Class 12 RD Sharma Mathematics Solutions - Chapter 19 Indefinite Integrals - Exercise 19.13 | Set 1

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 ∫ e^x/ √16-e^2x dx

Solution:

Let us assume I =∫ e^x/ √16-e^2x dx (i)

Put e^x = t

e^x 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(e^x/4) + c

Hence, I = sin^-1(e^x/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,

• Indefinite Integrals
• Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.11
• Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.12

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.