Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 18 Mar, 2021
Geek Week

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

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.




My Personal Notes arrow_drop_up
Recommended Articles
Page :