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# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.3 | Set 1

• Last Updated : 03 Mar, 2021

### 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

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