Class 12 RD Sharma Mathematics Solutions Chapter 19 Indefinite Integrals Exercise 19.8 | Set 1 (original) (raw)
Last Updated : 23 Jul, 2025
Chapter 19 of RD Sharma's Class 12 Mathematics textbook focuses on the indefinite integrals a crucial concept in calculus. This chapter delves into the various techniques for finding the integral of the functions without specified limits. Exercise 19.8 | Set 1 provides practice problems that enhance understanding of the applying integration methods to diverse functions thereby reinforcing theoretical knowledge and problem-solving skills.
Indefinite Integrals
The Indefinite integrals also known as antiderivatives represent a family of the functions whose derivative is the given function. They are used to find the general form of the functions that when differentiated yield the original function. An indefinite integral is denoted as the ∫f(x) dx and includes an arbitrary constant of the integration (C) since integration can only determine a function up to an additive constant. This concept is fundamental for the solving differential equations and for the applications in physics and engineering.
Class 12 RD Sharma Mathematics Solutions - 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/√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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Conclusion
Exercise 19.8 | Set 1 of Chapter 19 on indefinite integrals in RD Sharma's textbook provides the essential practice for the mastering integration techniques. By working through these problems students can strengthen their understanding of how to integrate various functions apply integration rules and solve practical problems. Mastery of indefinite integrals is critical for the advancing in calculus and for the applications in the various scientific fields.