# RD Sharma Class 12 Ex 19.3 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.3 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.3 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 19.3 Solutions Chapter 19 Indefinite Integrals

### Question 1. Integrate ∫(2x – 3)5  + √3x + 2 dx

Solution:

Let I = ∫(2x – 3)5 + √3x + 2 dx         -(1)

On integrating the equation(1), we get

Therefore, I =

### Question 2. Integrate

Solution:

Let I =

I = ∫(7x – 5)-3 + (5x – 4)-1/2 dx            -(1)

On integrating the equation(1), we get

Hence, I =

### Question 3. Integrate

Solution:

Let I =

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

On integrating the equation(1), we get

= log |2 – 3x| /(-3) + (2/3) × (3x – 2)1/2 + c

= (-1/3) log|2 – 3x| + (2/3)√3x – 2 + c

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

### Question 4. Integrate

Solution:

Let I =

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

On integrating the equation(1), we get

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

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

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

### Question 5. Integrate

Solution:

Let I =

= ∫(√x + 1 – √x) dx

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

On integrating the equation(1), we get

= (2/3)(x + 1)3/2  – (2/3)(x)3/2 + c

Hence, I = (2/3)(x + 1)3/2 – (2/3)(x)3/2 + c

### Question 6. Integrate

Solution:

Let I =

Now multiply with the conjugate

=

= 1/6 ∫(√2x + 3 – √2x – 3) dx

= 1/6 ∫(2x + 3)1/2 – (2x – 3)1/2 dx            -(1)

On integrating the equation(1), we get

Hence, I =

### Question 7. Integrate

Solution:

Let I =

=

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

On integrating the equation(1), we get

=

= (1/2) log|2x + 1| + (1/2)(2x + 1)-1 + c

Hence, I= (1/2) log|2x + 1| + (1/2)(2x + 1)-1 + c

### Question 8. Integrate

Solution:

Let I =

Now multiply with the conjugate,

= (1/a – b)∫(x + a)1/2 – (x + b)1/2 dx            -(1)

On integrating the equation(1), we get

= (1/a – b) ((2/3) (x + a)3/2 – (2/3) (x + b)3/2) + c

Hence, I = (1/a – b) (2/3) ((x + a)3/2 – (x + b)3/2) + c

### Question 9. Integrate ∫Sinx√1 + Cos2x dx

Solution:

Let I = ∫Sinx√1 + Cos2x dx

On  substituting the formula, we get

= ∫Sinx√(2Cos2x) dx

= ∫Sinx√2Cosx dx

= √2 ∫Sinx Cosx dx

Multiply and divide the above equation by 2

= √2/2∫2SinxCosx dx

= √2/2∫Sin2x dx            -(1)

On integrating the equation(1), we get

= √2/2 (-Cos2x/2) + c

Hence, I = (-1/2√2) Cos2x + c

### Question 10. Integrate

Solution:

Let I =

= ∫cot2(x/2) dx

= ∫cosec2(x/2) – 1 dx            -(1)

On integrating the equation(1), we get

Hence, I = -2cot(x/2) – x + c

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 ∫(e+ 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 + 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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