RD Sharma Class 12 Ex 19.8 Solutions Chapter 19 Indefinite Integrals

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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 ∫ e^3x /(e^3x + 1) dx

Solution: 

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

Put e^3x + 1 = t, then

3e^3x dx = dt

dx = dt/3e^3x 

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 |3e^3x + 1| + c

Hence, I = 1/3 log |3e^3x + 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 ∫ e^x + 1 / e^x + x dx

Solution:

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

Let e^x + x = t

d(e^x + x) = dt

(e^x + 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|e^x + x| + c

Hence, I = log|e^x + 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 + ce^x dx

Solution:

Let us assume I = ∫ a/ b + ce^x dx

= 

= ∫ a/ e^x(be^-x+c) dx  ………..(i)      

Let be^-x + c = t 

d(be^-x + c) = dt

-be^-x dx = dt

-b/e^x dx = dt

1/e^x 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/ e^x + 1 dx

Solution:

Let us assume I = ∫ 1/ e^x + 1 dx

= 

= ∫ 1/ e^x[1 + e^-x] dx   ………..(i)     

Let 1 + e^-x = t 

d(1 + e^-x) = dt

-e^-x dx = dt

1/e^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 + 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 ∫ e^2x / e^2x – 2 dx

Solution:

Let us assume I = ∫ e^2x / e^2x – 2 dx  ………..(i) 

Let e^2x – 2 = t 

d(e^2x – 2) = dt

e^2x 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|e^2x – 2| + c

Hence, I = 1/2 log|e^2x – 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 ∫ 10x⁹ + 10^x log_e 10/ (10^x + x¹⁰) dx
Solution:
Let us assume I= ∫ 10x⁹ + 10^x log_e 10/ (10^x + x¹⁰)dx  ………..(i) 
Put 10^x + x¹⁰ = t             
d(10^x + x¹⁰) = dt
(10^x log_e 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 |10^x + x¹⁰| + c
Hence, I = log |10^x + 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/ a² + b²sin²x dx
Solution:
Let us assume I = ∫ 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 ∫ 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 ∫ 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 ∫  e^x-1 + x^e-1/ e^x + x^e dx
Solution: 
Let us assume I = ∫ e^x-1 + x^e-1/ e^x + x^e dx    ………..(i) 
= 1/e ∫ e^x + ex^e-1/ e^x + x^e dx
= 1/e ∫ e^x + ex^e-1/ e^x + x^e dx
Put e^x + x^e= t      
d(e^x + x^e) = dt
(e^x + ex^e-1) dx = dt
(e^x + ex^e-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 |e^x + x^e| + c
Hence, I = 1/e log |e^x + x^e| + 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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