# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.28

• Last Updated : 11 Feb, 2021

### Question 1. Find Solution: Let considered x – 1 = t,

so that dx = dt

Thus, ### Question 2. Evaluate Solution:

Let I = ### Question 3. Evaluate Solution:

I = Hence, ### Question 4. Evaluate Solution:

Let I = Therefore, I = ="528" style="vertical-align: -9px;"/>

### Question 5. Solution:

I = Let us considered sinx = t

So, on differentiating, we get

cosx dx = dt

I = Therefore, I = ### Question 6. Evaluate Solution:

I = <

Let us considered ex = t

So, on differentiating, we get

exdx = dt

Therefore, I = Hence, I = ### Question 7. Evaluate Solution:

I =  Therefore, I = ### Question 8. Evaluate Solution:

Let us assume I =  Therefore, I = ### Question 9. Evaluate Solution:

Let us assume I =  Therefore, I = ### Question 10. Evaluate Solution:

Let us assume I =  "780" style="vertical-align: -25px;"/>

Therefore, I = ### Question 11. Evaluate Solution:

Let us assume I =  Therefore, I = ### Question 12. Evaluate Solution:

Let us assume x2 = t

On differentiating we get

2x dx = dt

Therefore, I = ="466" style="vertical-align: 15px;"/>

Hence, I = ### Question 13. Evaluate Solution:

I = Let us considered x3 = t

So, on differentiating, we get

3x2dx = dt

Therefore, I = Hence, I = ### Question 14. Evaluate Solution:

I = "182" style="vertical-align: -13px;"/>

Let us considered logx = t

So, on differentiating, we get

1/x dx = dt

Therefore, I = Hence, I = ### Question 15. Evaluate Solution:

I =  Therefore, I = ### Question 16. Evaluate Solution:

Let I =  I = My Personal Notes arrow_drop_up