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

### Evaluate the following integrals:

### Question 21. ∫(logx)^{2} x dx

**Solution:**

Given that, I = ∫(logx)

^{2}x dxUsing integration by parts,

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

We get

I = (logx)

^{2}∫xdx – ∫(2(logx)(1/x) ∫xdx)dx= x

^{2}/2(logx)^{2 }– 2∫(logx)(1/x)(x^{2}/2)dx= x

^{2}/2(logx)^{2 }– ∫x(logx)dx= x

^{2}/2(logx)^{2 }– [logx∫xdx – ∫ (1/x ∫xdx)dx]= x

^{2}/2(logx)^{2 }– [x^{2}2/2 logx – ∫(1/x × x^{2}/2)dx]= x

^{2}/2(logx)^{2 }– x^{2}/2 logx + 1/2 ∫xdx= x

^{2}/2(logx)^{2 }– x^{2}/2 logx + 1/4 x^{2 }+ cHence, I = x

^{2}/2 [(logx)^{2 }– logx + 1/2] + c

### Question 22. ∫e^{√x} dx

**Solution**:

Given that, I = ∫ e

^{√x}dxLet us assume, √x = t

x = t

^{2}dx = 2tdt

I = 2∫ e

^{t}tdtUsing integration by parts,

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

We get

I = 2[t∫e

^{t}dt – ∫(1∫e^{t}dt)dt]= 2[te

^{t }– ∫e^{t}dt]= 2[te

^{t }– e^{t}] + c= 2e

^{t}(t – 1) + cHence, 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}) dxLet us assume (1/(x + 2) = t

-1/((x + 2)

^{2}) dx = dtI = -∫log(1/t)dt

= -∫logt

^{-1}dt= -∫1 × logtdt

Using integration by parts,

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

We get

I = logt∫dt – ∫(1/t ∫dt)dt

= tlogt – ∫(1/t × t)dt

= tlogt – ∫dt

= tlogt – t + c

= 1/(x + 2) (log(x + 2)

^{-1 }– 1) + cHence, I = (-1)/(x + 2) – (log(x + 2))/(x + 2) + c

### Question 24. ∫(x + sinx)/(1 + cosx) dx

**Solution:**

Given that, I = ∫(x + sinx)/(1 + cosx) dx

= ∫x/(2cos

^{2}x/2) dx + ∫(2sinx/2 cosx/2)/(2cos^{2}x/2) dx= 1/2 ∫xsec

^{2}x/2 dx + ∫tanx/2 dxUsing integration by parts,

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

We get

= 1/2 [x∫sec

^{2}x/2 dx – ∫(1∫ sec²x/2 dx)dx] + ∫tanx/2 dx= 1/2 [2xtanx/2 – 2∫tanx/2 dx] + ∫tanx/2 dx + c

= xtanx/2 – ∫tanx/2 dx + ∫tanx/2 dx+c

Hence, I = xtanx/2 + c

### Question 25. ∫log_{10}xdx

**Solution:**

Given that, I = ∫log

_{10}xdx= ∫(logx)/(log10) dx

= 1/(log10) ∫1 × logxdx

Using integration by parts,

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

We get

= 1/(log10) [logx∫dx – ∫(1/x ∫dx)dx]

= 1/(log10) [xlogx – ∫(x/x)dx]

= 1/(log10)[xlogx – x]

Hence, I = (x/(log10)) × (logx – 1) + c

### Question 26. ∫cos√x dx

**Solution:**

Given that, I = ∫cos√x dx

Let us assume, √x = t

x = t

^{2}dx = 2tdt

= ∫2tcostdt

I = 2∫tcostdt

Using integration by parts,

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

We get

I = 2[t]costdt – ∫(1 ∫costdt)dt]

= 2[tsint – ∫sintdt]

= 2[tsint + cost] + c

Hence, I = 2[√x sin√x + cos√x] + c

### Question 27. ∫(xcos^{-1}x)/√(1 – x^{2}) dx

**Solution:**

Given that, I = ∫(xcos

^{-1}x)/√(1 – x^{2}) dxLet us assume, t = cos

^{-1}xdt = (-1)/√(1 – x

^{2}) dxAlso, cost = x

I = -∫tcostdt

Now, using integration by parts,

So, let

u = t;

du = dt

∫costdt = ∫dv

sint = v

Therefore,

I = -[tsint – ∫sintdt]

= -[tsint + cost] + c

On substituting the value t = cos

^{-1}x we get,I = -[cos

^{-1}xsint + x] + cHence, I = -[cos

^{-1}x√(1 – x²) + x] + c

### Question 28. ∫cosec^{3}xdx

**Solution:**

Given that, I =∫cosec

^{3}xdx=∫cosecx × cosec

^{2}xdxUsing integration by parts,

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

We get

= cosecx × ∫cosec

^{2}xdx + ∫(cosecxcotx]cosec^{2}xdx)dx= cosecx × (-cotx) + ∫cosecxcotx(-cotx)dx

= -cosecxcotx – ∫cosecxcot

^{2}xdx= -cosecxcotx – ∫cosecx(cosec

^{2}x – 1)dx= -cosecxcotx – ∫cosec

^{3}xdx + ∫cosecxdxI = -cosecxcotx – I + log|tanx/2| + c

_{1}2l = -cosecxcotx + log|tanx/2| + c

_{1}Hence, I = -1/2cosecxcotx + 1/2 log|tanx/2| + c

### Question 29. ∫sec^{-1}√x dx

**Solution:**

Given that, I = ∫sec

^{-1}√x dxLet us assume, √x = t

x = t

^{2}dx = 2tdt

I = ∫2tsec

^{-1}tdtUsing 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√(t^{2}-1))∫tdt)dt]= 2[t

^{2}/2 sec^{-1 }– ∫(t/(2t√(t^{2 }– 1)))dt]= t

^{2}sec^{-1}t – ∫t/√(t^{2 }– 1) dt= t

^{2}sec^{-1}t – 1/2∫2t/√(t^{2 }– 1) dt= t

^{2}sec^{-1}t – 1/2 × 2√(t^{2 }– 1) + cHence, I = xsec

^{-1}√x – √(x – 1) + c

### Question 30. ∫sin^{-1}√x dx

**Solution**:

Given that, I = ∫sin

^{-1}√x dxLet us assume, x = t

dx = 2tdt

∫sin

^{-1}√x dx = ∫sin^{-1}√(t^{2}) 2tdt= ∫sin

^{-1}t2tdt= sin

^{-1}t∫2tdt – (∫(dsin^{-1}t)/dt (∫2tdt)dt= sin

^{-1}t(t^{2}) – ∫1/√(1 – t^{2}) (t^{2})dtNow, lets solve ∫1/√(1 – t

^{2}) (t^{2})dt∫1/√(1 – t

^{2}) (t^{2})dt = ∫(t^{2 }– 1 + 1)/√(1 – t^{2}) dt= ∫(t

^{2 }– 1)/√(1 – t^{2}) dt + ∫1/√(1 – t^{2}) dtAs we know that, value of ∫1/√(1 – t

^{2}) dt = sin^{-1}tSo, the remaining integral to evaluate is

∫(t

^{2 }– 1)/√(1 – t^{2}) dt= ∫-√(1 – t^{2}) dtNow, substitute, t = sinu, dt = cosudu, we gte

∫-√(1 – t

^{2}) dt = ∫-cos^{2}udu = -∫[(1 + cos2u)/2]du= -u/2-(sin2u)/4

Now substitute back u = sin

^{-1}x 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))/4sin(2sin

^{-1}√x) = 2√x √(1 – x)Hence, I = xsin

^{-1}√x – (sin^{-1}√x)/2 – √(x(1 – x))/2 + c

### Question 31. ∫xtan^{2}xdx

**Solution:**

Given that, I =∫xtan

^{2}xdx= ∫x(sec

^{2}x – 1)dxUsing integration by parts,

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

We get

= ∫xsec8xdx – ∫xdx

= [x∫sec

^{2}xdx – ∫(1∫sec^{2}xdx)dx] – x^{2}/2= xtanx – ∫tanxdx – x

^{2}/2Hence, I = xtanx – log|secx| – x

^{2}/2 + c

### Question 32. ∫ x((sec2x – 1)/(sec2x + 1))dx

**Solution:**

Given that, I = ∫ x((sec2x – 1)/(sec2x + 1))dx

= ∫x((1 – cos2x)/(1 + cos2x))dx

= ∫x((sec

^{2}x)/(cos^{2}x))dx= ∫xtan

^{2}xdx= ∫x(sec

^{2}x – 1)dxUsing integration by parts,

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

We get

= ∫xsec

^{2}xdx – ∫dx= [x∫sec

^{2}xdx – ∫(1∫ sec^{2}xdx)dx] – x^{2}/2= xtanx – ∫tanxdx – x

^{2}/2= xtanx – log|secx| – x

^{2}/2 + cHence, I = xtanx – log|secx| – x

^{2}/2 + c

### Question 33. ∫(x + 1)e^{x}log(xe^{x})dx

**Solution:**

Given that, I = ∫(x + 1)e

^{x}log(xe^{x})dxLet us assume, xe

^{x }= t(1 × e

^{x }+ xe^{x})dx = dt(x + 1)e

^{x}dx = dtI = ∫logtdt

= ∫1 × logtdt

Using integration by parts,

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

We get

= logt∫dt – ∫(1/t∫dt)dt

= tlogt – ∫(1/t × t)dt

= tlogt – ∫dt

= tlogt – t + c

= t(logt – 1) + c

Hence, I = xe

^{x}(logxe^{x }– 1) + c

### Question 34. ∫sin^{-1}(3x – 4x^{3})dx

**Solution:**

Given that, I = ∫sin

^{-1}(3x – 4x^{3})dxLet us assume, x = sinθ

dx = cosθdθ

= ∫sin

^{-1}(3sinθ – 4sin^{3}θ)cosθdθ= ∫sin

^{-1}(sin3θ)cosθdθ= ∫3θcosθdθ

Using integration by parts,

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

We get

= 3[θ]cosθdθ – ∫(1∫cosθdθ)dθ]

= 3[θsinθ – ∫sinθdθ]

= 3[θsinθ + cosθ] + c

Hence, I = 3[xsin

^{-1}x + √(1 – x^{2})] + c)

### Question 35. ∫sin^{-1}(2x/(1 + x^{2}))dx.

**Solution:**

Given that, I = ∫sin

^{-1}(2x/(1 + x^{2}))dxLet us assume, x = tanθ

dx = sec

^{2}θdθsin

^{-1}(2x/(1 + x^{2})) = sin^{-1}((2tanθ)/(1 + tan²θ))= sin

^{-1}(sin2θ) = 2θ∫sin

^{-1}(2x/(1 + x^{2}))dx = ∫2θsec^{2}θdθ = 2∫θsec^{2}θdθUsing integration by parts,

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

We get

2[θ∫sec

^{2}θdθ – ∫{(d/dθ θ) ∫sec^{2}θdθ}dθ= 2[θtanθ – ∫tanθdθ]

= 2[θtanθ + log|cosθ|] + c

= 2[xtan

^{-1}x + log|1/√(1 + x^{2})|] + c= 2xtan

^{-1}x + 2log(1 + x^{2})^{1/2 }+ c= 2xtan

^{-1}x + 2[-1/2 log(1 + x^{2})] + c= 2xtan

^{-1}x – log(1 + x^{2}) + cHence, I = 2xtan

^{-1}x – log(1 + x^{2}) + c

### Question 36. ∫ tan^{-1}((3x – x^{3})/(1 – 3x^{2}))dx

**Solution:**

Given that, I = ∫tan

^{-1}((3x – x^{3})/(1 – 3x^{2}))dxLet us assume, x = tanθ

dx = sec

^{2}θdθI = ∫tan

^{-1}((3tanθ – tan^{3}θ)/(1 – 3tan^{2}x)) sec^{2}θdθ=∫tan

^{-1}(tan3θ)sec^{2}θdθ= ∫3θsec

^{2}θdθUsing integration by parts,

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

We get

= 3[θ∫ sec

^{2}θdθ – ∫(1∫sec^{2}θdθ)dθ]= 3[θtanθ – ∫tanθdθ]

= 3[θtanθ + logsecθ] + c

= 3[xtan

^{-1}x – log√(1 + x^{2})] + cHence, I = 3[xtan

^{-1}x – log√(1 + x^{2})] + c

### Question 37. ∫x^{2}sin^{-1}xdx

**Solution:**

Given that, I = ∫x

^{2}sin^{-1}xdxI = sin

^{-1}x∫x^{2}dx – ∫(1/√(1 – x^{2}) ∫x^{2}dx)dx= x

^{3}/3 sin^{-1}x – ∫x^{3}/(3√(1 – x^{2})) dxI = x

^{3}/3 sin^{-1}x – 1/3 I_{1 }+ c_{1 }…..(1)Let I

_{1 }= ∫x^{3}/√(1 – x^{2}) dxLet 1 – x

^{2 }= t^{2}-2xdx = 2tdt

-xdx = tdt

I

_{1 }= -∫(1 – t^{2})tdt/t= ∫(t

^{2 }– 1)dt= t

^{3}/3 – t + c_{2}= (1 – x

^{2})^{3/2}/3 – (1 – x^{2})^{1/2 }+ c_{2}Now, put the value of I

_{1}in eq(1), we getHence, I = x

^{3}/3 sin^{-1}x – 1/9 (1 – x^{2})^{3/2 }+ 1/3 (1 – x^{2})^{1/2 }+ c

### Question 38. ∫(sin^{-1}x)/x^{2}dx

**Solution:**

Given that, I =∫(sin

^{-1}x)/x^{2}dx= ∫(1/x

^{2})(sin^{-1}x)dxI = [sin

^{-1}x∫1/x^{2}dx – ∫(1/√(1 – x^{2}) ∫1/x^{2}dx)dx]= sin

^{-1}×(-1/x) – ∫1/√(1 – x^{2}) (-1/x)dxI = -1/x sin

^{-1}x + ∫1/(x√(1 – x^{2})) dxI = -1/x sin

^{-1}x + I_{1 }…….(1)Where,

I

_{1 }= ∫1/(x√(1 – x^{2})) dxLet 1 – x

^{2 }= t^{2}-2xdx = 2tdt

I

_{1 }= ∫x/(x^{2}√(1 – x^{2})) dx= -∫tdt/((1 – t

^{2}) √t)= -∫dt/((1 – t

^{2}))= ∫1/(t

^{2 }– 1) dt= 1/2 log|(t – 1)/(t + 1)|

= 1/2 log|(t – 1)/(t + 1)|

= 1/2 log|(√(1 – x

^{2}) – 1)/(√(1 – x^{2}) + 1)| + c_{1}Now, put the value of I

_{1}in eq(1), we getI = -(sin

^{-1}x)/x + 1/2 log|((√(1 – x^{2}) – 1)/(√(1 – x^{2}) + 1))((√(1 – x^{2}) – 1)/(√(1 – x^{2}) – 1))| + c= -(sin

^{-1}x)/x + 1/2 log|(√(1 – x^{2}) – 1)^{2}/(1 – x^{2 }– 1)| + c= -(sin

^{-1}x)/x + 1/2 log|(√(1 – x^{2}) – 1)^{2}/(-x^{2})| + c= -(sin

^{-1}x)/x + log|(√(1 – x^{2}) – 1)/(-x)| + cHence, I = -(sin

^{-1}x)/x + log|(1 – √(1 – x^{2}))/x| + c

### Question 39. ∫(x^{2} tan^{-1}x)/(1 + x^{2}) dx

**Solution:**

Given that, I = ∫(x

^{2}tan^{-1}x)/(1 + x^{2}) dxLet us assume, tan

^{-1}x = t [x = tant]1/(1 + x

^{2}) dx = dtI = ∫t × tan

^{2}tdt= ∫t(sec

^{2}t – 1)dt= ∫(tsec

^{2}t – t)dtUsing integration by parts,

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

We get

= ∫tsec

^{2}tdt – ∫tdt= [t∫sec

^{2}tdt – ∫(1)sec^{2}tdt)dt] – t^{2}/2= [t × tant – ∫tantdt] – t

^{2}/2= t tant – logsect – t

^{2}/2 + c= xtan

^{-1}x – log√(1 + x^{2}) – (tan^{2}x)/2 + cHence, I = xtan

^{-1}x – 1/2 log|1 + x^{2}| – (tan^{2}x)/2 + c

### Question 40. ∫cos^{-1}(4x^{3 }– 3x)dx

**Solution:**

Given that, I = ∫cos

^{-1}(4x^{3 }– 3x)dxLet us assume, x = cosθ

dx = -sinθdθ

I = -∫cos

^{-1}(4cos^{3}θ – 3cosθ)sinθdθ= – ∫cos

^{-1}(cos3θ)sinθdθ= -∫3θsinθdθ

Using integration by parts,

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

We get

= -3[θ]sinθdθ – ∫(1∫sinθdθ)dθ]

= -3[-θcosθ + ∫cosθdθ]

= 3θcosθ – 3sinθ + c

Hence, I = 3xcos

^{-1}x – 3√(1 – x^{2}) + c

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