Class 12 RD Sharma Mathematics Solutions Chapter 19 Indefinite Integrals Exercise 19.23 | Set 1 (original) (raw)

Last Updated : 23 Jul, 2025

Chapter 19 of RD Sharma's Class 12 Mathematics textbook focuses on the Indefinite Integrals a fundamental concept in calculus. Exercise 19.23 | Set 1 delves into the practical applications and problems related to the finding indefinite integrals which are essential for the understanding antiderivatives and their applications in the various fields of the mathematics and science.

Indefinite Integrals

The Indefinite integrals represent the family of the functions whose derivatives yield the original function. They are also known as antiderivatives and are crucial in the solving differential equations. The process of finding an indefinite integral involves determining a function whose derivative matches the given function with an added constant of the integration (C) to account for the general solution.

Class 12 RD Sharma Mathematics Solutions - Exercise 19.23 | Set 1

Question 1. Evaluate ∫ 1/ 5+4cosx dx

**Solution:

Let us assume I = ∫ 1/ 5+4cosx dx

Put cosx = 1-tan2(x/2)/ 1+tan2(x/2)

= ∫1/ 5+4{1-tan2(x/2)/ 1+tan2(x/2)} dx

= ∫ 1+tan2(x/2)/ 5(1+tan2x/2)+4(1-tan2x/2) dx

= ∫ sec2(x/2)/ 5+5tan2x/2+4-4tan2x/2 dx

= ∫ sec2(x/2)/ 9+tan2x/2 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ (3)2+t2

On integrate the above eq. then, we get

= 2 × 1/3tan-1t/3 +c

= 2/3 tan-1{tan(x/2)/3} +c

Hence, I = 2/3 tan-1{tan(x/2)/3} +c

Question 2. Evaluate ∫ 1/ 5-4sinx dx

**Solution:

Let us assume I = ∫ 1/ 5-4sinx dx

Put sinx = 2tan(x/2)/ 1+tan2(x/2)

= ∫1/ 5-4{2tan(x/2)/ 1+tan2(x/2)} dx

= ∫ 1+tan2(x/2)/ 5(1+tan2x/2)-4(2tanx/2) dx

= ∫ sec2(x/2)/ 5+5tan2x/2-8tanx/2 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ 5t2-8t+5

= 2/5 ∫ dt/ t2-(8/5)t+1

= 2/5 ∫ dt/ t2-2t(4/5)+(4/5)2-(4/5)2+1

= 2/5 ∫ dt/ (t-4/5)2+(3/5)2

On integrate the above eq. then, we get

= 2/5 × 1/(3/5)tan-1{t-(4/5)/ (3/5)} +c

= 2/3 tan-1(5t-4)/ 3 +c

Hence, I = 2/3 tan-1(5tanx/2-4)/ 3 +c

Question 3. Evaluate ∫ 1/ 1-2sinx dx

**Solution:

Let us assume I = ∫ 1/ 1-2sinx dx

Put sinx = 2tan(x/2)/ 1+tan2(x/2)

= ∫1/ 1-2{2tan(x/2)/ 1+tan2(x/2)} dx

= ∫ 1+tan2(x/2)/ 1(1+tan2x/2)-2(2tanx/2) dx

= ∫ sec2(x/2)/ tan2x/2-4tanx/2+1 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ t2-4t+1

= ∫ 2dt/ t2-2t(2)+(2)2-(2)2+1

= 2∫ dt/ (t-2)2+3

= 2∫ dt/ (t-2)2+(√3)2

On integrate the above eq. then, we get

= 2 × 1/2√3log|t-2-√3/ t-2+√3| +c

= 2 × 1/2√3log|tanx/2-2-√3/ tanx/2-2+√3| +c

Hence, I = 2 × 1/2√3log|tanx/2-2-√3/ tanx/2-2+√3| +c

Question 4. Evaluate ∫ 1/ 4cosx-1 dx

**Solution:

Let us assume I = ∫ 1/ 4cosx-1 dx

Put cosx = 1-tan2(x/2)/ 1+tan2(x/2)

= ∫1/ 4{1-tan2(x/2)/ 1+tan2(x/2)}-1 dx

= ∫ 1+tan2(x/2)/ 4(1-tan2x/2)-(1+tan2x/2) dx

= ∫ sec2(x/2)/ 4-4tan2x/2+1-tan2x/2 dx

= ∫ sec2(x/2)/ 3-5tan2x/2 dx (i)

Let √5tanx/2 = t

√5/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ dt/ (√3)2+t2

On integrate the above eq. then, we get

= 1/√15 log|√3+t/√3-t|

Hence, I = 1/√15 log|√3+√5tan(x/2)/√3-√5tan(x/2)|

Question 5. Evaluate ∫ 1/ 1-sinx+cosx dx

**Solution:

Let us assume I = ∫ 1/ 1-sinx+cosx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ 1-{2tan(x/2)/1+tan2x/2} + {1-tan2(x/2)/1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 1+tan(x/2)-2tan(x/2)+1-tan2(x/2) dx

= ∫ sec2(x/2)/ 2-2tanx/2 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= 2/2 ∫ dt/ 1-t

= ∫ dt/ 1-t

Integrate the above eq. then, we get

= -log|1-t| +c

Hence, I = -log|1-tanx/2| +c

Question 6. Evaluate ∫ 1/ 3+2sinx+cosx dx

**Solution:

Let us assume I = ∫ 1/ 3+2sinx+cosx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ 3+2{2tan(x/2)/1+tan2x/2} + {1-tan2(x/2)/1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 3+3tan2(x/2)+4tan(x/2)+1-tan2(x/2) dx

= ∫ sec2(x/2)/ 2tan2x/2+4tanx/2+4 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= 2/2 ∫ dt/ t2+2t+2

= ∫ dt/ t2+2t+1-1+2

= ∫ dt/ (t+1)2+12

Integrate the above eq. then, we get

= tan-1(t+1) +c

Hence, I = tan-1(tanx/2+1) +c

Question 7. Evaluate ∫ 1/ 13+3cosx+4sinx dx

**Solution:

Let us assume I = ∫ 1/ 13+3cosx+4sinx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ 13+3{1-tan2(x/2)/1+tan2(x/2)} + 4{2tan(x/2)/ 1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 13(1+tan2x/2)+3-3tan2(x/2)+8tan(x/2) dx

= ∫ sec2(x/2)/ 16+13tan2x/2-3tan2x/2+8tanx/2 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= 2/10 ∫ dt/ 16+10t2+8t

= 1/5 ∫ dt/ t2+(4/5)t+8/5

= 1/5 ∫ dt/ t+2t(2/5)2+(2/5)2-(2/5)2+8/5

= 1/5 ∫ dt/ (t+2/5)2+(6/5)2

Integrate the above eq. then, we get

= 1/5 × 1/(6/5)tan-1(t+(2/5)/ (6/5)) +c

= 1/6 tan-1(5t+2/ 6) +c

Hence, I = 1/6 tan-1(5tanx/2+2/ 6) +c

Question 8. Evaluate ∫ 1/ cosx-sinx dx

**Solution:

Let us assume I = ∫ 1/ cosx-sinx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ {1-tan2(x/2)/1+tan2x/2} - {2tan(x/2)/1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 1-tan2(x/2)-2tan(x/2) dx

= ∫ sec2(x/2)/ 1-tan2(x/2)-2tan(x/2) dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ 1-t2-2t

= -∫ 2dt/ t2+2t-1

= -∫ 2dt/ t2+2t+1-1-1

= -∫ 2dt/ (t+1)2-(√2)2

= ∫ 2dt/ (√2)2-(t+1)2

Integrate the above eq. then, we get

= 2/(2√2) log|√2+t+1/√2-t-1| +c

Hence, I = 1/√2 log|√2+tanx/2+1/ √2-tanx/2-1| +c

Read More,

• **Indefinite Integrals
• **Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.7
• **Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.3

Conclusion

Exercise 19.23 | Set 1 in Chapter 19 of RD Sharma's textbook provides the valuable practice in solving problems involving the indefinite integrals. By mastering these exercises students can enhance their understanding of the antiderivatives and their applications. This exercise is an essential step in the developing a solid foundation in calculus.