Class 12 RD Sharma Solutions Chapter 19 Indefinite Integrals Exercise 19.15 (original) (raw)
Last Updated : 23 Jul, 2025
Indefinite integrals are a crucial concept in calculus, representing the antiderivative of a function. They play a vital role in finding the area under curves, volumes of solids, and solving various problems in physics, engineering, and economics. In this exercise, we will focus on evaluating indefinite integrals using various techniques such as integration by substitution, integration by parts, and integration by partial fractions.
Important Formulas and Concepts: Indefinite Integrals
**Integration Techniques
- Integration by Substitution: ∫f(g(x))g'(x) dx = ∫f(u) du, where u = g(x)
- Integration by Parts: ∫u dv = uv - ∫v du
- Integration by Partial Fractions: ∫(px+q)/(x^2+ax+b) dx = ∫(A/(x+c) + B/(x+d)) dx
**Standard Integrals
- ∫e^x dx = e^x + C
- ∫sin(x) dx = -cos(x) + C
- ∫cos(x) dx = sin(x) + C
Class 12 RD Sharma Mathematics Solutions - Exercise 19.15
Question 1. Evaluate ∫ 1/(4x2 + 12x + 5) dx
**Solution:
Let I = ∫ 1/(4x2 + 12x + 5) dx
by taking 1/4 common from the above eq
= 1/4 ∫ 1/ x2 + 3x + 5/4 dx
= 1/4 ∫ 1/ x2 + 2x × (3/2)x + (3/2)2 - (3/2)2 + 5/4 dx
= 1/4 ∫ 1/ (x + 3/2)2 - 1 dx (i)
put (x+ 3/2) = t
dx = dt
put the above value in eq. (i)
= 1/4 ∫ 1/ t2 - (1)2 dt
Integrate the above eq. then, we get
= 1/4 × 1/2×(1) log |t-1/t+1| +c [since, ∫1/x2 - a2 dx = 1/2a log|x-a/x+a| +c]
put the value of t in the above eq.
= 1/8 log|x+ 3/2 - 1/x+ 3/2 + 1| + c
Hence, I = 1/8 log|2x+1/ 2x+5| + c
Question 2. Evaluate ∫1/x2 - 10x + 34 dx
**Solution:
Let I = ∫1/x2 - 10x + 34 dx
=∫1/x2 - 2x × 5 + (5)2 - (5)2 + 34 dx
=∫1/ (x - 5)2 + 9 dx (i)
substituting (x-1) = t
dx = dt
put the above value in eq. (i)
= ∫ 1/ t2 + (3)2 dt
Integrate the above eq. then, we get
= 1/3 tan-1 (t/3) + c [Since, ∫ 1/x2 + a2 dx = 1/a tan-1 (x/2) + c]
Put the value of t in the above eq.
Hence, I = 1/3 tan-1 (x-5/ 3) + c
Question 3. Evaluate ∫ 1/ 1-x-x2 dx
**Solution:
Let I = ∫ 1/ 1-x-x2 dx
= ∫ 1/ -(x2 - x - 1) dx
adding and subtracting 1/4 in the denominator to make it a perfect square
= ∫ 1/ -(x2 - x + 1/4 - 1 - 1/4) dx
= ∫ 1/ -([x2 - x + 1/4] - 1 - 1/4) dx
= ∫ 1/ -([x - 1/2]2 - 5/4) dx
= ∫ 1/ (5/4 - [x - 1/2]2) dx
= ∫ 1/ ([√5/2]2 - [x - 1/2]2) dx
Integrate the above eq. then, we get
= 1/2(√5/2) log|√5/2 + (x-1/2)/ √5/2 - (x-1/2)| + c [since ∫ 1/a2 + x2 dx = 1/2a log|x+a/x-a| +c]
Hence, I = 1/√5 log|√5/2 + (x-1/2) /√5/2 - (x-1/2)| + c
Question 4. Evaluate ∫ 1/2x2 - x - 1 dx
**Solution:
Let I = ∫ 1/2x2 - x - 1 dx
taking 1/2 common from the above eq.
=1/2 ∫ 1/ x2 - x/2 - 1/2 dx
=1/2 ∫ 1/ x2 - 2x × 1/4 + (1/4)2 - (1/4)2 - 1/2 dx
= 1/2 ∫ 1/ (x - 1/4)2 - 9/16 dx
put, x- 1/4 = t
dx = dt
= 1/2 ∫ 1/ t2 - (3/4)2 dt
Integrate the above eq. then, we get
= (1/2) 1/[2×(3/4)] log|t-(3/4) / t+(3/4)| + c [Since, ∫ 1/x2 -a2 dx = 1/2a log|x - a/ x+a| + c]
Put the value of t in above eq.
= 1/3 log|(x-1/4-3/4)/x-1/4+3/4| + c
Hence, I = 1/3log|x - 1/2x+1| + c
Question 5. Evaluate ∫ dx/x2 + 6x +13
**Solution:
Let I =∫ dx/x2 + 6x +13 (i)
We have x2 + 6x +13 = x2 + 6x + 32 - 32 +13 =(x + 3)2 + 4
Put the above value in eq. (i)
∫ 1/x2 + 6x +13 dx = ∫ 1/(x+3)2 + 22 dx
put x+3 = t and
dx = dt
= ∫ dt/ t2 + 22
Integrate the above eq. then, we get
= 1/2 tan-1 t/2 + c
Put the value of t in above eq.
Hence, I = 1/2 tan-1 x+3/2 + c
Practice Questions on Indefinite Integrals
**Question 1. ∫ √(x^2 + 1) dx
**Question 2. ∫ dx / √(x^2 - 1)
**Question 3. ∫ dx / √(1 - x^2)
**Question 4. ∫ x√(x^2 + 4) dx
**Question 5. ∫ x^2 / √(x^2 + 1) dx
**Read More:
Conclusion
1. Trigonometric Substitutions: Exercise 19.15 likely focuses on integrals that can be solved using trigonometric substitutions.
2. Common Substitutions:
- For √(a^2 - x^2), use x = a sin θ
- For √(a^2 + x^2), use x = a tan θ
- For √(x^2 - a^2), use x = a sec θ
3. Steps for Trigonometric Substitution:
a) Identify the appropriate substitution
b) Express dx in terms of θ
c) Replace the radical term using trigonometric identities
d) Integrate the resulting trigonometric function
e) Substitute back to express the answer in terms of x
4. Important Trigonometric Identities:
- sin^2 θ + cos^2 θ = 1
- tan^2 θ + 1 = sec^2 θ
- sec^2 θ - tan^2 θ = 1
5. Inverse Trigonometric Functions:
- sin^(-1)(x) is the inverse of sin(x)
- tan^(-1)(x) is the inverse of tan(x)
- sec^(-1)(x) is the inverse of sec(x)