Class 12 RD Sharma Mathematics Solutions - Chapter 19 Indefinite Integrals - Exercise 19.13 | Set 2

Exercise 19.13 | Set 2 builds upon the techniques introduced in Set 1, presenting more challenging integration problems involving √(ax + b). These problems often require a combination of techniques such as substitution, trigonometric substitution, and integration by parts.

Important Formulas and Concepts

Standard integral form:

∫ √(ax + b) dx = (2/3a) * (ax + b)^(3/2) + C

Substitution method:

Let u = ax + b, then du = a dx

∫ f(√(ax + b)) dx = (1/a) ∫ f(√u) du

Trigonometric substitution:

For √(ax + b), use x = (1/a)(tan²θ - b)

Integration by parts:

∫ u dv = uv - ∫ v du

Partial fractions:

Used for integrating complex rational functions

Class 12 RD Sharma Mathematics Solutions -Exercise 19.13 | 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/ x^2/3√x^2/3-4 dx

Solution:

Let us assume I =∫ 1/ x^2/3√x^2/3-4 dx (i)

Put x^1/3 = t

(1/3) x^1/3-1 dx = dt

(1/3) x^-2/3 dx = dt

dx/ x^2/3 = 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|x^1/3 +√(x^1/3)²-(2)²| + c

Hence, I = 3 log|x^1/3 +√x^2/3-4| + 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)² (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

1. ∫ dx / (x² + 9)

2. ∫ (2x + 1) dx / (x² + 4)

3. ∫ dx / (x² - 2x + 5)

4. ∫ x dx / (x² + 1)

5. ∫ (x + 2) dx / (x² + 4x + 5)

6. ∫ dx / (x² + 2x + 2)

7. ∫ (3x - 1) dx / (x² + 1)

8. ∫ dx / (4x² + 4x + 5)

9. ∫ (2x - 3) dx / (x² - 2x + 10)

10. ∫ x dx / (x² + 4x + 13)

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

Conclusion

Exercise 19.13 Set 2 in RD Sharma's Class 12 Chapter on Indefinite Integrals focuses on integrating rational functions with quadratic denominators. The key techniques used in this exercise include:

1. Completing the square in the denominator

2. Recognizing standard integral forms

3. Using substitution method

4. Partial fraction decomposition (in some cases)

The problems typically involve transforming the given integral into a standard form that can be solved using known formulas or by applying substitution.