+t2
Integrate the above eq. then, we get
= log|t +√(2)2+t2| + c [since ∫ 1/√x2+a2 dx =log|x +√x2+a2| + c]
= log|sinx +√(2)2+sin2x| + c
Hence, I = log|sinx +√4+sin2x| + c
Question 5. Evaluate ∫ sinx/ √4cos2x-1 dx
**Solution:
Let us assume I =∫ sinx/ √4cos2x-1 dx (i)
Put 2cosx = t
-2sinx dx = dt
sinx dx = -dt/2
Put the above value in eq. (i)
= -1/2 ∫ dt/ √t2-(1)2
Integrate the above eq. then, we get
= -1/2 log|t +√t2-(1)2| + c [since ∫ 1/√x2-a2 dx =log|x +√x2-a2| + c]
= -1/2 log|2cosx +√(2cosx)2-(1)2| + c
Hence, I = -1/2 log|2cosx +√4cos2x-1| + c
Question 6. Evaluate ∫ x/ √4-x4 dx
**Solution:
Let us assume I =∫ x/ √4-x4 dx (i)
Put x2 = t
2x dx = dt
x dx = dt/2
Put the above value in eq. (i)
=1/2 ∫ dt/ √(2)2-(t)2
Integrate the above eq. then, we get
= sin-1(t/2) + c [ since ∫1/ √a2 - x2 dx = sin-1(x/a) + c]
= sin-1(x2/2) + c
Hence, I = sin-1(x2/2) + c
Question 7. Evaluate ∫ 1/ x√4-9(logx)2 dx
**Solution:
Let us assume I =∫ 1/ x√4-9(logx)2 dx
=∫ 1/ x√4-(3logx)2 dx (i)
Put 3logx = t
3/x dx = dt
1/x dx = dt/3
Put the above value in eq. (i)
=1/3 ∫ dt/ √4-t2
=1/3 ∫ dt/ √(2)2-t2
Integrate the above eq. then, we get
=1/3 sin-1(t/2) + c [since ∫1/ √a2 - x2 dx = sin-1(x/a) + c]
=1/3 sin-1(3logx/2) + c
Hence, I =1/3 sin-1(3logx/2) + c
Question 8. Evaluate ∫ sin8x/ √9+sin44x dx
**Solution:
Let us assume I =∫ sin8x/ √9+sin44x dx (i)
Put sin24x = t
2sin4xcos4x (4)dx = dt
4sin8x dx = dt
sin8x dx = dt/4
Put the above value in eq. (i)
= 1/4 ∫ dt/ √9+t2
= 1/4 ∫ dt/ √(3)2+t2
Integrate the above eq. then, we get
= 1/4 log|t +√(3)2+t2| + c [since ∫ 1/√a2+x2 dx =log|x +√a2+x2| + c]
= 1/4 log|sin24x +√(3)2+sin44x| + c
Hence, I = 1/4 log|sin24x +√9+sin44x| + c
Question 9. Evaluate ∫ cos2x/ √sin22x+8 dx
**Solution:
Let us assume I =∫ cos2x/ √sin22x+8 dx (i)
Put sin2x = t
2cos2x dx = dt
cos2x dx = dt/2
Put the above value in eq. (i)
=1/2 ∫ dt/ √t2+8
=1/2 ∫ dt/ √t2+(2√2)2
Integrate the above eq. then, we get
= 1/2 log|t +√t2+(2√2)2| + c [since ∫ 1/√x2+a2 dx =log|x +√x2+a2| + c]
= 1/2 log|sin2x +√sin22x+(2√2)2| + c
Hence, I = 1/2 log|sin2x +√sin22x+8| + c
Question 10. Evaluate ∫ sin2x/ √sin4x+4sin2x-2 dx
**Solution:
Let us assume I =∫ sin2x/ √sin4x+4sin2x-2 dx (i)
Put sin2x = t
2sinxcosx dx = dt
sin2x dx = dt
Put the above value in eq. (i)
= ∫ dt/ √t2+4t-2
= ∫ dt/ √t2+2t(2)+(2)2-(2)2-2
= ∫ dt/ √(t+2)2-6 (ii)
Put t+2 =u
dt = du
Put the above value in eq. (ii)
= ∫ du/ √u2-6
= ∫ du/ √u2-(√6)2
Integrate the above eq. then, we get
= log|u +√u2-(√6)2| + c [since ∫ 1/√x2-a2 dx =log|x +√x2-a2| + c]
= log|t+2 +√(t+2)2-6| + c
= log|sin2x+2 +√(sin2x+2)2-6| + c
= log|sin2x+2 +√sin4x+4sin2x+4-6| + c
Hence, I = log|sin2x+2 +√sin4x+4sin2x-2| + c
Practice Questions on Indefinite Integrals
**Question 1. ∫ dx / (x - 3)
**Question 2. ∫ dx / (x² + 4)
**Question 3. ∫ (2x + 1) dx / (x² - 1)
**Question 4. ∫ dx / (x² - 2x + 2)
**Question 5. ∫ x dx / (x² + 9)
**Question 6. ∫ (x + 2) dx / (x² - x - 2)
**Question 7. ∫ dx / (x² + 2x + 5)
**Question 8. ∫ (3x - 1) dx / (x² + 4)
**Question 9. ∫ dx / (x³ - 1)
**Question 10. ∫ (x² + 1) dx / (x² - 4)
Also Read,
Conclusion
Exercise 19.13 Set 1 in RD Sharma's Class 12 Chapter on Indefinite Integrals focuses on integrating rational functions. The key techniques used in this exercise include:
**1. Partial fraction decomposition
**2. Completing the square for quadratic denominators
**3. Recognizing standard integral forms
**4. Using substitution method
The problems typically involve transforming the given integral into a simpler form or a combination of standard forms that can be solved using known formulas.