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

Question 1. Integrate âˆ«(2x – 3)5  + âˆš3x + 2 dx

Solution:

Let I = âˆ«(2x – 3)5 + âˆš3x + 2 dx         -(1)

On integrating the equation(1), we get

Therefore, I =

Question 2. Integrate

Solution:

Let I =

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

On integrating the equation(1), we get

Hence, I =

Question 3. Integrate

Solution:

Let I =

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

On integrating the equation(1), we get

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

= (-1/3) log|2 – 3x| + (2/3)âˆš3x – 2 + c

Hence, I = (-1/3) log|2 – 3x| + (2/3)âˆš3x – 2 + c

Question 4. Integrate

Solution:

Let I =

= âˆ«(x + 1)-3 dx + 2âˆ«(x + 1)-4 dx            -(1)

On integrating the equation(1), we get

= (x + 1)-2/(-2) + 2Ã—(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 =

= âˆ«(âˆšx + 1 – âˆšx) dx

= âˆ«(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 âˆ«(âˆš2x + 3 – âˆš2x – 3) dx

= 1/6 âˆ«(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 =

=

= âˆ«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)âˆ«(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 âˆ«Sinxâˆš1 + Cos2x dx

Solution:

Let I = âˆ«Sinxâˆš1 + Cos2x dx

On  substituting the formula, we get

= âˆ«Sinxâˆš(2Cos2x) dx

= âˆ«Sinxâˆš2Cosx dx

= âˆš2 âˆ«Sinx Cosx dx

Multiply and divide the above equation by 2

= âˆš2/2âˆ«2SinxCosx dx

= âˆš2/2âˆ«Sin2x dx            -(1)

On integrating the equation(1), we get

= âˆš2/2 (-Cos2x/2) + c

Hence, I = (-1/2âˆš2) Cos2x + c

Question 10. Integrate

Solution:

Let I =

= âˆ«cot2(x/2) dx

= âˆ«cosec2(x/2) – 1 dx            -(1)

On integrating the equation(1), we get

Hence, I = -2cot(x/2) – x + c

Previous
Next