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

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

**Solution:**

Let us assume I = ∫ 1/√1 – cos2x dx

∫ 1/√1 – cos2x dx = ∫1/√2sin

^{2}x 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/√2cos

^{2}x/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 =

= ∫ √(2cos

^{2}x/2sin^{2}x) dx= ∫ √cot

^{2}x 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 =

=

= ∫ √tan

^{2}x/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}| + cHence, 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 =

=

=

= ∫ √tan

^{2}(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, then3e

^{3x}dx = dtdx = 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| + cHence, 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 = td(e

^{x}+ x) = dt(e

^{x}+ 1) dx = dtPut all these values in equation(i), we get

= ∫ dt/t

Integrate the above equation then, we get

= log |t| + c

= log|e

^{x}+ x| + cHence, 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/ (acos^{2}x + bsin^{2}x) dx

**Solution:**

Let us assume I = ∫ sin2x/ (acos

^{2}x + bsin^{2}x) dx ………..(i)Let acos

^{2}x + bsin^{2}x = tOn differentiating both side with respect to x, we get

d(acos

^{2}x + bsin^{2}x) = 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|acos

^{2}x + bsin^{2}x| + cHence, I = 1/(b – a) log|acos

^{2}x + bsin^{2}x| + 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 = td(be

^{-x}+ c) = dt-be

^{-x}dx = dt-b/e

^{x}dx = dt1/e

^{x}dx = -dt/bPut 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| + cHence, 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}= td(1 + e

^{-x}) = dt-e

^{-x}dx = dt1/e

^{x}dx = -dtPut all these values in equation(i), we get

= -∫ dt/t

Integrate the above equation then, we get

= -log|t| + c

= -log|1 + e

^{-x}| + cHence, 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 = td(e

^{2x}– 2) = dte

^{2x}dx = dtPut 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| + cHence, I = 1/2 log|e

^{2x}– 2| + c

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