Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.3 | Set 2

Question 11. Integrate ∫\frac{1-cosx}{1+cosx}dx

Solution:

Let I = ∫\frac{1-cosx}{1+cosx}dx

On simplifying the above equation, we get

I = ∫\frac{2 sin^2(\frac{x}{2})}{2 cos^2(\frac{x}{2})} dx

= ∫\frac{sin^2 (\frac{x}{2})}{cos^2 (\frac{x}{2})} dx

= ∫ tan² x/2 dx

= ∫(sec^2\frac{x}{2} - 1) dx         -(1)

On integrating the equation(1), we get

= \frac{(tan \frac{x}{2})}{(\frac{1}{2})} - x +c

Hence, I = 2 tan x/2 - x + c 

Question 12. Integrate ∫ \frac{1}{1-sin\frac{x}{2}} dx

Solution:

 Let I = ∫ \frac{1}{1-sin\frac{x}{2}} dx

Now multiply with the conjugate,

= ∫ \frac{1}{(1-sin\frac{x}{2})} × \frac{(1 + sin \frac{x}{2})}{(1+ sin \frac{x}{2})} dx\\ = ∫\frac{(1+ sin\frac{x}{2})}{(1- sin^2 \frac{x}{2})} dx\\ = ∫\frac{(1+ sin\frac{x}{2})}{(cos^2 \frac{x}{2})} dx\\ = ∫\frac{1}{(cos^2\frac{x}{2})} dx + ∫\frac{(sin \frac{x}{2})}{(cos^2\frac{x}{2})} dx\\= ∫ sec^2 \frac{x}{2} dx + ∫ \frac{sec \frac{x}{2} tan \frac{x}{2}} dx

On integrating the equation, we get

= \frac{(tan \frac{x}{2})}{(\frac{1}{2})} + \frac{(sec \frac{x}{2})}{(\frac{1}{2})} + c

= 2 tan x/2 + 2 sec x/2 +c

Hence, I = 2 (tan x/2 + sec x/2) + c

Question 13. Integrate ∫ \frac{1}{(1 + cos 3x)} dx

Solution:

Let I = ∫ \frac{1}{(1 + cos 3x)} dx

Now multiply with the conjugate,

= ∫ 1/(1 + cos 3x) × (1 - cos 3x)/(1 - cos 3x) dx

= ∫ (1 - cos 3x)/ (1 - cos² 3x) dx

= ∫ (1 - cos 3x)/ (sin² 3x) dx

= ∫ (1/ sin² 3x) - (cos 3x/ sin² 3x) dx

= ∫ (cosec² 3x - cosec3x cot3x) dx                   -(1)

On integrating the equation(1), we get

= - cot 3x/3 + cosec 3x/3 + c

= (-1/3) × (cos 3x/ sin 3x) + (1/3) × (1/sin 3x) + c

= (1 - cos 3x) / 3 sin 3x + c

Therefore, I = (1 - cos 3x) / 3 sin 3x + c

Question 14. Integrate ∫(e^x + 1)² e^x dx

Solution:

Let I = ∫ (e^x + 1)² e^x dx                   -(1)

(e^x + 1) = t                   -(2)

On differentiating the above equation, we get

e^x dx = dt                   -(3)

Now, put the eq(2) and (3) in eq(1)

= ∫ (t²) dt                   -(4)

On integrating the equation(4), we get

= (t³ /3) + c

Therefore, I = (e^x + 1)³ /3 + c

Question 15. Integrate ∫ (e^x + (1 + e^x))² dx

Solution: 

Let I = ∫ (e^x + (1/e^x))² dx

= ∫ (e^2x + (1/e^2x) + 2)dx                   -(1)

On integrating the equation(1), we get

= (e^2x/2) - (1/2 e^-2x) + 2x + c

Therefore, I = (e^2x/2) - (1/2 e^-2x) + 2x + c

Question 16. Integrate ∫ \frac{(1 + cos 4x)}{(cot x - tan x)} dx

Solution: 

Let I = ∫ \frac{(1 + cos 4x)}{(cot x - tan x)} dx

= ∫\frac{(2 cos^2 2x)}{(\frac{cos x}{sin x}) - (\frac{sin x}{cos x})} dx \\ = ∫ \frac{(2 cos^2 2x)}{\frac{cos^2x - sin^2x}{cosx sinx}} dx \\ = ∫ \frac{(2 cos^2 2x. cosx.sinx)}{(cos^2x - sin^2x)} dx

On simplifying the above equation,

= ∫ cos²2x. (sin2x / cos2x) dx

= ∫ cos 2x. sin2x dx

= 1/2∫ sin (2x + 2x) + sin (2x - 2x) dx

= 1/2∫(sin 4x + sin 0) dx

= 1/2∫(sin 4x + 0) dx

= 1/2 ∫sin 4x dx                   -(1)

On integrating the equation(1), we get

 = (-1/2) ((cos 4x)/4) + c

Therefore, I = (-1/8) (cos 4x) + c

Question 17. Integrate ∫ \frac{1} {(\sqrt{x +3} - \sqrt{x + 2})} dx

Solution:

Let I = ∫ \frac{1} {(\sqrt{x +3} - \sqrt{x + 2})} dx

Now multiply with the conjugate,

= ∫ \frac{1}{(\sqrt{x +3} - \sqrt{x + 2})} × \frac{(\sqrt{x +3} + \sqrt{x + 2})}{(\sqrt{x +3} + \sqrt{x + 2})} dx\\ = ∫\frac{(\sqrt{x +3} + \sqrt{x + 2})}{(x+3-x-2)}  dx

= ∫(x +3)^1/2 + (x + 2)^1/2 dx                   -(1)

On integrating the equation(1), we get

= \frac{(x+3)^{\frac{3}{2}}}{(\frac{3}{2})} + \frac{(x+2)^{\frac{3}{2}}}{(\frac{3}{2})} + c

= (2/3)(x + 3)^3/2 +(2/3) (x + 2)^3/2 + c

Hence, I = (2/3){(x + 3)^3/2 + (x + 2)^3/2} +c

Question 18. Integrate ∫ tan²(2x - 3) dx

Solution:

Let I = ∫ tan²(2x - 3) dx

= ∫ sec² (2x - 3) - 1 dx                   -(1)

Now put,  2x - 3 = t                   -(2)

2dx = dt                   -(3)

Put eq(3) and (2) in eq(1)

= 1/2∫sec² t dt - ∫1dx                   -(4)

On integrating the equation(4), we get

= 1/2 tan t - x + c

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

Therefore, I = 1/2 tan(2x - 3) - x + c

Question 19. Integrate ∫ \frac{1}{(cos^2x(1+tanx)^2} dx

Solution:

Let I = ∫ \frac{1}{(cos^2x(1+tanx)^2} dx

= ∫ \frac{1}{cos^2x(1-\frac{sinx}{cosx}^2)} dx\\ = ∫ \frac{1}{(cosx - sinx)^2} dx\\ = ∫ \frac{1}{(1-sin2x)} dx\\ = ∫ \frac{1}{1+cos(\frac{π}{2} + 2x)} dx\\ = ∫ \frac{1}{2cos^2(\frac{π}{4} + x)}  dx\\ = \frac{1}{2} ∫ sec^2(\frac{π}{4} + x) dx

On integrating the equation, we get

= 1/2 tan(π/4 + x) + c

Therefore, I = 1/2 tan(π/4 + x) + c

Summary

Exercise 19.3 Set 2 in Chapter 19 on Indefinite Integrals focuses on integrating rational functions where the degree of the numerator is greater than or equal to the degree of the denominator. This set of problems requires students to apply polynomial long division before integration. After division, students will typically encounter a polynomial term plus a proper rational function. The polynomial term is integrated using standard integration rules, while the proper rational function is integrated using partial fraction decomposition. This exercise aims to enhance students' skills in handling complex rational functions, reinforcing their understanding of polynomial division, partial fraction decomposition, and integration techniques for different types of functions.