Here we provide RD Sharma Class 12 Ex 19.22 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.22 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 19 |

Exercise | 19.22 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Class 12 Ex 19.22 Solutions Chapter 19 Indefinite Integrals**

### Question 1. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos^{2}x, we get

Let us considered tan x = t

So, sec^{2}x 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 cos^{2}x, we get

Now, let us considered tan x = t

So, sec^{2}xdx = 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 cos^{2}x, we get

Now, 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 get

Now, let us considered tan x = t

So, sec^{2}x dx = dt

Again, 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 get

Now, let us assume 2tan x = t

So, 2sec^{2}x dx = dt

I = 1/2 ∫dt/(1 + t^{2})

= 1/2 tan^{-1}t + c

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

### Question 6. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos^{2}x, we get

Now, let us assume √3 tanx = t

So, √3 sec^{2}x dx = dt

Hence, 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 get

Now, 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 get

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

So, 2tanx sec^{2}x dx = dt

I = ∫dt/(t^{2 }+ 1)

= tan^{-1}t + c

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

### Question 9. Evaluate the integral:

**Solution:**

Let

On dividing numerator and denominator by cos^{2}x, we get

Now, let us assume 2 + tanx = t

So, sec^{2}x 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 cos^{2}x, we get

Now, 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 get

Now, let us assume √2tanx = t

So, √2sec^{2}dx = dt

I = 1/√2 ∫1/(1 + t^{2})

= 1/√2 tan^{-1}t + c

Hence, I = 1/√2 tan^{-1}(√2tanx) + c

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