# RD Sharma Class 12 Ex 19.8 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.8 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.8 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 19.8 Solutions Chapter 19 Indefinite Integrals

### Question 1. Evaluate ∫ 1/√1 – cos2x dx

Solution:

Let us assume I = ∫ 1/√1 – cos2x dx

∫ 1/√1 – cos2x dx = ∫1/√2sin2x dx

= ∫ 1/(√2 sinx) dx

= (1/√2) ∫ cosec x dx

Integrate the above equation then we get

= (1/√2) log|tan x/2| + c

Hence, I = (1/√2) log|tan x/2| + c

### Question 2. Evaluate ∫ 1/√1 + cos2x dx

Solution:

Let us assume I = ∫ 1/√1 + cos2x dx

∫ 1/√1 + cos2x dx = ∫1/√2cos2x/2 dx

= ∫ 1/(√2 cosx/2) dx

= (1/√2) ∫ sec x/2 dx

= (1/√2) ∫ cosec (π/2 + x/2) dx

Integrate the above equation then we get

= (2/√2) log|tan (π/4 + x/4| + c

Hence, I = √2 log|tan (π/4 + x/4| + c

### Question 3. Evaluate

Solution:

Let us assume I =

= ∫ √(2cos2x/2sin2x) dx

= ∫ √cot2x dx

= ∫ cotx dx

Integrate the above equation then we get

= log|sinx| + c                  [∫ cotx dx = log|sinx| + c]

Hence, I = log|sinx| + c

### Question 4. Evaluate

Solution:

Let us assume I =

=

= ∫ √tan2x/2 dx

= ∫ tanx/2 dx

Integrate the above equation then we get

= -2log|cosx/2| + c                    [∫ tanx dx = – log|cosx| + c]

Hence, I = -2log|cosx/2| + c

### Question 5. Evaluate ∫secx/sec2x dx

Solution:

Let us assume I = ∫secx/sec2x dx

∫secx/sec2x dx

= ∫cos2x/cosx dx

= ∫ 2cosx dx – ∫ secx dx

= 2∫ cosx dx – ∫ secx dx

Integrate the above equation then we get

= 2sinx – log|secx + tanx| + c

Hence, I = 2sinx – log|secx + tanx| + c

### Question 6. Evaluate ∫ cos2x/(cosx + sinx)2 dx

Solution:

Let us assume I = ∫ cos2x/(cosx + sinx)2 dx

∫ cos2x/(cosx + sinx)2 dx

= ∫ cos2x/(1 + sin2x) dx ………….(i)

Put 1 + sin2x = t

2cos2x dx = dt

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

= 1/2 ∫ 1/t dt

Integrate the above equation then we get

= 1/2 log|t| + c

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

= 1/2 log|(cosx + sinx)2| + c

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

### Question 7. Evaluate ∫ sin(x – a)/sin(x – b) dx

Solution:

Let us assume I = ∫ sin(x – a)/sin(x – b) dx

∫ sin(x – a)/sin(x – b) dx = ∫ sin(x – a + b – b)/sin(x – b) dx

= ∫ sin(x – b + b – a)/sin(x – b) dx

= ∫ cos(b – a) dx + ∫ cot(x – b)sin(b – a) dx

= cos(b – a) ∫dx + sin(b – a)∫ cot(x – b) dx

Integrate the above equation then we get

= xcos(b – a) + sin(b – a) log|sin(x – b)| + c

Hence, I = xcos(b – a) + sin(b – a) log|sin(x – b)| + c

### Question 8. Evaluate ∫ sin(x – a)/sin(x + a) dx

Solution:

Let us assume I = ∫ sin(x – a)/sin(x + a) dx

∫ sin(x – a)/sin(x + a) dx = ∫ sin(x – a + a – a)/sin(x + a) dx

= ∫ sin(x + a – 2a)/sin(x + a) dx

= ∫ cos(2a) dx – ∫ cot(x+a)sin(2a) dx

= cos(2a) ∫dx + sin(2a)∫ cot(x+a) dx

Integrate the above equation then we get

xcos(2a) + sin(2a) log|sin(x + a)| + c

Hence, I = xcos(2a) + sin(2a) log|sin(x + a)| + c

### Question 9. Evaluate ∫ 1 + tanx/1 – tanx dx

Solution:

Let us assume I = ∫ 1 + tanx/1 – tanx dx

∫ 1 + tanx/1 – tanx dx

= ∫ (cosx + sinx) / (cosx – sinx) dx ………….(i)

Put cosx – sinx dx = t

(-sinx – cosx) dx = dt

– (sinx + cosx) dx = dt

dx = – dt/(sinx + cosx)

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

= – ∫dt/t

Integrate the above equation then we get

= – log|t| + c

= – log|cosx – sinx| + c

Hence, I = – log|cosx – sinx| + c

### Question 10. Evaluate ∫ cosx/cos(x – a) dx

Solution:

Let us assume I = ∫ cosx/cos(x – a) dx

∫ cosx/cos(x – a) dx = ∫ cos(x + a – a)/cos(x – a) dx

= ∫ [cos(x – a)cosa – sin(x – a)sina]/cos(x – a) dx

= ∫ [cos(x – a)cosa]/cos(x – a) dx – ∫ [sin(x – a)sina]/cos(x – a) dx

= ∫ cosa dx – ∫ tan(x – a)sina dx

= cosa ∫dx – sina∫ tan(x – a) dx

Integrate the above equation then we get

= x cosa – sina log|sec(x – a)| + c

Hence, I = x cosa – sina log|sec(x – a)| + c

### Question 11. Evaluate

Solution:

Let us assume I =

= ∫ √tan2(π/4 – x) dx

= ∫ tan(π/4 – x) dx

Integrate the above equation then we get

= log|cos(π/4 – x)| + c

Hence, I = log|cos(π/4 – x)| + c

### Question 12. Evaluate ∫ e3x /(e3x + 1) dx

Solution:

Let us assume I = ∫ e3x /(e3x + 1) dx ………..(i)

Put e3x + 1 = t, then

3e3x dx = dt

dx = dt/3e3x

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

= 1/3 ∫dt/t

Integrate the above equation then, we get

= 1/3 log|t| + c

= 1/3 log |3e3x + 1| + c

Hence, I = 1/3 log |3e3x + 1| + c

### Question 13. Evaluate ∫ secxtanx/3secx + 5 dx

Solution:

Let us assume I = ∫ secxtanx/3secx + 5 dx ………..(i)

Put 3secx + 5 = t

3secxtanx dx = dt

dx = dt/3secxtanx

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

= 1/3 ∫ dt/t

Integrate the above equation then, we get

= 1/3 log|t| + c

= 1/3 log|3secx + 5| + c

Hence, I = 1/3 log|3secx + 5| + c

### Question 14. Evaluate ∫ 1 – cotx/1 + cotx dx

Solution:

Let us assume I = ∫ 1 – cotx/1 + cotx dx

=

………..(i)

Put sinx + cosx = t

cosx – sinx dx = dt

-(sinx – cosx) dx = dt

– dx = dt/sinx – cosx

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

= ∫ – dt/t

Integrate the above equation then, we get

= – log|t| + c

= – log|sinx + cosx| + c

Hence, I = – log|sinx + cosx| + c

### Question 15. Evaluate ∫ secxcosecx/log(tanx) dx

Solution:

Let us assume I = ∫ secxcosecx/log(tanx) dx ………..(i)

log(tanx) = t

secxcosecx dx = dt

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

= ∫ dt/t

Integrate the above equation then, we get

= log|t| + c

= log|log(tanx)| + c

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

### Question 16. Evaluate ∫1/ x(3+logx) dx

Solution:

Let us assume I = ∫1/ x(3+logx) dx ………..(i)

Let 3 + logx = t

d(3 + log x) = dt

\1/x dx = dt

dx = x dt

Putting 3 + logx =t and dx = xdt in equation (i), we get,

= ∫ dt/t

Integrate the above equation then, we get

= log |t| + c

= log|(3 + log x)| + c

Hence, I = log|(3 + log x)| + c

### Question 17. Evaluate ∫ ex + 1 / ex + x dx

Solution:

Let us assume I = ∫ e+ 1 / e+ x dx ………..(i)

Let ex + x = t

d(ex + x) = dt

(ex + 1) dx = dt

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

= ∫ dt/t

Integrate the above equation then, we get

= log |t| + c

= log|ex + x| + c

Hence, I = log|ex + x| + c

### Question 18. Evaluate ∫1/ (xlogx) dx

Solution:

Let us assume I =  ∫1/(xlogx) dx ………..(i)

Let logx = t

d(log x) = dt

1/x dx = dt

dx = x dt

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

= ∫ dt/t

Integrate the above equation then, we get

= log |t| + c

= log|(log x)| + c

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

### Question 19. Evaluate ∫ sin2x/ (acos2x + bsin2x) dx

Solution:

Let us assume I = ∫ sin2x/ (acos2x + bsin2x) dx ………..(i)

Let acos2x + bsin2x = t

On differentiating both side with respect to x, we get

d(acos2x + bsin2x) = dt

[a(2 cosx (-sinx)) + b(2sinxcosx)] dx = dt

[ -a (2 cosx sinx) + b(2 sinx cosx)] dx = dt

[ -a sin2x + bsin2x] dx = dt

sin2x (-a + b) dx = dt

sin2x (b – a) dx = dt

sin2x dx = dt/(b – a)

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

= 1/(b – a) ∫ dt/t

Integrate the above equation then, we get

= 1/(b – a) log |t| + c

= 1/(b – a) log|acos2x + bsin2x| + c

Hence, I = 1/(b – a) log|acos2x + bsin2x| + c

### Question 20. Evaluate ∫ cosx/ 2 + 3sinx dx

Solution:

Let us assume I = ∫ cosx/ 2 + 3sinx dx ………..(i)

Let 2 + 3sinx = t

d(2 + 3sinx) = dt

3cosx dx = dt

cosx dx = dt/3

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

= 1/3 ∫ dt/t

Integrate the above equation then, we get

= 1/3 log |t| + c

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

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

### Question 21. Evaluate ∫ 1 – sinx/ x + cosx dx

Solution:

Let us assume I = ∫ 1 – sinx/ x + cosx dx ………..(i)

Let x + cosx = t

d(x + cosx) = dt

(1 – sinx) dx = dt

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

= ∫ dt/t

Integrate the above equation then, we get

= log|t| + c

= log|x + cosx| + c

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

### Question 22. Evaluate ∫ a/ b + cex dx

Solution:

Let us assume I = ∫ a/ b + cex dx

= ∫ a/ ex(be-x+c) dx  ………..(i)

Let be-x + c = t

d(be-x + c) = dt

-be-x dx = dt

-b/ex dx = dt

1/ex dx = -dt/b

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

= -a/b ∫ dt/t

Integrate the above equation then, we get

= -a/b log|t| + c

= -a/b log|be-x + c| + c

Hence, I = -a/b log|be-x + c| + c

### Question 23. Evaluate ∫ 1/ ex + 1 dx

Solution:

Let us assume I = ∫ 1/ ex + 1 dx

= ∫ 1/ ex[1 + e-x] dx   ………..(i)

Let 1 + e-x = t

d(1 + e-x) = dt

-e-x dx = dt

1/ex dx = -dt

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

= -∫ dt/t

Integrate the above equation then, we get

= -log|t| + c

= -log|1 + e-x| + c

Hence, I = -log|1 + e-x| + c

### Question 24. Evaluate ∫ cotx/logsinx dx

Solution:

Let us assume I = ∫ cotx/logsinx dx  ………..(i)

Let logsinx = t

d(logsinx) = dt

cosx/sinx dx = dt

cotx dx = dt

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

= ∫ dt/t

Integrate the above equation then, we get

= log|t| + c

= log|log sinx| + c

Hence, I = log|log sinx| + c

### Question 25. Evaluate ∫ e2x / e2x – 2 dx

Solution:

Let us assume I = ∫ e2x / e2x – 2 dx  ………..(i)

Let e2x – 2 = t

d(e2x – 2) = dt

e2x dx = dt

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

= 1/2 ∫ dt/t

Integrate the above equation then, we get

= 1/2 log|t| + c

= 1/2 log|e2x – 2| + c

Hence, I = 1/2 log|e2x – 2| + c

Question 26. Evaluate ∫ 2cosx – 3sinx/ 6cosx + 4sinx dx
Solution:

Let us assume I = ∫ 2cosx – 3sinx/ 6cosx + 4sinx dx
= ∫ 2cosx – 3sinx/ 2(3cosx + 2sinx) dx
= ∫ 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 ∫ 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 ∫ cos2x + x + 1/ (x2 + sin2x + 2x) dx
Solution:
Let us assume I = ∫ 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 ∫ 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 ∫ 1/ cos(x + a) cos(x + b) dx
Solution:
Let us assume I = ∫ 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) ∫ tan(x + b) dx – ∫ 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 ∫ -sinx + 2cosx/(2sinx + cosx) dx
Solution:
Let us assume I = ∫ -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
= ∫ 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 ∫ cos4x – cos2x/ (sin4x – sin2x) dx
Solution:
Let us assume I = ∫ cos4x – cos2x/ (sin4x – sin2x) dx
= – ∫ 2sin3x sinx / 2cos3x sinx dx
= – ∫ 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 ∫ 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 ∫ secx/ log(secx + tanx) dx
Solution:
Let us assume I = ∫ 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
= ∫ 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 ∫ cosecx/ log|tanx/2| dx
Solution:
Let us assume I = ∫ 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
= ∫ 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 ∫ 1/ xlogxlog(logx) dx
Solution:
Let us assume I = ∫ 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
= ∫ 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 ∫ cosec2 x/ 1 + cot x dx
Solution:
Let us assume I = ∫ 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
= – ∫ 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 ∫ 10x+ 10x log10/ (10+ x10) dx
Solution:
Let us assume I= ∫ 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
= ∫ 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 ∫ 1 – sin2x/ x + cos2x dx
Solution:
Let us assume I = ∫ 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
= ∫ 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 ∫ 1 + tanx/ x + logsecx dx
Solution:
Let us assume I = ∫ 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
= ∫ 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 ∫ sin2x/ a+ b2sin2x dx
Solution:
Let us assume I = ∫ sin2x/ a+ b2sin2x dx  ………..(i)
Put a+ b2sin2x = t
d(a+ b2sin2x) = dt
b2(2sinxcosx) dx = dt
sin2x dx = dt/b2
Put all these values in equation(i), we get
= 1/b2 ∫ dt/t
Integrate the above equation then, we get
= 1/b2 log |t| + c
= 1/b2 log |a+ b2sin2x| + c
Hence, I = 1/b2 log |a+ b2sin2x| + c
Question 39. Evaluate ∫ x + 1/ x(x + logx) dx
Solution:
Let us assume I = ∫ 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
= ∫ 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 × 1/ √(1 – x2)) dx = dt
(1/ √(1 – x2)) dx = dt/3
Put all these values in equation(i), we get
= 1/3 ∫ 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 ∫ sec2x/ tanx + 2 dx
Solution:
Let us assume I = ∫ 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
= ∫ 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 ∫ 2cos2x + sec2x/ sin2x + tanx – 5 dx
Solution:
Let us assume I = ∫ 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
= ∫ 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 ∫ sin2x/ sin5xsin3x dx
Solution:
Let us assume I = ∫ sin2x/ sin5xsin3x dx
= ∫ sin(5x – 3x)/ sin5xsin3x dx
= ∫ (sin5x cos3x – cos5x sin3x)/ sin5xsin3x dx        [Using formula: sin(a-b) = sina cosb – cosa sinb]
= ∫ (sin5x cos3x)/ sin5xsin3x dx – ∫ (cos5x sin3x)/ sin5xsin3x dx
= ∫ cos3x/sin3x dx – ∫ cos5x/sin5x dx
= ∫ cot3x dx – ∫ 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 ∫ 1 + cotx/ x + logsinx  dx
Solution:
Let us assume I = ∫ 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
= ∫ 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 ∫ 1/ √x (√x + 1)  dx
Solution:
Let us assume I = ∫ 1/ √x (√x + 1) dx    ………..(i)
Put √x + 1 = t
d(√x + 1) = dt
(1/2√x) dx = dt
(1/√x) dx = 2dt
Put all these values in equation(i), we get
= 2∫ dt/t
Integrate the above equation then, we get
= 2 log |t| + c
= 2 log |√x + 1| + c
Hence, I = 2 log |√x + 1| + c
Question 46. Evaluate ∫ tan2x tan3x tan 5x dx
Solution:
Let us assume I = ∫ 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
= ∫ 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 ∫ {1 + tanx tan(x + θ)}  dx
Solution:
Let us assume I = ∫ {1 + tanx tan(x + θ)}  dx    ………..(i)
As we know that,
tan(a – b) = tan a – tan b/ (1+ tana tanb)
tan(x + θ – x) = tan (x + θ) – tan x/ (1+ tan(x + θ) tanx)
tan θ = tan (x + θ) – tan x/ (1+ tan(x + θ) tanx)
(1+ tan(x + θ) tanx) = tan (x + θ) – tan x/tan θ    ………..(ii)
By using equation (i) and (ii), we get
= ∫ tan (x + θ) – tan x/ tan θ dx
= 1/tan θ ∫ tan (x + θ) – tan x dx
Integrate the above equation then, we get
= 1/tan θ [-log|cos(x + θ)| + log |cosx|] + c
= 1/tan θ [log |cosx| – log|cos(x + θ)|] + c
Hence, I = 1/tan θ [log {cosx/ cos(x + θ)}] + c
Question 48. Evaluate ∫ sin2x/ sin(x – π/6)sin(x + π/6) dx
Solution:
Let us assume I = ∫ sin2x/ sin(x – π/6)sin(x + π/6) dx    ………..(i)
= ∫ sin2x/ sin2x – sin2π/6 dx
= ∫ 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
= ∫ dt/t
Integrate the above equation then, we get
= log |t| + c
= log |√x + 1| + c
Hence, I = log |sin2x – 1/4| + c
Question 49. Evaluate ∫  ex-1 + xe-1/ ex + xe dx
Solution:
Let us assume I = ∫ ex-1 + xe-1/ ex + xe dx    ………..(i)
= 1/e ∫ ex + exe-1/ ex + xe dx
= 1/e ∫ 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 ∫ 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 ∫ 1/sinx cos2x dx
Solution:
Let us assume I = ∫ 1/sinx cos2x dx
= ∫sin2x + cos2x/sinx cos2x dx
= ∫sin2x/sinx cos2x + cos2x/sinx cos2x dx
= ∫sinx/ cos2x + cosecx dx
= ∫secx tanx dx +∫ 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 ∫ 1/cos3x – cosx dx
Solution:
Let us assume I = ∫ 1/cos3x – cosx dx
∫ 1/cos3x – cosx dx = ∫ sin2x + cos2x / 4cos3x – 4cosx dx
= ∫ sin2x + cos2x / 4cos(cos2x – 1) dx
= -1/4 ∫ sin2x/ sin2xcosx dx + ∫ cos2x / sin2xcosx dx
= -1/4 ∫ 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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