Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.18 | Set 1

Chapter 19 of RD Sharma's Class 12 Mathematics textbook delves into the crucial topic of Indefinite Integrals. Exercise 19.18 | Set 1 specifically focuses on integrating functions involving trigonometric expressions. This set of problems challenges students to apply various integration techniques and trigonometric identities to solve complex integrals. Mastering these concepts is essential for advanced calculus and its applications in physics and engineering.

Important Formulas and Concepts: Indefinite Integrals

Basic Trigonometric Integrals:

∫ sin x dx = -cos x + C

∫ cos x dx = sin x + C

∫ tan x dx = -ln|cos x| + C

∫ cot x dx = ln|sin x| + C

∫ sec x dx = ln|sec x + tan x| + C

∫ cosec x dx = ln|cosec x - cot x| + C

Class 12 RD Sharma Mathematics Solutions -Exercise 19.18 | 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)

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 tan x = 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-(e)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|sin⁴4x +√(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^2 - 1)

Question 2. ∫ dx / (x^2 + 4x + 4)

Question 3. ∫ (2x - 1) / (x^2 - x - 2) dx

Question 4. ∫ dx / (x^3 + 1)

Question 5. ∫ (x + 1) / (x^2 - x - 2) dx

Also Read,

• Indefinite Integrals
• Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.12
• Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.16
• Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.17
• Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.18 | Set 2

Conclusion

1. Partial Fraction Decomposition: This is the primary technique for Exercise 19.18. It's used to integrate rational functions by breaking them into simpler fractions.

2. General Steps:

a) Ensure the numerator's degree is less than the denominator's.

b) Factor the denominator.

c) Write out the partial fraction decomposition.

d) Solve for the unknown coefficients.

e) Integrate each simple fraction.

3. Common Decomposition Forms:

- For (x - a): A / (x - a)

- For (x - a)^n: A1 / (x - a) + A2 / (x - a)^2 + ... + An / (x - a)^n

- For (x^2 + px + q): (Ax + B) / (x^2 + px + q)

4. Integration of Basic Forms:

- ∫ 1 / (x - a) dx = ln|x - a| + C

- ∫ 1 / (x^2 + a^2) dx = (1/a) arctan(x/a) + C

- ∫ 1 / (x^2 - a^2) dx = (1/2a) ln|(x-a)/(x+a)| + C