Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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

  • Last Updated : 16 May, 2021

Evaluate the following integrals:

Question 21. ∫(log⁡x)2 x dx

Solution:

Given that, I = ∫(log⁡x)2 x dx

Hey! Looking for some great resources suitable for young ones? You've come to the right place. Check out our self-paced courses designed for students of grades I-XII

Start with topics like Python, HTML, ML, and learn to make some games and apps all with the help of our expertly designed content! So students worry no more, because GeeksforGeeks School is now here!

 



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 asuume, √x = t

x = t2

dx = 2tdt

I = 2∫ et tdt

Using integration by parts,       

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

We get

I = 2[t∫etdt – ∫(1∫etdt)dt]

= 2[tet – ∫et dt]

= 2[tet – et] + c

= 2et (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)exlog⁡(xex)dx

Solution:

Given that, I = ∫(x + 1)exlog⁡(xex)dx

Let us assume, xex = t

(1 × ex + xex)dx = dt

(x + 1)exdx = 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 = xex (log⁡xex – 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




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!