# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.9 | Set 3

### Question 49.

**Solution:**

Given that I = ……..(i)

Let us consideredx + tan^{-1}x = t then,

On differentiating both side we get,d(x + tan

^{-1}x) = dt(1 + 1/(1 + x

^{2}))dx = dt((1 + x

^{2 }+ 1)/(1 + x^{2}))dx = dt((x

^{2 }+ 2))/((x^{2 }+ 1)) dx = dt

Now on puttingx + tan^{-1}x = t and ((x^{2 }+ 2)/(x^{2 }+ 1))dx = dt in equation (i), we getI = ∫5

^{t}dt= 5

^{t}/(log5) + c= /(log5) + c

Hence, I = /(log5) + c

### Question 50.

**Solution:**

Given that I = ……(i)

Let us consideredmsin^{-1}x = t then,

On differentiating both side we get,d(msin

^{-1}x) = dtm 1/√(1 – x

^{2}) dx = dtdx/√(1 – x

^{2}) = dt/m

Now on puttingmsin^{-1}x = t and dx/√(1 – x^{2}) = dt/m in equation (i), we getI = ∫e

^{t}dt/m= 1/m e

^{t}+c=

Hence, I =

### Question 51. ∫(cos√x)/√x dx

**Solution:**

Given that I = ∫(cos√x)/√x dx

Let us considered√x = t then,

On differentiating both side we get,1/(2√x) dx = dt

Now,

= ∫ (cos√x)/√x dx

= 2∫ costdt2

= 2sint + c

Hence, I = 2sin√x + c

### Question 52. ∫sin(tan^{-1}x)/(1 + x^{2}) dx

**Solution:**

Given that I = ∫sin(tan

^{-1}x)/(1 + x^{2}) dx ……(i)

Let us consideredtan^{-1 }= t then,

On differentiating both side we get,d(tan

^{-1}x) = dt1/(1 + x

^{2}) dx = dt

Now on puttingtan^{-1}x = t and dx/(1 + x^{2}) = dt in equation (i), we getI = ∫ sintdt

= -cost + c

= -cos(tan

^{-1}x) + cHence, I = -cos(tan

^{-1}x) + c

### Question 53. ∫(sin(logx))/x dx

**Solution:**

Given that I = ∫(sin(logx))/x dx ……..(i)

Let us consideredlogx = t then,

On differentiating both side we get,d(logx) = dt

1/x dx = dt

Now on puttinglogx = t and 1/x dx = dt in equation (i), we getI = ∫sintdt

= -cost + c

= -cos(logx) + c

Hence, I = -cos(logx) + c

### Question 54.

**Solution:**

Given that I =

Let us considered tan

^{-1}x = t, thenOn differentiating the above function we have,

1/(1 + x

^{2}) dx = dt= ∫e

^{mt }× dt== e

^{mt}/mOn Substituting the value of t, we

I = + c

### Question 55. ∫x/(√(x^{2 }+ a^{2}) + √(x^{2 }– a^{2})) dx

**Solution:**

Given that I = ∫x/(√(x

^{2 }+ a^{2}) + √(x^{2 }– a^{2})) dx= ∫ x/(√(x

^{2}+ a²) + √(x^{2}– a^{2})) × (√(x^{2}+ a^{2}) – √(x^{2}– a^{2}))/(√(x^{2 }+ a^{2}) – √(x^{2}– a^{2})) dx= ∫ x(√(x

^{2 }+ a^{2}) – √(x^{2 }– a^{2}))/(x^{2 }+ a^{2 }– x^{2 }+ a^{2}) dx= ∫ x/(2a

^{2}) (√(x^{2 }+ a^{2}) – √(x^{2 }– a^{2}))dxI = 1/(2a

^{2}) ∫x(√(x^{2 }+ a^{2}) – √(x^{2 }– a^{2}))dx ……(i)Let us considered x

^{2 }= t then,On differentiating the above function we have,

d(x

^{2}) = dt2xdx = dt

xdx = dt/2

Now on puttingx^{2 }= t and xdx = dt/2 in equation (i), we getI = 1/(2a

^{2}) ∫(√(t + a^{2}) – √(t – a^{2})) dx/2Hence, I = 1/(4a

^{2}) [2/3 (t + a^{2})^{3/2 }– 2/3 (t – a^{2})^{3/2}] + c

### Question 56. ∫x(tan^{-1}x^{2})/(1 + x^{4}) dx

**Solution:**

Given that I = ∫x(tan

^{-1}x^{2})/(1 + x^{4}) dx ……..(i)Let us considered tan

^{-1}x^{2 }= t then,On differentiating the above function we have,

d(tan

^{-1}x^{2}) = dt(1 × 2x)/(1 + (x

^{2})^{2}) dx = dt(1 × x)/(1 + x

^{4}) dx = dt/2

Now on puttingtan^{-1}x^{2 }= t and x/(1 + x^{4}) dx = dt/2 in equation (i), we getI = ∫ t dx/2

= 1/2 ∫tdt

= 1/2 × t

^{2}/2 + c= t

^{2}/4 + c – 1= (tan

^{-1}x^{2})^{2}/4 + cHence, I = 1/4 (tan

^{-1}x^{2})^{2 }+ c

### Question 57. ∫(sin^{-1}x)^{3}/√(1 – x^{2}) dx

**Solution:**

Given that I = ∫(sin

^{-1}x)^{3}/√(1 – x^{2}) dx ……(i)Let us considered sin

^{-1}x = t then,On differentiating the above function we have,

d(sin

^{-1}x) = dt1/√(1 – x

^{2}) dx = dt

Now on puttingsin^{-1}x = t and 1/√(1 – x^{2}) dx = dt in equation (i), we getI = ∫t

^{3}dt= t

^{4}/4 + cHence, I = 1/4 (sin

^{-1}x)^{4 }+ c

### Question 58.∫(sin(2 + 3logx))/x dx

**Solution:**

Given that I = ∫(sin(2 + 3logx))/x dx ……..(i)

Let us considered 2 + 3logx = t then,

On differentiating the above function we have,

d(2 + 3logx) = dt

3 1/x dx = dt

dx/x = dt/3

Now on putting2 + 3logx = t and dx/x = dt/3 in equation (i), we getI = ∫sint dt/3

= 1/3(-cost) + ct

= -1/3 cos(2 + 3logx) + c

Hence, I = -1/3 cos(2 + 3logx) + c

### Question 59.

**Solution:**

Given that I = ……(i)

Let us considered x

^{2}= t then,On differentiating the above function we have,

d(x

^{2}) = dt2xdx = dt

xdx = dt/2

Now on puttingx^{2 }= t and xdx = dt/2 in equation (i), we getI = ∫e

^{t}dt/2= 1/2 e

^{t}+c= 1/2

^{ }+ cHence, I = 1/2

^{ }+ c

### Question 60. ∫e^{2x}/(1 + e^{x}) dx

**Solution:**

Given that I = ∫e

^{2x}/(1 + e^{x}) dx …….(i)Let us considered 1 + e

^{x }= t then,On differentiating the above function we have,

d(1 + e

^{x}) = dte

^{x}dx = dtdx = dt/e

^{x}

Now on putting1 + e^{x }= t and dx = dt/e^{x }in equation (i), we getI = ∫e

^{2x}/t × dt/e^{x}= ∫e

^{x}/t dt= ∫ (t – 1)/t dt

= ∫ (t/t – 1/t)dt

= t – log|t| + c

= (1 + e

^{x}) – log|1 + e^{x}| + cHence, I = 1 + e

^{x }– log|1 + e^{x}| + c

### Question 61. ∫(sec^{2}√x)/√x dx

**Solution:**

Given that I = ∫(sec

^{2}√x)/√x dx ……(i)Let us considered √x = t then,

On differentiating the above function we have,

d(√x) = dt

1/(2√x) dx = dt

dx = 2√x dt

dx = 2tdt [√x = t])

Now on putting√x = t and dx = 2tdt in equation (i), we getI = ∫ (sec

^{2}t)/t × 2tdt= 2∫ sec

^{2}tdt= 2tant + c

= 2tan√x + c

Hence, I = 2tan√x + c

### Question 62. ∫tan^{3}2x sec2x dx

**Solution:**

Given that I = ∫tan

^{3}2x sec2x dx= tan

^{2}2xtan2x sec2x= (sec

^{2}2x – 1)tan2x sec2x= sec

^{2}2x.tan2xsec2x – tan2xsec2x= ∫ sec

^{2}2xtan2xsec2xdx – ∫tan2xsec2xdx= ∫ sec

^{2}2xtan2xsec2xdx – (sec2x)/2 + cLet us considered sec2x = t

2sec2xtan2xdx = dt

I = 1/2 ∫t

^{2}dt – (sec2x)/2 + cI = t

^{3}/6 – (sec2x)/2 + cHence, I = (sec2x)

^{3}/6 – (sec2x)/2 + c

### Question 63. ∫(x + √(x + 1))/(x + 2) dx

**Solution:**

Given that I = ∫(x+√(x+1))/(x+2) dx …….(i)

Let us considered x + 1 = t

^{2}then,On differentiating the above function we have,

d(x + 1) = d(t

^{2})dx = 2tdt

Now on putting x + 1 = t

^{2}and dx = 2tdt in equation (i), we getI = ∫ (x + √(t

^{2}))/(x + 2) 2tdt= 2∫((t

^{2 }– 1) + t)/((t^{2 }– 1) + 2) × tdt [x + 1 = t^{2}]= 2∫(t

^{2 }+ t – 1)/(t^{2 }+ 1) tdt= 2∫ (t

^{3 }+ t^{2 }– t)/(t^{2}+ 1) dt= 2[∫ t

^{3}/(t^{2 }+ 1) dt + ∫ t^{2}/(t^{2 }+ 1) dt – ∫ t/(t^{2 }+ 1) dt]I = 2[∫t

^{3}/(t^{2 }+ 1) dt + ∫t^{2}/(t^{2 }+ 1) dt – ∫t/(t^{2 }+ 1) dt] ……(ii)Let I

_{1 }= ∫t^{3}/(t^{2 }+ 1) dtI

_{2 }= ∫t^{2}/(t^{2 }+ 1) dtand I

_{3 }= ∫t/(t^{2 }+ 1) dtNow, I

_{1 }= ∫t^{3}/(t^{2 }+ 1) dt= ∫(t – t/(t

^{2 }+ 1))dt= t

^{2}/2 – 1/2 log(t^{2 }+ 1)_{ }I

_{1 }= t^{2}/2 – 1/2 log(t^{2 }+ 1) + c_{1 }……..(iii)Since, I

_{2 }= ∫t^{2}/(t^{2 }+ 1) dt= ∫ (t

^{2 }+ 1 – 1)/(t^{2 }+ 1) dt= ∫(t

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

^{2 }+ 1) dtI

_{2 }= t – tan^{-1}(t^{2}) + c_{2}………….(iv)and,

I

_{3 }= ∫t/(t^{2 }+ 1) dt= 1/2 log(1 + t

^{2}) + c_{3}……..(v)Using equations (ii), (iii), (iv) and (v), we get

I = 2[t

^{2}/2 – 1/2 log(t^{2 }+ 1) + c_{1 }+ t-tan^{-1}(t^{2}) + c_{2 }– 1/2 log(1 + t^{2}) + c_{3}]= 2[t

^{2}/2 + t-tan^{-1}(t^{2}) – log(1 + t^{2}) + c_{1 }+ c_{2 }+ c_{3}]= 2[t

^{2}/2 + t – tan^{-1}(t^{2}) – log(1 + t^{2}) + c_{4}[Putting c_{1 }+ c_{2 }+ c_{3 }= c_{4}]= t

^{2 }+ 2t – 2tan^{-1}(t^{2}) – 2log(1 + t^{2}) + 2c_{4}= (x + 1) + 2√(x + 1) – 2tan

^{-1}(√(x + 1)) – 2log(1 + x + 1) + 2c_{4}= (x + 1) + 2√(x + 1) – 2tan

^{-1}(√(x + 1)) – 2log(x + 2) + c [Putting 2c_{4 }= c]Hence, I = (x + 1) + 2√(x + 1) – 2tan

^{-1}(√(x + 1)) – 2log(x + 2) + c

### Question 64.

**Solution:**

Given that I = …….(1)

Let us considered = t then

On differentiating the above function we have,

d() = dt

× × 5

^{x}× (log5)^{3}dx = dtdx = dt/((log5)^3 ))

Now on putting = t and dx = dt/((log5)^3 )) in equation (i), we get

I = ∫dt/((log5)

^{3})= 1/((log5)

^{3}) ∫dt= t/((log5)^3) + c

Hence, I = /((log5)

^{3}) + c)

### Question 65. ∫1/(x√(x^{4 }– 1)) dx

**Solution:**

Given that I = ∫1/(x√(x

^{4 }– 1)) dx ……….(i)Let us considered x

^{2}= t then,On differentiating the above function we have,

d(x

^{2}) = dt2xdx = dt

dx = dt/2x

Now on putting x

^{2}= t and dx = dt/2x in equation (i), we getI = ∫ 1/(x√(t

^{2 }– 1)) × dt/2x= 1/2 ∫ 1/(x

^{2}√(t^{2 }– 1)) dt= 1/2 ∫ 1/(t√(t

^{2 }– 1)) dt= 1/2 sec

^{-1}t + c= 1/2 sec

^{-1}x^{2 }+ cHence, I = 1/2 sec

^{-1}x^{2}+ c

### Question 66. ∫√(e^{x }– 1) dx

**Solution:**

Given that I = ∫√(e

^{x }– 1) dx ……..(i)Let us considered e

^{x }– 1 = t^{2}then,On differentiating the above function we have,

d(e

^{x }– 1) = dt(t^{2})e

^{x}dx = 2tdtdx = 2t/e

^{x }dtdx = 2t/(t

^{2}+ 1) dt [e^{x }– 1 = t^{2}]Now on putting e

^{x }– 1 = t² and dx = 2tdt/(t^{2}+ 1) in equation (i), we getI = ∫√(t

^{2}) × 2tdt/(t^{2}+ 1)= 2∫(t × t)/(t

^{2}+ 1) dt= 2∫t

^{2}/(t^{2}+ 1) dt= 2∫(t

^{2}+ 1 – 1)/(t^{2}+ 1) dt= 2∫[(t

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

^{2}+ 1) dt= 2t – 2tan

^{-1}(t) + c= 2√(e

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

^{x }– 1) – 2tan^{-1}√(e^{x }– 1) + c

### Question 67. ∫ 1/(x + 1)(x^{2 }+ 2x + 2) dx

**Solution:**

Given that I = ∫1/(x + 1)(x

^{2 }+ 2x + 2) dx= ∫1/(x + 1)((x + 1)

^{2 }+ 1) dxLet us considered x + 1 = tan u then, [tanu = Perpendicular/Base = (x + 1)/1]

On differentiating the above function we have,

dx = sec

^{2}u du [Hypotenuses = √(x^{2 }+ 2x + 2)]I = ∫sec

^{2}u/tanu(tan^{2}u + 1) du= ∫ cosu/sinu du

= log| sinu |+c

= log| sin(x + 1)| + c [As we know, sin(x + 1) = P/H = (x + 1)/√(x

^{2 }+ 2x + 2)]Hence, I = log| x + 1/√(x

^{2 }+ 2x + 2)| + c

### Question 68. ∫x^{5}/√(1 + x^{3}) dx

**Solution:**

Given that I = ∫ x

^{5}/(√(1 + x^{3})) dx …….(i)Let us considered 1 + x

^{3 }= t^{2}, thenOn differentiating the above function we have,

d(1 + x

^{3}) = d(t^{2})3x

^{2}dx = 2t * dtdx = 2t dt/3x

^{2}Now on putting 1 + x

^{3 }= t^{2}and dx = 2tdt/3x^{2}in equation (i), we getI = ∫ x

^{5}/√t^{2}* 2t/3x^{2 }dt= ∫x

^{5}/t * 2t/3x^{2 }dt= 2/3∫x

^{3}dt= 2/3 ∫( t

^{2 }– 1) dt= 2/3[t

^{3}/3 – 2t/3] + cHence, I = 2/9(1 + x

^{3})^{3/2 }– 2 √(1 + x^{2})/3 + c

### Question 69. ∫4x^{3} √(5 – x^{2}) dx

**Solution:**

Given that I = ∫4x

^{3}√(5 – x^{2}) dx ……(i)Let us considered 5 – x

^{2 }= t^{2 }then,On differentiating the above function we have,

d(5 – x

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

dx = (-t)/x dt

Now on putting 5 – x

^{2 }= t^{2}and dx = (-t)/x dt in equation (i), we getI = ∫4x

^{3}√(t^{2}) × (-t)/x dt= -4∫ x

^{2}t × tdt= -4∫(5 – t

^{2}) t^{2}dt [5 – x^{2 }= t^{2}]= -4∫(5t

^{2 }– t^{5})dt= -20×t

^{3}/3 + 4 t^{5}/5 + c= (-20)/3 × t

^{3 }+ 4/5 × t^{5 }+ c= (-20)/3 × (5 – x

^{2})^{3/2 }+ 4/5 × (5 – x^{2})^{5/2 }+ cI = (-20)/3 × (5 -x

^{2})^{3/2 }+ 4/5 × (5 – x^{2})^{5/2 }+ c

### Question 70. ∫1/(√x + x) dx

**Solution:**

Given that I = ∫1/(√x + x) dx ……..(i)

Let us considered √x = t then,

On differentiating the above function we have,

d(√x) = dt

1/(2√x) dx = dt

dx = 2√x dt

Now on putting √x = t and 2√x dt = dx in equation (i), we get

I = ∫1/(t + t

^{2}) 2t × dt [Since √x = t and x = t^{2}]= ∫2t/(t(1 + t)) dt

= 2∫t/(1 + t) dt

= 2log|1 + t| + c

= 2log|1 + √x| + c

Hence, I = 2log|1 + √x|+c

### Question 71. ∫1/(x^{2} (x^{4 }+ 1)^{3/4}) dx

**Solution:**

Given that I = 1/(x

^{2}(x^{4}+1)^{3/4})Multiplying and dividing by x

^{-3}, we obtain(x

^{-3}/(x^{2}x^{-3}(x^{4}+ 1)^{3/4}) = (x^{-3}(x^{4 }+ 1)^{-3/4}/(x^{2}x^{-3}))= (x

^{4 }+ 1)^{-3/4}/(x^{5}(x^{4})^{-3/4}= 1/x

^{5}((x^{4 }+ 1)/x^{4})^{-3/4}= 1/x

^{5}(1 + 1/x^{4})^{-3/4}Let us considered 1/x

^{4}= t-4/x

^{5 }dx = dt1/x

^{5 }dx = -dt/4I = ∫ 1/x

^{5}(1 + 1/x^{4})^{-3/4}dx= -1/4 ∫(1 + t)

^{-3/4}dt= -1/4 [(1 + t)

^{1/4})/(1/4)] + c= -1/4(1 + 1/x

^{4})^{1/4}/(1/4) + cHence, I = -(1 + 1/x

^{4})^{1/4 }+ c

### Question 72. ∫(sin^{5}x)/(cos^{4}x) dx

**Solution**:

Given that I = ∫(sin

^{5}x)/(cos^{4}x) dx ……(i)Let us considered cosx = t then,

On differentiating the above function we have,

d(cosx) = dt

-sinxdx = dt

dx = -dt/(sinx)

Now on putting cosx = t and dx = -dt/(sinx) in equation (i), we get

I = ∫(sin

^{5}x)/t^{4 }× -dt/(sinx)= -∫(sin

^{4}x)/t^{4}dt= -∫(1 – cos

^{2}x)^{2}/t^{4 }dt= -∫(1 – t

^{2})^{2}/t^{4 }dt= -∫(1 + t

^{4 }– 2t^{2})/t^{4}dt= -∫(1/t

^{4}+ t^{4}/t^{4}– (2t^{2})/t^{4})dt= -∫(t

^{-4 }+ 1 – 2t^{-2})dt= -[t

^{-3}/(-3) + t – 2 t^{-1}/(-1)] + c= 1/3 * 1/t

^{3}– t – 1/t + cHence, I = 1/3 * 1/cos

^{3}x – cosx -2/cosx + c

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