Class 12 RD Sharma Solutions Chapter 19 Indefinite Integrals Exercise 19.18 | Set 2 (original) (raw)
Last Updated : 23 Jul, 2025
Class 12 RD Sharma Solutions for Chapter 19 on Indefinite Integrals, Exercise 19.18 | Set 2, focuses on advanced integration techniques. This exercise challenges students to apply various methods of integration, including substitution, integration by parts, and partial fractions. By working through these problems, students can enhance their problem-solving skills and deepen their understanding of calculus concepts.
Important Formulas and Concepts: Indefinite Integrals
**Integration by Substitution:
**∫ f(g(x)) * g'(x) dx = ∫ f(u) du, where u = g(x)
**Integration by Parts:
**∫ u dv = uv - ∫ v du
**Partial Fractions:
**Used when integrating rational functions. Decompose the fraction into simpler terms before integrating.
**Trigonometric Identities:
**sin²x + cos²x = 1
**tan²x + 1 = sec²x
**cot²x + 1 = csc²x
**Standard Integral Forms:
**∫ (1/x) dx = ln|x| + C
**∫ e^x dx = e^x + C
**∫ sin(x) dx = -cos(x) + C
**∫ cos(x) dx = sin(x) + C
**∫ tan(x) dx = ln|sec(x)| + C
Class 12 RD Sharma Mathematics Solutions -Exercise 19.18 | Set 2
Question 11. Evaluate ∫ sin2x/ √cos4x-sin2x+2 dx
**Solution:
Let us assume I =∫ sin2x/ √cos4x-sin2x+2 dx
=∫sin2x/ √cos4x-(1-cos2x)+2 dx (i)
Put cos2x = t
-2sinxcosx dx = dt
sin2x dx = -dt
Put the above value in eq. (i)
= -∫ dt/ √t2-(1-t)+2
= -∫ dt/ √t2-1+t+2
= -∫ dt/ √t2+t+1
= -∫ dt/ √t2+t+(1/4)+(3/4)
= -∫ dt/ √(t+1/2)2+ 3/4 (ii)
Put t+1/2 =u
dt = du
Put the above value in eq. (ii)
= -∫ du/ √u2+ 3/4
= -∫ du/ √u2+3/4
Integrate the above eq. then, we get
= -log|u +√u2+3/4| + c [since ∫ 1/√x2+a2 dx =log|x +√x2+a2| + c]
= -log|t+1/2 +√(t+1/2)2+3/4| + c
= -log|t+1/2 +√(t2+t+1)| + c
= -log|(cos2x+1/2) +√(cos4x+cos2x+1| + c
Hence, I = -log|(cos2x+1/2) +√(cos4x+cos2x+1| + c
Question 12. 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-(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(sinx/2) + c
Hence, I = sin-1(sinx/2) + c
Question 13. Evaluate ∫ 1/ x2/3√x2/3-4 dx
**Solution:
Let us assume I =∫ 1/ x2/3√x2/3-4 dx (i)
Put x1/3 = t
(1/3) x1/3-1 dx = dt
(1/3) x-2/3 dx = dt
dx/ x2/3 = 3dt
Put the above value in eq. (i)
= 3 ∫ dt/ √t2-(2)2
Integrate the above eq. then, we get
= 3 log|t +√t2-(2)2| + c [since ∫ 1/√x2-a2 dx =log|x +√x2-a2| + c]
= 3 log|x1/3 +√(x1/3)2-(2)2| + c
Hence, I = 3 log|x1/3 +√x2/3- 4| + c
Question 14. Evaluate ∫ 1/ √(1-x2)[9+(sin-1x)2 dx
**Solution:
Let us assume I =∫ 1/ √(1-x2)[9+(sin-1x)2 dx (i)
Put sin-1x = t
dx/√1-x2 = dt
Put the above value in eq. (i)
= ∫ dt/ √(3)2 + t
Integrate the above eq. then, we get
= log|t +√(3)2+t2| + c [since ∫ 1/√a2+x2 dx =log|x +√a2+x2| + c]
= log|sin-1x +√(3)2+(sin-1x)2| + c
Hence, I = log|sin-1x +√9+(sin-1x)2| + c
Question 15. Evaluate ∫ cosx/ √sin2x-2sinx-3 dx
**Solution:
Let us assume I =∫ cosx/ √sin2x-2sinx-3 dx (i)
Put sinx = t
cosx dx = dt
Put the above value in eq. (i)
= ∫ dt/ √t2-2t-3
= ∫ dt/ √t2-2t+(1)2-(1)2-3
= ∫ dt/ √(t-1)2-(2)3 (ii)
Put t-1 =u
dt = du
Put the above value in eq. (ii)
= ∫ du/ √u2-(2)2
Integrate the above eq. then, we get
= log|u +√u2-(2)2| + c [since ∫1/√x2-a2 dx =log|x +√x2-a2| + c]
= log|t-1 +√(t-1)2-4| + c
= log|t-1 +√t2-2t+1-4| + c
= log|t-1 +√t2-2t-3| + c
= log|sinx-1 +√sin2x-2sinx-3| + c
Hence, I = log|sinx-1 +√sin2x-2sinx-3| + c
Question 16. Evaluate ∫ √cosecx-1 dx
**Solution:
Let us assume I =∫ √cosecx-1 dx
= ∫ √1/sinx -1 dx
=∫ √1-sinx /sinx dx
=∫ √(1-sinx)+(1 + sinx) /(1+sinx)sinx dx
=∫ √(1+sinx-sinx-sin2x) /sin2x+sinx dx
=∫ √cos2x /sin2x+sinx dx
=∫ cosx /√sin2x+sinx dx (i)
Let sinx = t
cosx dx = dt
Put the above value in eq. (i)
= ∫ dt/√t2+t
= ∫ dt/√t2+2t(1/2)+(1/2)2-(1/2)2
= ∫ dt/√(t+1/2)2-(1/2)2 (ii)
Let t+1/2 = u
dt = du
Put the above value in eq. (ii)
= ∫ dt/√(u)2-(1/2)2
Integrate the above eq. then, we get
= log|u +√u2-(1/2)2| + c [since ∫ 1/√x2-a2 dx =log|x +√x2-a2| + c]
= log|t+1/2 +√(t+1/2)2-(1/2)2| + c
= log|t+1/2 +√t2+t| + c
Hence, I = log|sinx+1/2 +√sin2x+sinx| + c
Question 17. Evaluate ∫ sinx-cosx/ √sin2x dx
**Solution:
Let us assume I =∫ sinx-cosx/ √sin2x dx
∫ sinx-cosx/ √sin2x dx = ∫ sinx-cosx/ √(sinx+cosx)2 -1 dx
= ∫ sinx-cosx/ √(sinx+cosx)2 -1 dx (i)
Let sinx+cosx = t
cosx-sinx dx = dt
Put the above value in eq. (i)
= -∫ dt/ √t2-(1)2
Integrate the above eq. then, we get
= - log|t +√t2-(1)2| + c [since ∫ 1/√x2-a2 dx =log|x +√x2-a2| + c]
= - log|sinx+cosx +√sin2x| + c
Hence, I = - log|sinx+cosx +√sin2x| + c
Practice Questions on Indefinite Integrals
**Question 1. ∫ (x^2 + 1)/(x^3 + x) dx
**Question 2. ∫ x / (x^2 + 1) dx
**Question 3. ∫ dx / (x^2 - 4x + 5)
**Question 4. ∫ (x^2 - 1) / (x^2 + 1) dx
**Question 5. ∫ dx / (x^3 - 1)
Also Read,
Conclusion
1. Partial Fraction Decomposition: This is likely the main focus of Exercise 19.18. It's used to integrate rational functions by breaking them into simpler fractions.
2. Steps for Partial Fraction Decomposition:
a) Ensure the numerator's degree is less than the denominator's.
b) Factor the denominator.
c) Set up the partial fraction decomposition based on the factors.
d) Solve for the constants.
e) Integrate each resulting simple fraction.
3. Types of Factors:
- Linear factors: A/(x - a)
- Repeated linear factors: A/(x - a) + B/(x - a)^2 + ...
- Quadratic factors: (Ax + B)/(x^2 + px + q)
4. Integration of resulting fractions:
- ∫ 1/(x - a) dx = ln|x - a| + C
- ∫ 1/(x^2 + 1) dx = arctan(x) + C
- ∫ 1/(x^2 - 1) dx = (1/2)ln|(x-1)/(x+1)| + C