# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.9 | Set 2

### Question 25. ∫(1 + cosx)/((x + sinx)^{3}) dx

**Solution:**

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Let us consideredx + sinx = t then,

On differentiating both side we get,d(x + sinx) = dt

(1 + cosx)dx = dt

Now on puttingx + sinx = t and (1 + cosx)dx = dt in equation (i), we getI = ∫ dt/t

^{3}= ∫ t

^{-3}dt= t

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

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

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

^{2}) + c

### Question 26. ∫(cosx – sinx)/(1 + sin2x) dx

**Solution:**

Given that I = (cosx – sinx)/(1 + sin2x)

= (cosx – sinx)/((sin

^{2}x + cos^{2}x) + 2sinxcosx) [Because sin^{2}x + cos^{2}x = 1 and sin2x = 2sinxcosx]

Let us consideredsinx + cosx = t

On differentiating both side we get,(cosx – sinx)dx = dt

Now,

= ∫(cosx – sinx)/(1 + sin2x) dx

= ∫(cosx – sinx)/((sinx + cosx)

^{2}) dx= ∫dt/t

^{2}= ∫t

^{-2}dt= -t

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

Hence, I = (-1)/(sinx + cosx) + c

### Question 27. ∫(sin2x)/(a + bcos2x)^{2 }dx

**Solution:**

Given that I = ∫(sin2x)/((a + bcos2x)

^{2}) dx ……(i)

Let us considereda + bcos2x = t then,

On differentiating both side we get,(a + bcos2x) = dt

b(-2sin2x)dx = dt

sin2x dx = -dt/2b

Now on puttinga + bcos2x = t and sin2xdx = -dt/2b in equation (i), we getI = ∫1/t

^{2 }× (-dt)/2b= (-1)/2b ∫ t

^{-2}dt= -1/2b (-1t

^{-1}) + c= 1/2bt + c

= 1/(2b(a + bcos2x)) + c

Hence, I = 1/(2b(a + bcos2x)) + c

### Question 28. ∫(logx^{2})/x dx

**Solution:**

Given that I = ∫(logx

^{2})/x dx ……..(i)

Let us consideredlogx = t then,

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

1/x dx = dt

dx/x = dt

Now, I = ∫(logx

^{2})/x dx= ∫(2logx)/x dx

= 2∫(logx)/x dx …….(ii)

Now on puttinglogx = t and dx/x = dt in equation (ii), we getI = 2∫tdt

= (2t

^{2})/2 + c= t

^{2 }+ cI = (logx)

^{2 }+ c

### Question 29. ∫(sinx)/(1 + cosx)^{2} dx

**Solution:**

Given that I = ∫(sinx)/((1 + cosx)

^{2}) dx …..(i)

Let us considered1 + cosx = t then,

On differentiating both side we get,d(1 + cosx) = dt

-sinxdx = dt

sinxdx = -dt

Now on putting1 + cosx = t and sindx = -dt in equation (i), we getI = ∫(-dt)/t

^{2}= -∫t

^{-2}dt= -(-1t

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

= 1/(1 + cosx) + c

Hence, I = 1/(1 + cosx) + c

### Question 30. ∫cotx log sinx dx

**Solution:**

Given that I = ∫cotx log sinx dx

Let us considered log sinx = t

1/(sinx).cosxdx = dt

cotx dx = dt

∫cotx log sinx dx = ∫tdt

= t

^{2}/2 + c= 1/2(logsinx)

^{2 }+ c

### Question 31. ∫secx.log(secx + tanx)dx

**Solution:**

Given that I = ∫secx.log(secx + tanx)dx ……..(i)

Let us considered log(secx + tanx) = t then,

On differentiating both side we get,d[log(secx + tanx)] = dt

secx dx = dt [Since, d/dx(log(secx + tanx)) = secx]

Now on puttinglog(secx + tanx) = t and secx dx = dt in equation (i), we getI = ∫tdt

= t

^{2}/2 + c= 1/2[log(secx + tanx)]

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

^{2 }+ c

### Question 32. ∫cosecx log(cosecx – cotx)dx

**Solution:**

Given that I = ∫cosecx log(cosecx – cotx)dx ……(i)

Let us considered log(cosecx – cotx) = t then,

On differentiating both side we get,dx[log(cosecx – cotx)] = dt

cosecx dx = dt [ Since, d/dx(log(cosecx – cotx)) = cosecx]

Now on puttinglog(cosecx – cotx) = t and cosecxdx = dt in equation (i), we getI = ∫tdt

= t

^{2}/2 + cHence, I = 1/2[log(cosecx – cotx)]

^{2 }+ c

### Question 33. ∫x^{3}cosx^{4} dx

**Solution:**

Given that I = ∫x

^{3}cosx^{4}dx …….(i)Let us considered x

^{4 }= t then,

On differentiating both side we get,dx(x

^{4}) = dt4x

^{3}dx = dtx

^{3 }= dt/4

Now on puttingx^{4 }= t and x^{3}dx = dt/4 in equation (i), we getI = ∫ cost dt/4

= 1/4sint + c

Hence, I = 1/4sinx

^{4 }+ c

### Question 34. ∫x^{3} sinx^{4} dx

**Solution:**

Given that I = ∫x

^{3}sinx^{4}dx …..(i)Let us considered x

^{4 }= t then,

On differentiating both side we get,d(x

^{4}) = dt4x

^{3}dx = dtx

^{3 }= dt/4

Now on puttingx^{4 }= t and x^{3}dx = dt/4 in equation (i), we getI = ∫sint dt/4

= 1/4 ∫sint dt

= -1/4 cost + c

Hence, I = -1/4 cosx

^{4 }+ c

### Question 35. ∫(xsin^{-1}x^{2})/√(1 – x^{4}) dx

**Solution:**

Given that I = ∫(xsin

^{-1}x^{2})/√(1 – x^{4}) dx …….(i)Let us considered sin

^{-1}x^{2 }= t then,

On differentiating both side we get,d(sin

^{-1}x^{2}) = dt2x × 1/√(1 – x

^{4}) dx = dtx/√(1 – x

^{4}) dx = dt/2

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

= 1/2 × t

^{2}/2 + c= 1/4 (sin

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

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

### Question 36. ∫x^{3}sin(x^{4 }+ 1)dx

**Solution:**

Given that I = ∫x

^{3}sin(x^{4 }+ 1)dx ……..(i)Let us considered x

^{4 }+ 1 = t then,

On differentiating both side we get,d(x

^{4 }+ 1) = dtx

^{3}dx = dt/4

Now on puttingx^{4 }+ 1 = t and x^{3}dx = dt/4 in equation (i), we getI = ∫ sint dt/4

= -1/4 cost + c

= -1/4 cos(x

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

^{4 }+ 1) + c

### Question 37. ∫(x + 1)e^{x}/(cos^{2}(xe^{x}) dx

**Solution:**

Given that I = ∫((x + 1)e

^{x})/(cos^{2}(xe^{x})) dx ……(i)Let us considered xe

^{x }= t then,

On differentiating both side we get,d(xe

^{x}) = dt(e

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

^{x}dx = dt

Now on puttingxe^{x }= t and (x + 1)e^{x}dx = dt in equation (i), we getI = ∫dt/(cos

^{2}t)= ∫ sec

^{2}tdt= tant + c

= tan(xe

^{x}) + cHence, I = tan(xe

^{x}) + c

### Question 38.

**Solution:**

Given that I = ……..(i)

Let us considered = t then,

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

3x

^{2}dx = dtx

^{2}dx = dt/3

Now on putting= t and x^{2}dx = dt/3 in equation (i), we getI = ∫cost dt/3

= (sint)/3 + c

Hence, I = sin()/3 + c

### Question 39. ∫2xsec^{3}(x^{2 }+ 3)tan(x^{2 }+ 3)dx

**Solution:**

Given that I = ∫2xsec

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

^{2 }+ 3) = t then,

On differentiating both side we get,d[sec(x

^{2 }+ 3)] = dt2xsec(x

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

Now on puttingsec(x^{2 }+ 3) = t and 2xsec(x^{2 }+ 3)tan(x^{2 }+ 3)dx = dt in equation (i), we getI = ∫t

^{2}dt= t

^{3}/3 + c= 1/3 [sec(x

^{2 }+ 3)]^{3 }+ cHence, I = 1/3 [sec(x

^{2 }+ 3)]^{3 }+ c

### Question 40. ∫(1 + 1/x)(x + logx)^{2} dx

**Solution**:

Given that I = ((x + 1)(x + logx)

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

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

^{2}Let us considered (x + logx) = t

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

Now,

I = ∫(1 + 1/x)(x + logx)

^{2}dx= ∫t

^{2}dt= t

^{3}/3 + cHence, I = 1/3(x + logx)

^{3 }+ c

### Question 41. ∫tanx sec^{2}x√(1 – tan^{2}x) dx

**Solution:**

Given that I = ∫tanx sec

^{2}x√(1 – tan^{2}x) dx ………(i)Let us considered 1 – tan

^{2}x = t then,

On differentiating both side we get,d(1 – tan

^{2}x) = dt-2tanx sec

^{2}x dx = dttanx sec

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

Now on putting1 – tan^{2}x = t and tanx sec^{2}x dx = -dt/2 in equation (i), we getI = ∫√t × (-dt)/2

=-1/2 ∫t

^{1/2 }dt=-1/2×t

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

^{3/2 }+ cHence, I = -1/3 [1 – tan

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

### Question 42.∫logx (sin(1 + (logx)^{2})/x dx

**Solution:**

Given that I = ∫logx (sin(1 + (logx)

^{2})/x dx ……..(i)

Now on putting1 + (logx)^{2 }= t and (logx)/x dx = dt/2 in equation (i), we getI = ∫sint × dt/2

= 1/2 ∫ sintdt

= -1/2 cost + c

= -1/2 cos[1 + (logx)

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

^{2}] + c

### Question 43.∫ 1/x^{2} × (cos^{2}(1/x))dx

**Solution:**

Given that I = ∫ 1/x

^{2}× (cos^{2}(1/x))dx ……(i)Let us considered 1/x = t then,

On differentiating both side we get,d(1/x) = dt

(-1)/x

^{2}dx = dt1/x

^{2 }dx = -dt

Now on putting1/x = t and 1/x^{2}dx = -dt in equation (i), we getI = ∫cos

^{2}t(-dt)= -∫cos

^{2}tdt= -∫(cos2t + 1)/2 dt

= -1/2 ∫cos2t dt – 1/2 ∫dt

= -1/2 × (sin2t)/2 – 1/2 t + c

= -1/4 sin2t – 1/2 t + c

= -1/4 sin2 × 1/x – 1/2 × 1/x + c

Hence, I = -1/4 sin(2/x) – 1/2 (1/x) + c

### Question 44. ∫sec^{4}x tanx dx

**Solution:**

Given that I = ∫sec

^{4}x tanx dx ……(i)Let us considered tanx = t then,

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

sec

^{2}xdx = dtdx = dt/sec

^{2}x

Now on puttingtanx = t and dx = dt/(sec^{2}x) in equation (i), we getI = ∫sec

^{4}x tanx dt/(sec^{2}x)= ∫ sec

^{2}x tdt= ∫ (1 + tan

^{2}x)tdt= ∫(1 + t

^{2})tdt= ∫(t + t

^{3})dt= t

^{2}/2 + t^{4}/4 + c= (tan

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

^{2}x + 1/4 tan^{4}x + c

### Question 45. ∫(e^{√x} cos(e^{√x} ))/√x dx

**Solution:**

Given that I = ∫(e

^{√x}cos(e^{√x}))/√x dx …….(i)Let us considered e

^{√x }= t then,

On differentiating both side we get,d(e

^{√x}) = dte

^{√x}(1/(2√x))dx = dte

^{√x}/√x dx = 2dt

Now on puttinge^{√x }= t and e^{√x}/√x dx = 2dt in equation (i), we getI = ∫ cost × 2dt

= 2∫ costdt

= 2sint + c

= 2sin(e

^{√x}) + cI = 2sin(e

^{√x}) + c

### Question 46. ∫(cos^{5}x)/(sinx) dx

**Solution:**

Given that I = ∫(cos

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

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

cosx dx = dt

dx = dt/(cosx)

Now on puttingsinx = t and dx = dt/(cosx) in equation (i), we getI = ∫(cos

^{5}x)/t × dt/(cosx)= ∫(cos

^{4}x)/t dt= ∫(1 – sin

^{2}x)^{2}/t dt= ∫(1 – t

^{2})^{2}/t dt= ∫(1 + t

^{4 }– 2t^{2})/t dt= ∫1/t dt + ∫t

^{4}/t dt – 2∫t^{2}/t dt= log|t| + t

^{4}/4 – (2t^{2})/2 + c= log|sinx| + (sin

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

^{4}x – sin^{2}x + log|sinx| + c

### Question 47. ∫(sin√x)/√x dx

**Solution:**

Given that I = ∫(sin√x)/√x dx

Let us considered √x = t then,

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

1/√x dx = 2dt

Now,

I = ∫(sin√x)/√x dx

= 2 ∫sint dt

= -2 cost + c

Hence, I = -2cos√x + c

### Question 48. ∫((x + 1)e^{x})/(sin^{2}(xe^{x})) dx

**Solution:**

Given that I = ∫((x + 1)e

^{x})/(sin^{2}(xe^{x})) dx …….(i)Let us considered xe

^{x }= t then,d(xe

^{x}) = dt(xe

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

^{x}dx = dt

Now on puttingxe^{x }= t and (x + 1)e^{x}dx = dt in equation (i), we getI = ∫dt/(sin

^{2}t)= ∫cosec

^{2}t dt= -cot + c

Hence, I = -cot(xe

^{x}) + c