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• RD Sharma Class 12 Solutions for Maths

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

### Question 11. Integrate

Solution:

Let I =

On simplifying the above equation, we get

I =

= ∫ tan2 x/2 dx

On integrating the equation(1), we get

Hence, I = 2 tan x/2 – x + c

### Question 12. Integrate

Solution:

Let I =

Now multiply with the conjugate,

On integrating the equation, we get

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

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

### Question 13. Integrate

Solution:

Let I =

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

Solution:

Let I =

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

Solution:

Let I =

Now multiply with the conjugate,

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

On integrating the equation(1), we get

= (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

Solution:

Let I =

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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