Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.17

Evaluate the following integrals:

Question 1. ∫dx/√(2x – x²)

Solution:

We have,

Let I = ∫dx/√(2x – x²)

= ∫dx/√(1 – 1 + 2x – x²)

= ∫dx/√{1 – (x² – 2x + 1)}

= ∫dx/√{1² – (x – 1)²}

Let x – 1 = q …(1)

= ∫dx/√{1² – (q)²}

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= sin^-1(q) + C    

Now put the value of q from eq(1), we get

= sin^-1(x – 1) + C    

Question 2. ∫dx/√(8 + 3x – x²)

Solution:

We have,

Let I = ∫dx/√(8 + 3x – x²)

Here, (8 + 3x – x²) can be written as 8 – (x² – 3x + 9/4 – 9/4)

= (8 + 9/4) – (x – 3/2)²

= (41/4) – (x – 3/2)²

= 

Let x – 3/2 =  q …(1)

= 

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= 

Now put the value of q from eq(1), we get

= 

= sin^-1(2x – 3/√41) + C

Question 3. ∫dx/√(5 – 4x – 2x²)

Solution:

We have,

Let I = ∫dx/√(5 – 4x – 2x²)

= ∫dx/√{2(5/2 – 2x – x²)}

= (1/√2)∫dx/√{5/2 – (x² – 2x + 1 – 1)}

= (1/√2)∫dx/√{(5/2 + 1) – (x² – 2x + 1)}

= (1/√2)∫dx/√{7/2 – (x – 1)²}

=

Let x + 1 = q …(1)

= 

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= 

Now put the value of q from eq(1), we get

= 

= 

Question 4. ∫dx/√(3x² + 5x + 7)

Solution:

We have,

Let I = ∫dx/√(3x² + 5x + 7)

= ∫dx/√{3(x² + 5x/3 + 7/3)}

= (1/√3)∫dx/√{5/2-(x2-2x+1-1)}

= 

= 

= 

= 

= 

Question 5. ∫dx/√{(x – α)(β – x)}

Solution:

We have,

Let I = ∫dx/√{(x – α)(β – x)}

= ∫dx/√(-x² +αx + βx – αβ)

= ∫dx/√{-x² + x(α + β) – αβ}

= 

= 

= 

Let x – (α + β)/2 = q …(1)

= 

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= 

Now put the value of q from eq(1), we get

= 

= 

Question 6. ∫dx/√(7 – 3x – 2x²)

Solution:

We have,

Let I = ∫dx/√(7 – 3x – 2x²)

= ∫dx/√{2(7/2 – 3x/2 – x²)}

= 

= 

= 

= 

Let x + 3/2 = q …(1)

= 

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= 

Now put the value of q from eq(1), we get

= 

= 

Question 7. ∫dx/√(16 – 6x – x²)

Solution:

We have,

Let I = ∫dx/√(16 – 6x – x²)

= ∫dx/√{16 – (x² + 2.3x + 9 – 9)}

= ∫dx/√{25 – (x² + 2.3x + 9)}

= 

Let x + 3/2 = q …(1)

= 

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= 

Now put the value of q from eq(1), we get

= 

Question 8. ∫dx/√(7 – 6x – x²)

Solution:

We have,

Let I = ∫dx/√(7 – 6x – x²)

= ∫dx/√{7-(x² + 2.3x + 9 – 9)}

= ∫dx/√{16 – (x² + 2.3x + 9)}

= 

Let x + 3/2 = q …(1)

= 

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= 

Now put the value of q from eq(1), we get

= 

Question 9. ∫dx/√(5x² – 2x)

Solution:

We have,

Let I = ∫dx/√(5x² – 2x)

= ∫dx/√{5(x² – 2x/5)}

= 

= 

= 

=