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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

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

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