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

Indefinite integrals are a fundamental concept in calculus, and they have numerous applications in various fields, including physics, engineering, and economics. In this exercise, we will focus on solving indefinite integrals using various techniques, such as substitution, integration by parts, and integration by partial fractions.

Important Formulas and Concepts

Indefinite Integral: ∫f(x)dx = F(x) + C

Substitution Method: ∫f(g(x))g'(x)dx = F(g(x)) + C

Integration by Parts: ∫u(x)v'(x)dx = u(x)v(x) - ∫v(x)u'(x)dx

Integration by Partial Fractions: ∫(P(x)/Q(x))dx = ∫(A/(x-a))dx + ∫(B/(x-b))dx + ...

Class 12 RD Sharma Mathematics Solutions -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)}

= 

= 

= 

= 

Practice Questions:

1. ∫ x^2 * e^x dx

2. ∫ x * sin(x) dx

3. ∫ ln(x) * x^2 dx

4. ∫ x * cos(2x) dx

5. ∫ e^x * sin(x) dx

Also Read,

Also Read,

• Indefinite Integrals
• Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.12
• Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.16
• Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.18 | Set 1
• Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.18 | Set 2

Conclusion

1. Integration by Parts: This is likely the main focus of Exercise 19.17. It's used when integrating products of functions.

2. Formula: ∫ u dv = uv - ∫ v du

3. LIATE Rule: A helpful mnemonic for choosing u when using integration by parts:

L - Logarithmic functions

I - Inverse trigonometric functions

A - Algebraic functions

T - Trigonometric functions

E - Exponential functions

4. Steps for Integration by Parts:

a) Choose u and dv based on the LIATE rule

b) Calculate du and v

c) Apply the formula

d) Simplify and solve the resulting integral

5. Multiple Applications: Some problems may require applying integration by parts more than once.