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