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

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  • Last Updated : 03 Mar, 2021

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 ∫(ex + 1)2 ex dx

Solution:

Let I = ∫ (ex + 1)2 ex dx                   -(1)

(ex + 1) = t                   -(2)

On differentiating the above equation, we get

ex 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 = (ex + 1)3 /3 + c

Question 15. Integrate ∫ (ex + (1 + ex))2 dx

Solution: 

Let I = ∫ (ex + (1/ex))2 dx

= ∫ (e2x + (1/e2x) + 2)dx                   -(1)

On integrating the equation(1), we get

= (e2x/2) – (1/2 e-2x) + 2x + c

Therefore, I = (e2x/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


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