Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.25 | Set 2

In this article, we will delve into the solutions for Exercise 19.25 | Set 2 from Chapter 19 of the RD Sharma Class 12 Mathematics textbook. This chapter focuses on indefinite integrals a crucial concept in calculus that forms the foundation for understanding more advanced topics in the integration. These solutions aim to provide students with a comprehensive understanding of the problems and methodologies required to solve them effectively helping to solidify their knowledge and prepare for board exams.

Indefinite Integrals

The Indefinite integrals also known as antiderivatives are functions that reverse the process of differentiation. They are used to determine the original function from its derivative. The general form of an indefinite integral is expressed as ∫f(x)dx where f(x) is the integrand, and the result includes a constant of the integration denoted as C. Indefinite integrals play a vital role in calculus especially in finding the areas under curves and solving differential equations.

Evaluate the following integrals:

Question 21. ∫(log⁡x)² x dx

Solution:

Given that, I = ∫(log⁡x)² x dx

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = (log⁡x)²∫xdx - ∫(2(log⁡x)(1/x) ∫xdx)dx

= x²/2(log⁡x)² - 2∫(log⁡x)(1/x)(x²/2)dx

= x²/2(log⁡x)² - ∫x(log⁡x)dx

= x²/2(log⁡x)² - [log⁡x∫xdx - ∫ (1/x ∫xdx)dx]

= x²/2(log⁡x)² - [x²2/2 log⁡x - ∫(1/x × x²/2)dx]

= x²/2(log⁡x)² - x²/2 log⁡x + 1/2 ∫xdx

= x²/2(log⁡x)² - x²/2 log⁡x + 1/4 x² + c

Hence, I = x²/2 [(log⁡x)² - log⁡x + 1/2] + c

Question 22. ∫e^√x dx

Solution:

Given that, I = ∫ e^√x dx

 Let us assume, √x = t

x = t²

dx = 2tdt

I = 2∫ e^t tdt

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = 2[t∫e^tdt - ∫(1∫e^tdt)dt]

= 2[te^t - ∫e^t dt]

= 2[te^t - e^t] + c

= 2e^t (t - 1) + c

Hence, I = 2e^√x(√x - 1) + c

Question 23. ∫(log⁡(x + 2))/((x + 2)²) dx

Solution:

Given that, I = ∫(log⁡(x + 2))/((x + 2)²) dx

 Let us assume (1/(x + 2) = t

-1/((x + 2)²) dx = dt

I = -∫log⁡(1/t)dt

= -∫log⁡t^-1 dt

= -∫1 × log⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡t∫dt - ∫(1/t ∫dt)dt

= tlog⁡t - ∫(1/t × t)dt

= tlog⁡t - ∫dt

= tlog⁡t - t + c

= 1/(x + 2) (log⁡(x + 2)^-1 - 1) + c

Hence, I = (-1)/(x + 2) - (log⁡(x + 2))/(x + 2) + c

Question 24. ∫(x + sin⁡x)/(1 + cos⁡x) dx

Solution:

Given that, I = ∫(x + sin⁡x)/(1 + cos⁡x) dx

= ∫x/(2cos²x/2) dx + ∫(2sin⁡x/2 cos⁡x/2)/(2cos²⁡x/2) dx

= 1/2 ∫xsec²⁡x/2 dx + ∫tan⁡x/2 dx

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 1/2 [x∫sec²x/2 dx - ∫(1∫ sec²x/2 dx)dx] + ∫tan⁡x/2 dx

= 1/2 [2xtan⁡x/2 - 2∫tan⁡x/2 dx] + ∫tan⁡x/2 dx + c

= xtan⁡x/2 - ∫tan⁡x/2 dx + ∫tan⁡x/2 dx+c

Hence, I = xtan⁡x/2 + c

Question 25. ∫log₁₀xdx

Solution:

Given that, I = ∫log₁₀⁡xdx

= ∫(log⁡x)/(log⁡10) dx

= 1/(log⁡10) ∫1 × log⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 1/(log⁡10) [log⁡x∫dx - ∫(1/x ∫dx)dx]

= 1/(log⁡10) [xlog⁡x - ∫(x/x)dx]

= 1/(log⁡10)[xlog⁡x - x]

Hence, I = (x/(log⁡10)) × (log⁡x - 1) + c

Question 26. ∫cos⁡√x dx

Solution:

Given that, I = ∫cos⁡√x dx

Let us assume, √x = t

x = t²

dx = 2tdt

= ∫2tcos⁡tdt

I = 2∫tcos⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = 2[t]cos⁡tdt - ∫(1 ∫cos⁡tdt)dt]

= 2[tsin⁡t - ∫sin⁡tdt]

= 2[tsin⁡t + cos⁡t] + c

Hence, I = 2[√x sin⁡√x + cos√x] + c

Question 27. ∫(xcos^-1x)/√(1 - x²) dx

Solution:

Given that, I = ∫(xcos^-1x)/√(1 - x²) dx

Let us assume, t = cos^-1⁡x

dt = (-1)/√(1 - x²) dx

Also, cost = x

I = -∫tcos⁡tdt

Now, using integration by parts,       

So, let

u = t;

du = dt

∫cos⁡tdt = ∫dv

sin⁡t = v

Therefore,

 I = -[tsint - ∫sin⁡tdt]

= -[tsint + cos⁡t] + c

On substituting the value t = cos^-1x we get,

 I = -[cos^-1⁡xsin⁡t + x] + c

Hence, I = -[cos^-1⁡x√(1 - x²) + x] + c

Question 28. ∫cosec³xdx

Solution:

Given that, I =∫cosec³xdx

 =∫cosec⁡x × cosec²xdx

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

= cosec⁡x × ∫cosec²xdx + ∫(cosec⁡xcot⁡x]cosec²xdx)dx

= cosecx × (-cot⁡x) + ∫cosec⁡xcot⁡x(-cot⁡x)dx

= -cosec⁡xcot⁡x - ∫cosecx⁡cot²⁡xdx

= -cosec⁡xcot⁡x - ∫cosec⁡x(cosec²x - 1)dx

= -cosec⁡xcot⁡x - ∫cosec³xdx + ∫cosecxd⁡x

I = -cosec⁡xcot⁡x - I + log⁡|tan⁡x/2| + c₁

2l = -cosec⁡xcot⁡x + log⁡|tan⁡x/2| + c₁

Hence, I = -1/2cosecx⁡cot⁡x + 1/2 log⁡|tan⁡x/2| + c

Question 29. ∫sec^-1√x dx

Solution:

Given that, I = ∫sec^-1⁡√x dx

 Let us assume, √x = t

x = t²

dx = 2tdt

I = ∫2tsec^-1⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 2[sec^-1⁡t∫tdt - ∫(1/(t√(t²-1))∫tdt)dt]

= 2[t²/2 sec^-1 - ∫(t/(2t√(t² - 1)))dt]

= t² sec^-1⁡t - ∫t/√(t² - 1) dt

= t² sec^-1⁡t - 1/2∫2t/√(t² - 1) dt

= t² sec^-1⁡t - 1/2 × 2√(t² - 1) + c

Hence, I = xsec^-1⁡√x - √(x - 1) + c

Question 30. ∫sin^-1√x dx

Solution:

Given that, I = ∫sin^-1⁡√x dx 

Let us assume, x = t

dx = 2tdt

∫sin^-1√x dx = ∫sin^-1⁡√(t²) 2tdt

= ∫sin^-1⁡t2tdt

= sin⁡^-1t∫2tdt - (∫(dsin^-1⁡t)/dt (∫2tdt)dt

= sin^-1⁡t(t²) - ∫1/√(1 - t²) (t²)dt

Now, lets solve ∫1/√(1 - t²) (t²)dt

∫1/√(1 - t²) (t²)dt = ∫(t² - 1 + 1)/√(1 - t²) dt

= ∫(t² - 1)/√(1 - t²) dt + ∫1/√(1 - t²) dt

As we know that, value of ∫1/√(1 - t²) dt = sin^-1⁡t

So, the remaining integral to evaluate is 

∫(t² - 1)/√(1 - t²) dt= ∫-√(1 - t²) dt

Now, substitute, t = sin⁡u, dt = cos⁡udu, we gte

∫-√(1 - t²) dt = ∫-cos²udu = -∫[(1 + cos⁡2u)/2]du

= -u/2-(sin⁡2u)/4

Now substitute back u = sin^-1x and t = √x, we get

= -(sin^-1√x)/2 - (sin⁡(2sin^-1⁡√x))/4

∫sin^-1√x dx = xsin^-1⁡√x-(sin^-1⁡√x)/2 - (sin⁡(2sin^-1√x))/4

sin⁡(2sin^-1⁡√x) = 2√x √(1 - x)

Hence, I = xsin^-1⁡√x - (sin^-1⁡√x)/2 - √(x(1 - x))/2 + c

Question 31. ∫xtan²⁡xdx

Solution:

Given that, I =∫xtan²⁡xdx

 = ∫x(sec²x - 1)dx

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

 = ∫xsec8xdx - ∫xdx

 = [x∫sec²xdx - ∫(1∫sec²xdx)dx] - x²/2

 = xtan⁡x - ∫tan⁡xdx - x²/2

Hence, I = xtan⁡x - log⁡|sec⁡x| - x²/2 + c

Question 32. ∫ x((sec⁡2x - 1)/(sec⁡2x + 1))dx

Solution:

Given that, I = ∫ x((sec⁡2x - 1)/(sec⁡2x + 1))dx

= ∫x((1 - cos⁡2x)/(1 + cos⁡2x))dx

= ∫x((sec²⁡x)/(cos²⁡x))dx

= ∫xtan²xdx

= ∫x(sec²⁡x - 1)dx

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

= ∫xsec²xdx - ∫dx

= [x∫sec²⁡xdx - ∫(1∫  sec²xdx)dx] - x²/2

= xtan⁡x - ∫tan⁡xdx - x²/2

= xtan⁡x - log|secx| - x²/2 + c

Hence, I = xtan⁡x - log|secx| - x²/2 + c

Question 33. ∫(x + 1)e^xlog⁡(xe^x)dx

Solution:

Given that, I = ∫(x + 1)e^xlog⁡(xe^x)dx

Let us assume, xe^x = t

(1 × e^x + xe^x)dx = dt

(x + 1)e^xdx = dt

I = ∫log⁡tdt

= ∫1 × log⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

= log⁡t∫dt - ∫(1/t∫dt)dt

= tlog⁡t - ∫(1/t × t)dt

= tlog⁡t - ∫dt

= tlog⁡t - t + c

= t(log⁡t - 1) + c

Hence, I = xe^x (log⁡xe^x - 1) + c

Question 34. ∫sin^-1(3x - 4x³)dx

Solution:

Given that, I = ∫sin^-1(3x - 4x³)dx

Let us assume, x = sin⁡θ

dx = cos⁡θdθ

= ∫sin^-1⁡(3sin⁡θ - 4sin³⁡θ)cos⁡θdθ

= ∫sin^-1(sin⁡3θ)cos⁡θdθ

= ∫3θcos⁡θdθ

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 3[θ]cos⁡θdθ - ∫(1∫cos⁡θdθ)dθ]

= 3[θsin⁡θ - ∫sin⁡θdθ]

= 3[θsin⁡θ + cos⁡θ] + c

Hence, I = 3[xsin^-1⁡x + √(1 - x²)] + c)

Question 35. ∫sin^-1(2x/(1 + x²))dx.

Solution:

Given that, I = ∫sin^-1(2x/(1 + x²))dx

Let us assume, x = tan⁡θ

dx = sec²⁡θdθ

sin^-1⁡(2x/(1 + x²)) = sin^-1⁡((2tan⁡θ)/(1 + tan²⁡θ))

= sin^-1⁡(sin⁡2θ) = 2θ

∫sin^-1⁡(2x/(1 + x²))dx = ∫2θsec²θdθ = 2∫θsec²⁡θdθ

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

2[θ∫sec²⁡θdθ - ∫{(d/dθ θ) ∫sec²θdθ}dθ

= 2[θtan⁡θ - ∫tan⁡θdθ]

= 2[θtan⁡θ + log⁡|cos⁡θ|] + c

= 2[xtan^-1⁡x + log⁡|1/√(1 + x²)|] + c

= 2xtan^-1⁡x + 2log⁡(1 + x²)^1/2 + c

= 2xtan^-1⁡x + 2[-1/2 log⁡(1 + x²)] + c

= 2xtan^-1⁡x - log⁡(1 + x²) + c

Hence, I = 2xtan^-1⁡x - log⁡(1 + x²) + c

Question 36. ∫ tan^-1((3x - x³)/(1 - 3x²))dx

Solution:

Given that, I = ∫tan^-1⁡((3x - x³)/(1 - 3x²))dx

 Let us assume, x = tan⁡θ

dx = sec²⁡θdθ

I = ∫tan^-1⁡((3tan⁡θ - tan³θ)/(1 - 3tan²⁡x)) sec²⁡θdθ

=∫tan^-1⁡(tan⁡3θ)sec²θdθ

= ∫3θsec²θdθ

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 3[θ∫ sec²⁡θdθ - ∫(1∫sec²θdθ)dθ]

= 3[θtan⁡θ - ∫tan⁡θdθ]

= 3[θtan⁡θ + log⁡sec⁡θ] + c

= 3[xtan^-1x - log⁡√(1 + x²)] + c

Hence, I = 3[xtan^-1x - log⁡√(1 + x²)] + c

Question 37. ∫x²sin^-1xdx

Solution:

Given that, I = ∫x²sin^-1⁡xdx

I = sin^-1x∫x² dx - ∫(1/√(1 - x²) ∫x² dx)dx

= x³/3 sin^-1⁡x - ∫x³/(3√(1 - x²)) dx

I = x³/3 sin^-1⁡x - 1/3 I₁ + c₁ .....(1)

Let I₁ = ∫x³/√(1 - x²) dx

Let 1 - x² = t²

-2xdx = 2tdt

-xdx = tdt

I₁ = -∫(1 - t²)tdt/t

= ∫(t² - 1)dt

= t³/3 - t + c₂

= (1 - x²)^3/2/3 - (1 - x²)^1/2 + c₂

Now, put the value of I₁ in eq(1), we get

Hence, I = x³/3 sin^-1⁡x - 1/9 (1 - x²)^3/2 + 1/3 (1 - x²)^1/2 + c

Question 38. ∫(sin^-1x)/x²dx

Solution:

Given that, I =∫(sin^-1⁡x)/x²dx

 = ∫(1/x²)(sin^-1⁡x)dx

I = [sin^-1⁡x∫1/x²dx - ∫(1/√(1 - x²) ∫1/x²dx)dx]

= sin^-1×(-1/x) - ∫1/√(1 - x²) (-1/x)dx

I = -1/x sin^-1x + ∫1/(x√(1 - x²)) dx

I = -1/x sin^-1⁡x + I₁  .......(1)

Where,

I₁ = ∫1/(x√(1 - x²)) dx

Let 1 - x² = t²

-2xdx = 2tdt

I₁ = ∫x/(x²√(1 - x²)) dx

= -∫tdt/((1 - t²) √t)

= -∫dt/((1 - t²))

= ∫1/(t² - 1) dt

= 1/2 log⁡|(t - 1)/(t + 1)| 

= 1/2 log⁡|(t - 1)/(t + 1)|

= 1/2 log⁡|(√(1 - x²) - 1)/(√(1 - x²) + 1)| + c₁

Now, put the value of I₁ in eq(1), we get

I = -(sin^-1x)/x + 1/2 log⁡|((√(1 - x²) - 1)/(√(1 - x²) + 1))((√(1 - x²) - 1)/(√(1 - x²) - 1))| + c 

= -(sin^-1⁡x)/x + 1/2 log⁡|(√(1 - x²) - 1)²/(1 - x² - 1)| + c

= -(sin^-1⁡x)/x + 1/2 log⁡|(√(1 - x²) - 1)²/(-x²)| + c

= -(sin^-1⁡x)/x + log⁡|(√(1 - x²) - 1)/(-x)| + c

Hence, I = -(sin^-1x)/x + log⁡|(1 - √(1 - x²))/x| + c

Question 39. ∫(x² tan^-1x)/(1 + x²) dx

Solution:

Given that, I = ∫(x² tan^-1⁡x)/(1 + x²) dx

Let us assume, tan^-1⁡x = t     [x = tan⁡t]

1/(1 + x²) dx = dt

I = ∫t × tan²tdt

= ∫t(sec²⁡t - 1)dt

= ∫(tsec²t - t)dt

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

= ∫tsec²tdt - ∫tdt

= [t∫sec²tdt - ∫(1)sec²⁡tdt)dt] - t²/2

= [t × tan⁡t - ∫tan⁡tdt] - t²/2

= t tan⁡t - log⁡sec⁡t - t²/2 + c

= xtan^-1⁡x - log⁡√(1 + x²) - (tan²x)/2 + c

Hence, I = xtan^-1⁡x - 1/2 log⁡|1 + x²| - (tan²⁡x)/2 + c

Question 40. ∫cos^-1(4x³ - 3x)dx

Solution:

Given that, I = ∫cos^-1⁡(4x³ - 3x)dx

 Let us assume, x = cos⁡θ

dx = -sin⁡θdθ

I = -∫cos^-1(4cos⁡³θ - 3cos⁡θ)sin⁡θdθ

= - ∫cos^-1(cos⁡3θ)sin⁡θdθ

= -∫3θsin⁡θdθ

Using integration by parts,       

∫u v dx = v∫ u dx - ∫{d/dx(v) × ∫u dx}dx + c 

We get

= -3[θ]sin⁡θdθ - ∫(1∫sin⁡θdθ)dθ]

= -3[-θcos⁡θ + ∫cos⁡θdθ]

= 3θcos⁡θ - 3sin⁡θ + c

Hence, I = 3xcos^-1x - 3√(1 - x²) + c

Read More:

• Mathematics | Indefinite Integrals
• Integration Formulas

Summary

Exercise 19.25 | Set 2 typically deals with integrating rational functions where the denominator is of the form x⁴ - 1. The key points to remember are:

• These integrals often require partial fraction decomposition.
• The denominator x⁴ - 1 can be factored as (x² - 1)(x² + 1) or (x - 1)(x + 1)(x² + 1).
• After partial fraction decomposition, you'll usually end up with terms of the form A/(x - 1), B/(x + 1), and C/(x² + 1).
• These terms can be integrated using standard integration techniques.