# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.8 | Set 2

Last Updated : 18 Mar, 2021

### Question 26. Evaluate âˆ« 2cosx – 3sinx/ 6cosx + 4sinx dx

Solution:

Let us assume I = âˆ« 2cosx – 3sinx/ 6cosx + 4sinx dx

= âˆ« 2cosx – 3sinx/ 2(3cosx + 2sinx) dx

= âˆ« 2cosx – 3sinx/ 2(3cosx + 2sinx) dx ………..(i)

Let 3cosx + 2sinx = t

d(3cosx + 2sinx) = dt

(-3sinx + 2cosx) dx = dt

(2cosx – 3sinx) dx = dt

Put all these values in equation(i), we get

= 1/2 âˆ« dt/t

Integrate the above equation then, we get

= 1/2 log|t| + c

= 1/2 log|3cosx + 2sinx| + c

Hence, I = 1/2 log|3cosx + 2sinx| + c

### Question 27. Evaluate âˆ« cos2x + x + 1/ (x2 + sin2x + 2x) dx

Solution:

Let us assume I = âˆ« cos2x + x + 1/ (x2 + sin2x + 2x) dx                               (i)

Let x2 + sin2x + 2x = t

d(x2 + sin2x + 2x) = dt

(2x + 2cos2x + 2) dx = dt

2(x + cos2x + 1) dx = dt

(x + cos2x + 1) dx = dt/2

Put all these values in equation(i), we get

= 1/2 âˆ« dt/t

Integrate the above equation then, we get

= 1/2 log|t| + c

= 1/2 log|x2 + sin2x + 2x| + c

Hence, I = 1/2 log|x2 + sin2x + 2x| + c

### Question 28. Evaluate âˆ« 1/ cos(x + a) cos(x + b) dx

Solution:

Let us assume I = âˆ« 1/ cos(x + a) cos(x + b) dx

On multiplying and dividing the above equation by sin[(x + b) – (x + a)], we get

= 1/ sin(b – a) âˆ« tan(x + b) dx – âˆ« tan(x + a) dx

Integrate the above equation then, we get

= 1/ sin(b – a) [log(sec(x + b)) – log(sec(x + a))] + c

Hence, I = 1/ sin(b – a) [log(sec(x + b)/sec(x + a))] + c

### Question 29. Evaluate âˆ« -sinx + 2cosx/(2sinx + cosx) dx

Solution:

Let us assume I = âˆ« -sinx + 2cosx/(2sinx + cosx) dx ………..(i)

Let 2sinx + cosx = t

d(2sinx + cosx) = dt

(2cosx – sinx) dx = dt

(-sinx + 2cosx) dx = dt

Put all these values in equation(i), we get

= âˆ« dt/t

Integrate the above equation then, we get

= log|t| + c

= log|2sinx + cosx| + c

Hence, I = log|2sinx + cosx| + c

### Question 30. Evaluate âˆ« cos4x – cos2x/ (sin4x – sin2x) dx

Solution:

Let us assume I = âˆ« cos4x – cos2x/ (sin4x – sin2x) dx

= – âˆ« 2sin3x sinx / 2cos3x sinx dx

= – âˆ« sin3x / cos3x dx ………..(i)

Let cos3x = t

d(cos3x) = dt

-3sin3x dx = dt

– sin3x dx = dt/3

Put all these values in equation(i), we get

= 1/3 âˆ« dt/t

Integrate the above equation then, we get

= 1/3 log|t| + c

= 1/3 log|cos3x| + c

Hence, I = 1/3 log|cos3x| + c

### Question 31. Evaluate âˆ« secx/ log(secx + tanx) dx

Solution:

Let us assume I = âˆ« secx/ log(secx + tanx) dx  ………..(i)

Let log(secx + tanx) = t

d(log(secx + tanx) = dt

secx dx = dt

Put all these values in equation(i), we get

= âˆ« dt/t

Integrate the above equation then, we get

= log |t| + c

= log |log(secx + tanx)| + c

Hence I = log |log(secx + tanx)| + c

### Question 32. Evaluate âˆ« cosecx/ log|tanx/2| dx

Solution:

Let us assume I = âˆ« cosecx/ log|tanx/2| dx    ………..(i)

Let log|tanx/2| = t

d(log|tanx/2|) = dt

cosec x dx = dt

Put all these values in equation(i), we get

= âˆ« dt/t

Integrate the above equation then, we get

= log |t| + c

= log |log tanx/2| + c

Hence, I = log |log tanx/2| + c

### Question 33. Evaluate âˆ« 1/ xlogxlog(logx) dx

Solution:

Let us assume I = âˆ« 1/ xlogxlog(logx) dx   ………..(i)

Put log(logx) = t

d(log(logx)) = dt

1/ xlogx dx = dt

Put all these values in equation(i), we get

= âˆ« dt/t

Integrate the above equation then, we get

= log |t| + c

= log |log(logx)| + c

Hence, I = log |log(logx)| + c

### Question 34. Evaluate âˆ« cosec2 x/ 1 + cot x dx

Solution:

Let us assume I = âˆ« cosec2 x/ 1+cot x dx   ………..(i)

Put 1 + cotx = t          then,

d(1 + cotx) = dt

– cosec2 x dx = dt

Put all these values in equation(i), we get

= – âˆ« dt/t

Integrate the above equation then, we get

= – log |t| + c

= – log |1 + cotx| + c

Hence, I = – log |1 + cotx| + c

### Question 35. Evaluate âˆ« 10x9 + 10x loge 10/ (10x + x10) dx

Solution:

Let us assume I= âˆ« 10x9 + 10x loge 10/ (10x + x10)dx  ………..(i)

Put 10x + x10 = t

d(10x + x10) = dt

(10x loge 10 + 10x9) dx = dt

Put all these values in equation(i), we get

= âˆ« dt/t

Integrate the above equation then, we get

= log |t| + c

= log |10x + x10| + c

Hence, I = log |10x + x10| + c

### Question 36. Evaluate âˆ« 1 – sin2x/ x + cos2x dx

Solution:

Let us assume I = âˆ« 1 – sin2x/ x + cos2x dx  ………..(i)

Put x + cos2x = t

d(x + cos2x) = dt

(1 – 2sinxcosx) dx = dt

(1 – sin2x) dx = dt

Integrate the above equation then, we get

= âˆ« dt/t

Integrate the above equ then, we get

= log |t| + c

= log |x + cos2x| + c

Hence, I = log |x + cos2x| + c

### Question 37. Evaluate âˆ« 1 + tanx/ x + logsecx dx

Solution:

Let us assume I = âˆ« 1 + tanx/ x + logsecx dx  ………..(i)

Put x + logsecx = t

d(x+logsecx) = dt

(1 + tanx) dx = dt

Put all these values in equation(i), we get

= âˆ« dt/t

Integrate the above equation then, we get

= log |t| + c

= log |x + log secx| + c

Hence, I = log |x + log secx| + c

### Question 38. Evaluate âˆ« sin2x/ a2 + b2sin2x dx

Solution:

Let us assume I = âˆ« sin2x/ a2 + b2sin2x dx  ………..(i)

Put a2 + b2sin2x = t

d(a2 + b2sin2x) = dt

b2(2sinxcosx) dx = dt

sin2x dx = dt/b2

Put all these values in equation(i), we get

= 1/b2 âˆ« dt/t

Integrate the above equation then, we get

= 1/b2 log |t| + c

= 1/b2 log |a2 + b2sin2x| + c

Hence, I = 1/b2 log |a2 + b2sin2x| + c

### Question 39. Evaluate âˆ« x + 1/ x(x + logx) dx

Solution:

Let us assume I = âˆ« x + 1/ x(x + logx) dx   ………..(i)

Put x + logx = t

d(x + logx) = dt

(1 + 1/x) dx = dt

(x +1)/ x dx = dt

Put all these values in equation(i), we get

= âˆ« dt/t

Integrate the above equation then, we get

= log |t| + c

= log |x + logx| + c

Hence, I = log |x + logx| + c

### Question 40. Evaluate

Solution:

Let us assume I =     ………..(i)

Put 2 + 3sin-1x = t

d(2 + 3sin-1x) = dt

(3 Ã— 1/ âˆš(1 – x2)) dx = dt

(1/ âˆš(1 – x2)) dx = dt/3

Put all these values in equation(i), we get

= 1/3 âˆ« dt/t

Integrate the above equation then, we get

= 1/3 log |t| + c

= 1/3 log |2 + 3sin-1x| + c

Hence, I = 1/3 log |2 + 3sin-1x| + c

### Question 41. Evaluate âˆ« sec2x/ tanx + 2 dx

Solution:

Let us assume I = âˆ« sec2x/ tanx + 2 dx ………..(i)

Put tanx + 2 = t

d(tanx + 2) = dt

(sec2x) dx = dt

Put all these values in equation(i), we get

= âˆ« dt/t

Integrate the above equation then, we get

= log |t| + c

= log |tanx + 2| + c

Hence, I = log |tanx + 2| + c

### Question 42. Evaluate âˆ« 2cos2x + sec2x/ sin2x + tanx – 5 dx

Solution:

Let us assume I = âˆ« 2cos2x + sec2x/ sin2x + tanx – 5 dx   ………..(i)

Put sin2x + tanx – 5 = t

d(sin2x + tanx – 5) = dt

(2cos2x + sec2x) dx = dt

Put all these values in equation(i), we get

= âˆ« dt/t

Integrate the above equation then, we get

= log |t| + c

= log |sin2x + tanx – 5| + c

Hence, I = log |sin2x + tanx – 5| + c

### Question 43. Evaluate âˆ« sin2x/ sin5xsin3x dx

Solution:

Let us assume I = âˆ« sin2x/ sin5xsin3x dx

= âˆ« sin(5x – 3x)/ sin5xsin3x dx

= âˆ« (sin5x cos3x – cos5x sin3x)/ sin5xsin3x dx        [Using formula: sin(a-b) = sina cosb – cosa sinb]

= âˆ« (sin5x cos3x)/ sin5xsin3x dx – âˆ« (cos5x sin3x)/ sin5xsin3x dx

= âˆ« cos3x/sin3x dx – âˆ« cos5x/sin5x dx

= âˆ« cot3x dx – âˆ« cot5x dx

Integrate the above equation then, we get

= 1/3 log|sin3x| – 1/5 log|sin5x| + c

Hence, I = 1/3 log|sin3x| – 1/5 log|sin5x| + c

### Question 44. Evaluate âˆ« 1 + cotx/ x + logsinx  dx

Solution:

Let us assume I = âˆ« 1 + cotx/ x + logsinx dx    ………..(i)

Put x + logsinx = t

d(x + logsinx) = dt

(1 + cotx) dx = dt

Put all these values in equation(i), we get

= âˆ« dt/t

Integrate the above equation then, we get

= log |t| + c

= log |x + log sinx| + c

Hence, I = log |x + log sinx| + c

### Question 45. Evaluate âˆ« 1/ âˆšx (âˆšx + 1)  dx

Solution:

Let us assume I = âˆ« 1/ âˆšx (âˆšx + 1) dx    ………..(i)

Put âˆšx + 1 = t

d(âˆšx + 1) = dt

(1/2âˆšx) dx = dt

(1/âˆšx) dx = 2dt

Put all these values in equation(i), we get

= 2âˆ« dt/t

Integrate the above equation then, we get

= 2 log |t| + c

= 2 log |âˆšx + 1| + c

Hence, I = 2 log |âˆšx + 1| + c

### Question 46. Evaluate âˆ« tan2x tan3x tan 5x dx

Solution:

Let us assume I = âˆ« tan2x tan3x tan 5x dx   ………..(i)

Now,

tan(5x) = tan(2x + 3x)

tan(5x) = tan2x + tan3x/ (1 – tan2x tan3x)      [By using formula: tan(a + b) = tan a + tan b/ (1- tana tanb)]

tan(5x)(1 – tan2x tan3x) = tan2x + tan3x

(tan5x-tan2x tan3x tan5x) = tan2x + tan3x

tan2x tan3x tan5x = tan5x – tan2x – tan3x      ………..(ii)

Using equation (i) and equation (ii), we get

= âˆ« tan5x – tan2x – tan3x dx

Integrate the above equation then, we get

= 1/5 log|sec5x| – 1/2log|sec2x| -1/3log|sec3x| + c

Hence, I = 1/5 log|sec5x| – 1/2log|sec2x| – 1/3log|sec3x| + c

### Question 47. Evaluate âˆ« {1 + tanx tan(x + Î¸)}  dx

Solution:

Let us assume I = âˆ« {1 + tanx tan(x + Î¸)}  dx    ………..(i)

As we know that,

tan(a – b) = tan a – tan b/ (1+ tana tanb)

tan(x + Î¸ – x) = tan (x + Î¸) – tan x/ (1+ tan(x + Î¸) tanx)

tan Î¸ = tan (x + Î¸) – tan x/ (1+ tan(x + Î¸) tanx)

(1+ tan(x + Î¸) tanx) = tan (x + Î¸) – tan x/tan Î¸    ………..(ii)

By using equation (i) and (ii), we get

= âˆ« tan (x + Î¸) – tan x/ tan Î¸ dx

= 1/tan Î¸ âˆ« tan (x + Î¸) – tan x dx

Integrate the above equation then, we get

= 1/tan Î¸ [-log|cos(x + Î¸)| + log |cosx|] + c

= 1/tan Î¸ [log |cosx| – log|cos(x + Î¸)|] + c

Hence, I = 1/tan Î¸ [log {cosx/ cos(x + Î¸)}] + c

### Question 48. Evaluate âˆ« sin2x/ sin(x – Ï€/6)sin(x + Ï€/6) dx

Solution:

Let us assume I = âˆ« sin2x/ sin(x – Ï€/6)sin(x + Ï€/6) dx    ………..(i)

= âˆ« sin2x/ sin2x – sin2Ï€/6 dx

= âˆ« sin2x/ sin2x – 1/4 dx

Put sin2x – 1/4 = t

d(sin2x – 1/4) = dt

(2sinx cosx) dx = dt

(sin2x) dx = dt

Put all these values in equation(i), we get

= âˆ« dt/t

Integrate the above equation then, we get

= log |t| + c

= log |âˆšx + 1| + c

Hence, I = log |sin2x – 1/4| + c

### Question 49. Evaluate âˆ«  ex-1 + xe-1/ ex + xe dx

Solution:

Let us assume I = âˆ« ex-1 + xe-1/ ex + xe dx    ………..(i)

= 1/e âˆ« ex + exe-1/ ex + xe dx

= 1/e âˆ« ex + exe-1/ ex + xe dx

Put ex + xe= t

d(ex + xe) = dt

(ex + exe-1) dx = dt

(ex + exe-1) dx = dt

Put all these values in equation(i), we get

= 1/e âˆ« dt/t

Integrate the above equation then, we get

= 1/e log |t| + c

= 1/e log |ex + xe| + c

Hence, I = 1/e log |ex + xe| + c

### Question 50. Evaluate âˆ« 1/sinx cos2x dx

Solution:

Let us assume I = âˆ« 1/sinx cos2x dx

= âˆ«sin2x + cos2x/sinx cos2x dx

= âˆ«sin2x/sinx cos2x + cos2x/sinx cos2x dx

= âˆ«sinx/ cos2x + cosecx dx

= âˆ«secx tanx dx +âˆ« cosecx dx

Integrate the above equation then, we get

= sec x + log|tanx/2| + c

Hence, I = sec x + log|tanx/2| + c

### Question 51. Evaluate âˆ« 1/cos3x – cosx dx

Solution:

Let us assume I = âˆ« 1/cos3x – cosx dx

âˆ« 1/cos3x – cosx dx = âˆ« sin2x + cos2x / 4cos3x – 4cosx dx

= âˆ« sin2x + cos2x / 4cos(cos2x – 1) dx

= -1/4 âˆ« sin2x/ sin2xcosx dx + âˆ« cos2x / sin2xcosx dx

= -1/4 âˆ« secx + cosecx cotx dx

Integrate the above equation then, we get

= -1/4 [log|secx + tanx| – cosecx] + c

Hence, I = 1/4 [cosecx + log|secx + tanx|] + c

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