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Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.8 | Set 1

Last Updated : 18 Mar, 2021
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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 ∫ \sqrt{\frac{1 + cos2x}{1 - cos2x}} dx

Solution:

Let us assume I = ∫ \sqrt{\frac{1 + cos2x}{1 - cos2x}} dx

= ∫ √(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 ∫ \sqrt{\frac{1 - cos2x}{1 + cos2x}} dx

Solution:

Let us assume I = ∫ \sqrt{\frac{1 - cos2x}{1 + cos2x}} dx

= ∫ \sqrt{\frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}}} dx

= ∫ √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 

∫ \frac{\frac{1}{cosx}}{\frac{1}{cos2x}} dx

= ∫cos2x/cosx dx

∫ \frac{2cos^2x - 1}{cosx} dx

∫ \frac{2cos^2x}{cosx} dx - ∫ \frac{1}{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 

∫ \frac{cos2x}{(cos^2x + sin^2x + 2sinxcosx)} 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

∫ \frac{sin(x - b)cos(b - a) + cos(x - b)sin(b - a)}{sin(x - b)} dx

∫ \frac{sin(x-b)cos(b-a)}{sin(x-b)} dx + ∫ \frac{cos(x-b)sin(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

∫ \frac{sin(x + a)cos(2a) - cos(x + a)sin(2a)}{sin(x + a)} dx

∫ \frac{sin(x+a)cos2a)}{sin(x+a)} dx - ∫\frac{cos(x+a)sin(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 

∫ \frac{1 + (\frac{sinx}{cosx})}{1 - (\frac{sinx}{cosx})} dx

∫ \frac{\frac{(cosx + sinx)}{cosx}}{\frac{(cosx-sinx)}{cosx}} 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 ∫ \sqrt{\frac{1-sin2x}{1+sin2x}}dx

Solution:

Let us assume I = ∫ \sqrt{\frac{1-sin2x}{1+sin2x}}dx

∫ \sqrt{\frac{1 - cos(\frac{π}{2} - 2x)}{1 + cos(\frac{π}{2} - 2x)}} dx

∫ \sqrt{\frac{2sin^2(\frac{π}{4} - x)}{2cos^2(\frac{π}{4} - x)}} dx

= ∫ √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 

∫ \frac{1-(\frac{cosx}{sinx})}{1+(\frac{cosx}{sinx})} dx

=∫ \frac{\frac{(sinx-cosx)}{sinx}}{\frac{(sinx+cosx)}{sinx}} dx

∫ \frac{sinx-cosx }{ sinx+cosx} 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 = ∫ ex + 1 / ex + 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

∫ \frac{a}{e^x(\frac{b}{e^x}+c)} 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

∫\frac{1}{e^x[1 + \frac{1}{e^x}]} 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



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