Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.22

Question 1. Evaluate the integral:

Solution:

Let 

On dividing numerator and denominator by cos²x, we get

Let us considered tan x = t

So, sec²x dx = dt 

Again, let us considered 3t = u

3dt = du

= (3/2) × (1/2) × tan^-1(u/2) + c

= (1/6)tan^-1(3t/2) + c

Hence, I = (1/6)tan^-1(3tanx/2) + c

Question 2. Evaluate the integral:

Solution:

Let 

On dividing numerator and denominator by cos²x, we get

Now, let us considered tan x = t

So, sec²xdx = dt

Again, let us considered 2t = u

2dt = du

= (1/2) × (1/√5) × tan^-1(u/√5) + c

= (1/2√5) × tan^-1(2t/√5) + c

Hence, I = (1/2√5) × tan^-1(2tanx/√5) + c

Question 3. Evaluate the integral:

Solution:

Let 

On dividing numerator and denominator by cos²x, we get

Now, let us considered tan x = t

So, sec²x dx = dt

Question 4. Evaluate the integral:

Solution:

Let 

On dividing numerator and denominator by cos²x, we get

Now, let us considered tan x = t

So, sec²x dx = dt

Again, let us considered √3t = u

So, √3dt = du

Question 5. Evaluate the integral:

Solution:

Let 

On dividing numerator and denominator by cos²x, we get

Now, let us assume 2tan x = t

So, 2sec²x dx = dt

I = 1/2 ∫dt/(1 + t²)

= 1/2 tan^-1t + c

Hence, I = 1/2 tan^-1(2tanx) + c

Question 6. Evaluate the integral:

Solution:

Let 

On dividing numerator and denominator by cos²x, we get

Now, let us assume √3 tanx = t

So, √3 sec²x dx = dt

Hence, I = (1/√15)tan^-1(√3tanx/√5) + c

Question 7. Evaluate the integral:

Solution:

Let 

On dividing numerator and denominator by cos²x, we get

Now, let us assume tanx = t

So, sec²x dx = dt

Question 8. Evaluate the integral:

Solution:

Let 

On dividing numerator and denominator by cos⁴x, we get

Now, let us assume tan²x = t

So, 2tanx sec²x dx = dt

I = ∫dt/(t² + 1)

= tan^-1t + c

I = tan^-1(tan²x) + c

Question 9. Evaluate the integral:

Solution:

Let 

On dividing numerator and denominator by cos²x, we get

Now, let us assume 2 + tanx = t

So, sec²x dx = dt

I = ∫dt/t

= log|t| + c

I = log|2 + tanx| + c 

Question 10. Evaluate the integral:

Solution:

Let 

On dividing numerator and denominator by cos²x, we get

Now, let us assume tanx = t

So, sec²x dx = dt

Question 11. Evaluate the integral:

Solution:

Let 

On dividing numerator and denominator by cos²x, we get

Now, let us assume √2tanx = t

So, √2sec²dx = dt

I = 1/√2 ∫1/(1 + t²)

= 1/√2 tan^-1t + c

Hence, I = 1/√2 tan^-1(√2tanx) + c

Summary

Exercise 19.22 of RD Sharma's Class 12 Solutions focuses on advanced techniques for solving indefinite integrals. This exercise likely combines various methods such as integration by parts, trigonometric substitutions, partial fractions decomposition, and other specialized approaches for complex integrands. The problems are designed to challenge students' understanding of when and how to apply different integration techniques, often requiring a combination of methods to solve a single problem.