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# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.22

• Last Updated : 11 Feb, 2021

### 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) Ã— (1/2) Ã— 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) Ã— (1/âˆš5) Ã— tan-1(u/âˆš5) + c

= (1/2âˆš5) Ã— tan-1(2t/âˆš5) + c

Hence, I = (1/2âˆš5) Ã— tan-1(2tanx/âˆš5) + 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 âˆš3t = u

So, âˆš3dt = 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 âˆ«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 âˆš3 tanx = t

So, âˆš3 sec2x dx = dt

Hence, I = (1/âˆš15)tan-1(âˆš3tanx/âˆš5) + 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 = âˆ«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 = âˆ«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 âˆš2tanx = t

So, âˆš2sec2dx = dt

I = 1/âˆš2 âˆ«1/(1 + t2)

= 1/âˆš2 tan-1t + c

Hence, I = 1/âˆš2 tan-1(âˆš2tanx) + c

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