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

### Question 1. Evaluate âˆ« 1/âˆš1 – cos2x dx

Solution:

Let us assume I = âˆ« 1/âˆš1 – cos2x dx

âˆ« 1/âˆš1 – cos2x dx = âˆ«1/âˆš2sin2x dx

= âˆ« 1/(âˆš2 sinx) dx

= (1/âˆš2) âˆ« cosec x dx

Integrate the above equation then we get

= (1/âˆš2) log|tan x/2| + c

Hence, I = (1/âˆš2) log|tan x/2| + c

### Question 2. Evaluate âˆ« 1/âˆš1 + cos2x dx

Solution:

Let us assume I = âˆ« 1/âˆš1 + cos2x dx

âˆ« 1/âˆš1 + cos2x dx = âˆ«1/âˆš2cos2x/2 dx

= âˆ« 1/(âˆš2 cosx/2) dx

= (1/âˆš2) âˆ« sec x/2 dx

= (1/âˆš2) âˆ« cosec (Ï€/2 + x/2) dx

Integrate the above equation then we get

= (2/âˆš2) log|tan (Ï€/4 + x/4| + c

Hence, I = âˆš2 log|tan (Ï€/4 + x/4| + c

### Question 3. Evaluate

Solution:

Let us assume I =

= âˆ« âˆš(2cos2x/2sin2x) dx

= âˆ« âˆšcot2x dx

= âˆ« cotx dx

Integrate the above equation then we get

= log|sinx| + c                  [âˆ« cotx dx = log|sinx| + c]

Hence, I = log|sinx| + c

### Question 4. Evaluate

Solution:

Let us assume I =

=

= âˆ« âˆštan2x/2 dx

= âˆ« tanx/2 dx

Integrate the above equation then we get

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

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

### Question 5. Evaluate âˆ«secx/sec2x dx

Solution:

Let us assume I = âˆ«secx/sec2x dx

âˆ«secx/sec2x dx

= âˆ«cos2x/cosx dx

= âˆ« 2cosx dx – âˆ« secx dx

= 2âˆ« cosx dx – âˆ« secx dx

Integrate the above equation then we get

= 2sinx – log|secx + tanx| + c

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

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

Solution:

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

âˆ« cos2x/(cosx + sinx)2 dx

= âˆ« cos2x/(1 + sin2x) dx ………….(i)

Put 1 + sin2x = t

2cos2x dx = dt

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

= 1/2 âˆ« 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 âˆ« sin(x – a)/sin(x – b) dx

Solution:

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

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

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

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

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

Solution:

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

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

= âˆ« sin(x + a – 2a)/sin(x + a) dx

= âˆ« cos(2a) dx – âˆ« cot(x+a)sin(2a) dx

= cos(2a) âˆ«dx + sin(2a)âˆ« 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 âˆ« 1 + tanx/1 – tanx dx

Solution:

Let us assume I = âˆ« 1 + tanx/1 – tanx dx

âˆ« 1 + tanx/1 – tanx dx

= âˆ« (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

= – âˆ«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 âˆ« cosx/cos(x – a) dx

Solution:

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

âˆ« cosx/cos(x – a) dx = âˆ« cos(x + a – a)/cos(x – a) dx

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

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

= âˆ« cosa dx – âˆ« tan(x – a)sina dx

= cosa âˆ«dx – sinaâˆ« 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 =

= âˆ« âˆštan2(Ï€/4 – x) dx

= âˆ« tan(Ï€/4 – x) dx

Integrate the above equation then we get

= log|cos(Ï€/4 – x)| + c

Hence, I = log|cos(Ï€/4 – x)| + c

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

Solution:

Let us assume I = âˆ« 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 âˆ«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 âˆ« secxtanx/3secx + 5 dx

Solution:

Let us assume I = âˆ« 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 âˆ« 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 âˆ« 1 – cotx/1 + cotx dx

Solution:

Let us assume I = âˆ« 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

= âˆ« – 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 âˆ« secxcosecx/log(tanx) dx

Solution:

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

log(tanx) = t

secxcosecx 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)| + c

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

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

Solution:

Let us assume I = âˆ«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,

= âˆ« 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 âˆ« ex + 1 / ex + x dx

Solution:

Let us assume I = âˆ« 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

= âˆ« 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 âˆ«1/ (xlogx) dx

Solution:

Let us assume I =  âˆ«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

= âˆ« 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 âˆ« sin2x/ (acos2x + bsin2x) dx

Solution:

Let us assume I = âˆ« 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) âˆ« 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 âˆ« cosx/ 2 + 3sinx dx

Solution:

Let us assume I = âˆ« 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 âˆ« 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 âˆ« 1 – sinx/ x + cosx dx

Solution:

Let us assume I = âˆ« 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

= âˆ« 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 âˆ« a/ b + cex dx

Solution:

Let us assume I = âˆ« a/ b + cex dx

= âˆ« 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 âˆ« 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 âˆ« 1/ ex + 1 dx

Solution:

Let us assume I = âˆ« 1/ ex + 1 dx

= âˆ« 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

= -âˆ« 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 âˆ« cotx/logsinx dx

Solution:

Let us assume I = âˆ« 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

= âˆ« 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 âˆ« e2x / e2x – 2 dx

Solution:

Let us assume I = âˆ« 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 âˆ« 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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