# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.9 | Set 1

### Evaluate the following integrals:

### Question 1. ∫(logx)/x dx

**Solution:**

Given that, I = ∫(logx)/x dx

Let us considered logx = t

Now differentiating both side we get,

d(logx) = dt

1/x dx = dt

dx = xdt

Then, put logx = t and dx = xdt, we get

I = ∫ t/x × (x)dt

= ∫ tdt

= t

^{2}/2 + c= (logx)

^{2}/2 + cHence, I = (logx)

^{2}/2 + c

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

**Solution:**

Given that I = ∫(log(1 + 1/x))/(x(1 + x)) dx ……(i)

Let us considered log(1 + 1/x) = t then

On differentiating both side we get,

d[log(1 + 1/x)]=dt

1/(1 + 1/x) × (-1)/x

^{2}dx = dt1/((x + 1)/x) × (-1)/x

^{2}dx = dt(-x)/(x

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

Now, on putting log(1 + 1/x) = t and dx/(x(x + 1)) = -dt in equation (i), we get

I = ∫t × -dt

= -t

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

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

^{2 }+ c

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

**Solution:**

Given that I = ∫((1 + √x)

^{2})/√x dxLet us considered (1 + √x) = t then,

On differentiating both side we get,

d(1 + √x) = dt

1/(2√x) dx = dt

dx = dt × 2√x

Now on putting (1 + √x) = t and dx = dt × 2√x, we get

I = ∫t

^{2}/√x × dt × 2√x= 2∫t

^{2}dt= 2 × t

^{3}/3 + c= 2/3[1 + √x]

^{3 }+ cHence, I = 2/3(1 + √x)

^{3 }+ c

### Question 4. ∫√(1 + e^{x}) e^{x} d

**Solution:**

Given that I = ∫√(1 + e

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

^{x }= t then,On differentiating both side we get,

d(1 + e

^{x}) = dte

^{x}dx = dtdx = dt/e

^{x}Now on putting 1 + e

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

^{x}× dt/e^{x}= ∫ t

^{1/2}dt= 2/3t

^{3/2 }+ c= 2/3 (1 + e

^{x})^{3/2}+c

### Question 5. ∫∛(cos^{2}x) sinx dx

**Solution:**

Given that I = ∫∛(cos

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

On differentiating both side we get,

d(cosx) = dt

-sinxdx = dt

dx = -dt/(sinx))

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

I = ∫∛(t

^{2}) sinx × (-dt)/(sinx)= -∫t

^{2/3}sinx dt/(sinx)= -∫ t

^{2/3}dt= -3/5 × t

^{5/3}Hence, I = -3/5(cosx)

^{5/3 }+ c

### Question 6. ∫e^{x}/(1 + e^{x})^{2} dx

**Solution:**

Given that I = ∫e

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

^{x }= t then,On differentiating both side we get,

d(1 + e

^{x}) = dte

^{x}dx = dtdx = dt/e

^{x}Now on putting 1 + e

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

^{x}/t^{2}× dt/e^{x}= ∫dt/t

^{2}= ∫t

^{-2}dt= -t

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

= -1/(1 + e

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

^{x}) + c

### Question 7. ∫cot^{3}x cosec^{2}x dx

**Solution:**

Given that I = ∫cot

^{3}x cosec^{2}x dx …….(i)Let us considered cotx = t then,

On differentiating both side we get,

d(cotx) = dt

-cosec

^{2}x dx = dtdx = -dt/cosec

^{2}xNow on putting cotx = t and dx = -dt/(cosec

^{2}x) in equation (i), we getI = ∫ t

^{3 }cosec^{2}x × (-dt)/(cosec^{2}x)= -∫ t

^{3}dt= -t

^{4}/4 + c= -(cot

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

^{4}x)/4 + c

### Question 8.

**Solution:**

Given that I =[Tex] [/Tex] …….(i)

Let us considered sin

^{-1}x = t then,On differentiating both side we get,

d(sin

^{-1}x) = dt1/√(1 – x

^{2})dx = dtdx = √(1 – x

^{2}) dt)Now on putting sin

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

^{t})^{2}/√(1 – x^{2}) × √(1 – x^{2}) dt= ∫e

^{2t}dt= e

^{2t}/2 + c= + c

Hence, I = + c

### Question 9. ∫(1 + sinx)/√(x – cosx) dx

**Solution:**

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

Let us considered x – cosx = t, then

On differentiating both side we get,

d(x – cosx) = dt

[1 – (-sinx)]dx = dt

(1 + sinx)dx = dt

Now on putting x – cosx = t and (1 + sinx)dx = dt in equation (i), we get

I = ∫ dt/√t

= ∫ t

^{-1/2}dt= 2t

^{1/2 }+ c= 2(x – cosx)

^{1/2 }+ cHence, I = 2√(x – cosx) + c

### Question 10. ∫1/(√(1 – x^{2}) (sin^{-1}x)^{2}) dx

**Solution:**

Given that I = ∫1/(√(1 – x

^{2}) (sin^{-1}x)^{2}) dx …..(i)Let us considered sin

^{-1}x = t then,On differentiating both side we get,

d(sin

^{-1}x) = dt1/√(1 – x

^{2}) dx = dtNow on putting sin

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

^{2}= ∫t

^{-2}dt= -t

^{-1 }+ c= (-1)/t + c

= (-1)/(sin

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

^{-1}x) + c

### Question 11. ∫(cotx)/√(sinx) dx

**Solution:**

Given that I = ∫(cotx)/√(sinx) dx …….(i)

Let us considered sinx = t then,

On differentiating both side we get,

d(sinx) = dt

cosxdx = dt

Now, I = ∫(cotx)/√(sinx) dx

= ∫(cosx)/(sinx√(sinx)) dx

= ∫ cosx/(sinx)

^{3/2}dx= ∫ cosx/(sinx)

^{3/2}dx …….(ii)Now on putting sinx = t and cosxdx = dt in equation (ii), we get

I = ∫ dt/t

^{3/2}= ∫ t

^{-3/2}dt= -2t

^{-1/2 }+ c= -2/√t + c

= -2/√(sinx) + c

I = -2/√sinx + c

### Question 12. ∫(tanx)/√(cosx) dx

**Solution:**

Given that I = ∫(tanx)/√(cosx) dx

I = ∫sinx/cosx√(cosx) dx

= ∫ sinx/(cosx)

^{3/2}dx= ∫sinx/(cosx)

^{3/2}dx ……..(i)Let us considered cosx = t then,

On differentiating both side we get,

d(cosx) = dt

-sinxdx = dt

sinxdx = -dt

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

I = ∫(-dt)/t

^{-3/2}= -∫t

^{-3/2}dt= -[-2t

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

^{1/2 }+ c= 2/√(cosx) + c)

Hence, I = 2/√(cosx) + c

### Question 13. ∫cos^{3}x/√(sinx) dx

**Solution:**

Given that I = ∫cos

^{3}x/√(sinx) dx= ∫(cos

^{2}xcosx)/√(sinx) dx= ∫((1 – sin

^{2}x)cosx)/√(sinx) dx= ∫((1 – sin

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

On differentiating both side we get,

d(sinx) = dt

cosxdx = dt

Now on putting sinx = t and cosxdx = dt in equation (i), we get

I = ∫(1 – t

^{2})/√t dt= ∫(t

^{-1/2}-t^{2}x t^{-1/2})dt=∫(t

^{-1/2 }– t^{3/2})dt= 2t

^{1/2 }– 2/5 t^{5/2 }+ c= 2(sinx)

^{1/2 }– 2/5(sinx)^{5/2 }+ cHence, I = 2√(sinx) – 2/5(sinx)

^{5/2 }+ c

### Question 14. ∫(sin^{3}x)/√(cosx) dx

**Solution**:

Given that I = ∫(sin

^{3}x)/√(cosx) dx= ∫(sin

^{2}xsinx)/√(cosx) dx=∫((1 – cos

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

On differentiating both side we get,

d(cosx) = dt

-sinxdx = dt

sinxdx = -dt

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

I = ∫((1 – t

^{2}))/√t × -dt= ∫(t

^{2 }– 1)/√t dt= ∫(t

^{2}/t^{1/2}– 1/t^{1/2})dx= ∫(t

^{2-1/2 }– t^{-1/2})dt= ∫(t

^{3/2 }– t^{1/2})dt= 2/5 t

^{5/2 }– 2t^{1/2 }+ c= 2/5 cos

^{5/2}x – 2cos^{1/2}x + cHence, I = 2/5 cos

^{5/2}x – 2√(cosx) + c

### Question 15. ∫1/(√(tan^{-1}x) (1 + x^{2})) dx

**Solution:**

Given that I = ∫1/(√(tan

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

^{-1}x = t, thenOn differentiating both side we get,

d(tan

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

^{2}) dx = dtNow on putting tan

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

= ∫t

^{-1/2}dt= 2t

^{1/2 }+ c= 2√tan

^{-1}x + cHence, I = 2√tan

^{-1}x + c

### Question 16. ∫√(tanx)/(sinxcosx) dx

**Solution:**

Given that I = ∫√(tanx)/(sinxcosx) dx

= ∫(√(tanx)×cosx)/(sinxcosx×cosx) dx

= ∫√(tanx)/(tanxcos

^{2}x) dx= ∫(sec

^{2}xdx)/√(tanx) dxLet us considered tanx = t, then

On differentiating both side we get,

sec

^{2}xdx = dtNow

I = ∫ dt/√t

= 2√t + c

Hence, I = 2√tanx + c

### Question 17. 1/x × (logx)^{2} dx

**Solution:**

Given that I = ∫1/x × (logx)

^{2}dx …..(i)Let us considered logx = t then,

On differentiating both side we get,

d(logx) = dt

1/x dx = dt

Now on putting logx = t and 1/x dx = dt in equation (i), we get

I = ∫t

^{2}dt= t

^{3}/3 + c= (logx)

^{3}/3 + cHence, I = (logx)

^{3}/3 + c

### Question 18. ∫sin^{5}x cosx dx

**Solution:**

Given that I = ∫sin

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

On differentiating both side we get,

d(sinx) = dt

cosxdx = dt

Now on putting sinx = t and cosxdx = dt in equation (i), we get

I = ∫ t

^{5}dt= t

^{6}/6 + c= (sin

^{6}x)/6 + cHence, I = 1/6 (sin

^{6}x) + c

### Question 19. ∫tan^{3/2}x sec^{2}x dx

**Solution:**

Given that I = ∫tan

^{3/2}xsec^{2}xdx ……(i)Let us considered tanx = t then,

On differentiating both side we get,

d(tanx) = dt

sec

^{2}xdx = dtNow on putting tanx = t and sec

^{2}xdx = dt in equation (i), we getI = ∫ t

^{3/2}dt= 2/5 t

^{5/2 }+ c= 2/5(tanx)

^{5/2 }+ cHence, I = 2/5 tan

^{5/2}x + c

### Question 20. ∫(x^{3})/(x^{2 }+ 1)^{2}dx

**Solution:**

Given that I = ∫(x

^{3})/(x^{2 }+ 1)^{2}dx …….(i)Let us considered 1 + x

^{2 }= t then,On differentiating both side we get,

d(1 + x

^{2}) = dt2xdx = dt

xdx = dt/2

Now on putting 1 + x

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

^{2}/t^{3}× dt/2= 1/2∫(t – 1)/t

^{3}dt [1 + x^{2 }= t]= 1/2∫[(t/t

^{3}– 1/t^{3})dt]= 1/2∫(t

^{-2 }– t^{-3})dt= 1/2 [-1t

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

^{2})] + c= -1/2t + 1/(4t

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

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

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

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

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

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

^{2})/(4(x^{2 }+ 1)^{2}) + c

### Question 21. ∫(4x + 2)√(x^{2 }+ x + 1) dx

**Solution:**

Given that I = ∫(4x + 2)√(x

^{2 }+ x + 1) dxLet us considered x

^{2 }+ x + 1 = t then,On differentiating both side we get,

(2x + 1)dx = dt

Now,

I = ∫ (4x + 2)√(x

^{2 }+ x + 1) dx= ∫2√t dt

= 2∫√t dt

= 2t

^{3/2}/(3/2) + cHence, I = 4/3 (x

^{2 }+ x + 1)^{3/2 }+ c

### Question 22. ∫(4x + 3)/√(2x^{2 }+ 3x + 1) dx

**Solution:**

Given that l = ∫(4x + 3)/√(2x

^{2 }+ 3x + 1) dx ……(i)Let us considered 2x

^{2 }+ 3x + 1 = t then,On differentiating both side we get,

d(2x

^{2 }+ 3x + 1) = dt(4x + 3)dx = dt

Now on putting 2x

^{2 }+ 3x + 1 = t and (4x + 3)dx = dt in equation (i), we getI = ∫dt/√t

= ∫t

^{-1/2}dt= 2t

^{1/2 }+ c= 2√t + c

Hence, I = 2√(2x

^{2 }+ 3x + 1) + c

### Question 23. ∫1/(1 + √x) dx

**Solution:**

Given that I = ∫1/(1 + √x) dx …….(i)

Let us considered x = t

^{2}then,On differentiating both side we get,

dx = d(t

^{2})dx = 2tdt

Now on putting x = t

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

^{2})) dt= ∫2t/(1 + t) dt

= 2∫t/(1 + t) dt

= 2∫(1 + t – 1)/(1 + t) dt

= 2⌋[(1 + t)/(1 + t) – 1/(1 + t)]dt

= 2∫dt – 2∫1/(1 + t) dt

= 2t – 2log|(1 + t)| + c

= 2√x – 2log|(1 + √x)| + c

Hence, I = 2√x – 2log|(1 + √x)| + c

### Question 24.

**Solution:**

Given that I = …….(i)

Let us considered cos

^{2}x = t then,On differentiating both side we get,

d(cos

^{2}x) = dt-2cosx sinx dx = dt

-sin2x dx = dt

sin2x dx = -dt

Now on putting cos

^{2}x = t and sin2x dx = -dt in equation (i), we getI = ∫e

^{t}(-dt)= -e

^{t }+ c= – + c

Hence, I = – + c

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