# 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 cos

^{2}x, we getLet us considered tan x = t

So, sec

^{2}x dx = dtAgain, let us considered 3t = u

3dt = du

= (3/2) × (1/2) × tan

^{-1}(u/2) + c= (1/6)tan

^{-1}(3t/2) + cHence, I = (1/6)tan

^{-1}(3tanx/2) + c

### Question 2. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos

^{2}x, we getNow, let us considered tan x = t

So, sec

^{2}xdx = dtAgain, let us considered 2t = u

2dt = du

= (1/2) × (1/√5) × tan

^{-1}(u/√5) + c= (1/2√5) × tan

^{-1}(2t/√5) + cHence, I = (1/2√5) × tan

^{-1}(2tanx/√5) + c

### Question 3. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos

^{2}x, we getNow, let us considered tan x = t

So, sec

^{2}x dx = dt

### Question 4. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos

^{2}x, we getNow, let us considered tan x = t

So, sec

^{2}x dx = dtAgain, let us considered √3t = u

So, √3dt = du

### Question 5. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos

^{2}x, we getNow, let us assume 2tan x = t

So, 2sec

^{2}x dx = dtI = 1/2 ∫dt/(1 + t

^{2})= 1/2 tan

^{-1}t + cHence, I = 1/2 tan

^{-1}(2tanx) + c

### Question 6. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos

^{2}x, we getNow, let us assume √3 tanx = t

So, √3 sec

^{2}x dx = dtHence, I = (1/√15)tan

^{-1}(√3tanx/√5) + c

### Question 7. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos

^{2}x, we getNow, let us assume tanx = t

So, sec

^{2}x dx = dt

### Question 8. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos

^{4}x, we getNow, let us assume tan

^{2}x = tSo, 2tanx sec

^{2}x dx = dtI = ∫dt/(t

^{2 }+ 1)= tan

^{-1}t + cI = tan

^{-1}(tan^{2}x) + c

### Question 9. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos

^{2}x, we getNow, let us assume 2 + tanx = t

So, sec

^{2}x dx = dtI = ∫dt/t

= log|t| + c

I = log|2 + tanx| + c

### Question 10. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos

^{2}x, we getNow, let us assume tanx = t

So, sec

^{2}x dx = dt

### Question 11. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos

^{2}x, we getNow, let us assume √2tanx = t

So, √2sec

^{2}dx = dtI = 1/√2 ∫1/(1 + t

^{2})= 1/√2 tan

^{-1}t + cHence, I = 1/√2 tan

^{-1}(√2tanx) + c