Class 12 RD Sharma Mathematics Solutions Chapter 19 Indefinite Integrals Exercise 19.16 (original) (raw)
Last Updated : 23 Jul, 2025
Indefinite integrals are a fundamental concept in calculus, representing the antiderivative of a function. They are used to find the area under curves, volumes of solids, and solve various problems in physics, engineering, and economics.
Important Formulas and Concepts: Indefinite Integrals
**Basic Integration Formulas
- ∫x^n dx = (x^(n+1))/(n+1) + C
- ∫e^x dx = e^x + C
- ∫sin(x) dx = -cos(x) + C
- ∫cos(x) dx = sin(x) + C
**Integration by Substitution
- ∫f(g(x))g'(x) dx = ∫f(u) du, where u = g(x)
Class 12 RD Sharma Mathematics Solutions - Exercise 19.16
Question 1. Evaluate ∫ sec2x/ 1 - tan2x dx
**Solution:
Let us assume I = ∫ sec2x/ 1 - tan2x dx .....(i)
Now, put tan x = t
sec2x dx = dt
So, put all these values in eq(i)
= ∫ dt/ 12 - t2
On integrating the above equation then, we get
= 1/ 2(1) log|1 + t/1 - t| + c
Since, ∫ 1/ a2 - x2 dx = 1/ 2a log|a + x/a - x| + c]
Hence, I = 1/2 log|1 + tanx/1 - tanx| + c
Question 2. Evaluate ∫ ex/ 1 + e2x dx
**Solution:
Let us assume I = ∫ ex/ 1 + e2x dx .....(i)
Now, put ex = t
ex dx = dt
So, put all these values in eq(i)
= dt/ 1 + t2
On integrating the above equation then, we get
= tan-1t + c
Since, ∫ 1/ 1 + x2dx = tan-1x + c
Hence, I = tan-1ex + c
Question 3. Evaluate ∫ cosx/ sin2x + 4sinx + 5 dx
**Solution:
Let us assume I = ∫ cosx/ sin2x + 4sinx + 5 dx .....(i)
Now, put sinx = t
cosx dx = dt
So, put all these values in eq(i)
= ∫ dt/ t2 + 4t + 5
= ∫ dt/ t2 + 2t(2) + (2)2 - (2)2 + 5
= ∫ dt/ (t + 2)2 + 1 .....(ii)
Again, Put t + 2 = u
dt = du
Now, put all these values in eq(ii)
= ∫ du/ u2 + 1
On integrating the above equation then, we get
= tan-1u + c
Since, ∫1/ x2 + 1 dx = tan-1x + c
= tan-1(t + 2) + c
Hence, I = tan-1(sinx + 2) + c
Question 4. Evaluate ∫ ex/e2x + 5ex + 6 dx
**Solution:
Let us assume I = ∫ ex/e2x + 5ex + 6 dx .....(i)
Now, put ex = t
ex dx = dt
So, put all these values in eq(i)
= ∫ dt/ t2 + 5t + 6
= ∫ dt/ t + 2t(5/2) + (5/2)2 - (5/2)2 + 6
= ∫ dt/ (t + 5/2)2 - 1/4 .....(ii)
Put t + 5/2 = u
dt = du
Now, put the above value in eq(ii)
= ∫ du/ u2 - (1/2)2
On integrating the above equation then, we get
= 2/2 log|u - (1/2)/u + (1/2)| + c
Since, ∫ 1/ x2 - a2 dx = 1/ 2alog|x - a/x + a| + c
= log|2u - 1/2u + 1| + c
= log|2(t + 5/2) - 1/2(t + 5/2) + 1| + c
Hence, I = log|ex + 2/ex + 3| + c
Question 5. Evaluate ∫ e3x/ 4e6x - 9 dx
**Solution:
Let us assume I = ∫ e3x/ 4e6x- 9 dx .....(i)
Now, put e3x = t
3e3x dx = dt
e3x dx = dt/3
Now, put the above value in eq(i)
= 1/3 ∫ dt/ 4t2 - 9
= 1/12 ∫ dt/ t2 - (3/2)2
On integrating the above equation then, we get
= 1/12 x 1/ 2(3/2) log|t - 3/2/t + 3/2| + c
Since, ∫1/ x2 - a2 dx = 1/2a log|x - a/x + a| + c]
= 1/36 log|2t - 3/2t + 3| + c
Hence, I = 1/36 log|2e3x - 3/2e3x + 3| + c
Question 6. Evaluate ∫ dx/ex + e-x
**Solution:
Let us assume I = ∫ dx/ex + e-x
= ∫ dx/ex + 1/ex
= ∫ exdx/ (ex)2 + 1 .....(i)
Now, put ex = t
exdx = dt
Now, put the above value in eq(i)
= ∫ dt/ t2 + 1
On integrating the above equation then, we get
= tan-1t + c
Since ∫ 1/ 1 + x2 dx = tan-1x + c
Hence, I = tan-1(ex) + c
Question 7. Evaluate ∫ x/ x4 + 2x2 + 3 dx
**Solution:
Let us assume I = ∫ x/ x4 + 2x2 + 3 dx .....(i)
Now, put x2 = t
2x dx = dt
x dx = dt/2
Now, put the above value in eq(i)
= 1/2 ∫ dt/ t2 + 2t + 3
= 1/2 ∫ dt/ t2 + 2t + 1 - 1 + 3
= 1/2 ∫ dt/ (t + 1)2 + 2 .....(ii)
Now put t + 1 = u
dt = du
So, put the above value in eq(ii)
= 1/2 ∫ du/ u2 + (√2)2
On integrating the above equation then, we get
= 1/2 x 1/√2 tan-1(u/√2) + c
Since ∫1/ x2 + a2dx = 1/a tan-1(x/a) + c
= 1/2√2 tan-1(t + 1/ √2) + c
Hence, I = 1/2√2 tan-1(x2 + 1/ √2) + c
Question 8. Evaluate ∫ 3x5/ 1 + x12 dx
**Solution:
Let us assume I = ∫ 3x5/ 1 + x12 dx
= ∫ 3x5/ 1 + (x6)2dx .....(i)
Now, put x6 = t
6x5dx = dt
x5dx = dt/6
Now, put the above value in eq(i)
= 3/6 ∫ dt/ 1 + t2
On integrating the above equation then, we get
= 1/2 tan-1(t) + c
Since ∫ 1/ x2 + 1 dx = tan-1x + c
Hence, I = 1/2 tan-1(x6) + c
Question 9. Evaluate ∫ x2/ x6 - a6 dx
**Solution:
Let us assume I = ∫ x2/ x6 - a6 dx
= ∫ x2/ (x3)2 - (a3)2 dx .....(i)
Now, put x3 = t
3x2 dx = dt
x2 dx = dt/3
Now, put the above value in eq(i)
= 1/3 ∫ dt/ t2 - (a3)2
On integrating the above equation then, we get
= 1/3 x 1/2a3 log|t - a3/t + a3| + c
Since ∫1/ x2 - a2 dx = 1/2a log|x - a/x + a| + c
= 1/6a3 log|x3 - a3/x3 + a3| + c
Hence, I = 1/6a3 log|x3 - a3/x3 + a3| + c
Question 10. Evaluate ∫ x2/ x6 + a6 dx
**Solution:
Let us assume I = ∫ x2/ x6 + a6 dx
= ∫ x2/ (x3)2 + (a3)2 dx .....(i)
Now, put x3 = t
3x2 dx = dt
x2 dx = dt/3
Now, put the above value in eq(i)
= 1/3 ∫ dt/ t2 + (a3)2
On integrating the above equation then, we get
= 1/3 x (1/a3) tan-1(t/a3) + c
Since, ∫1/ x2 + a2 dx = 1/a tan-1(x/a) + c
Hence, I = 1/3a3 tan-1(x3/a3) + c
Question 11. Evaluate ∫ 1/ x(x6 + 1) dx
**Solution:
Let us assume I = ∫ 1/ x(x6 + 1) dx
= ∫ x5/ x6(x6 + 1) dx .....(i)
Now, put x6 = t
6x5 dx = dt
x5 dx = dt/6
Now, put the above value in eq(i)
= 1/6 ∫dt/ t(t + 1)
= 1/6 ∫dt/ t2 + t
= 1/6 ∫dt/ t2 + 2t(1/2) + (1/2)2 - (1/2)2
= 1/6 ∫dt/ (t + 1/2)2 - (1/2)2 .....(ii)
Let t + 1/2 = u
dt = du
So, put the above value in eq(ii)
= 1/6 ∫du/ (u)2 - (1/2)2
On integrating the above equation then, we get
= 1/6 x 1/ 2(1/2) log|u - (1/2)/u + (1/2)| + c
Since ∫ 1/ x2 - a2dx = 1/2a log|x - a/x + a| + c
= 1/6 log|{(t + 1/2) - 1/2}/(t + 1/2) + 1/2| + c
Hence, I = 1/6 log|x6/ x6 + 1| + c
Question 12. Evaluate ∫ x/ (x4 - x2 + 1) dx
**Solution:
Let us assume I = ∫ x/ (x4 - x2 + 1) dx .....(i)
Let x2 = t
2x dx = dt
x dx = dt/2
Now, put the above value in eq(i)
= 1/2 ∫dt/ t2 - t + 1
= 1/2 ∫dt/ t2 - 2t(1/2) + (1/2)2 - (1/2)2 + 1
= 1/2 ∫dt/ (t - 1/2)2 + (3/4) .....(ii)
Let t - 1/2 = u
dt = du
So, put the above value in eq(ii)
= 1/2 ∫du/ (u)2 + (√3/2)2
On integrating the above equation then, we get
= 1/2 x 1/(√3/2) tan-1(u/(√3/2)) + c
Since, ∫ 1/ x2 + a2dx = 1/a tan-1(x/a) + c
= 1/√3 tan-1(t - 1/2/ (√3/2)) + c
Hence, I = 1/√3 tan-1(2x2 - 1/ √3) + c
Question 13. Evaluate ∫ x/ (3x4 - 18x2 + 11) dx
**Solution:
Let us assume I = ∫ x/ (3x4 - 18x2 + 11) dx
= 1/3 ∫ x/ (x4 - 6x2 + 11/3) dx .....(i)
Let x2 = t
2x dx = dt
x dx = dt/2
So, put the above value in eq(i)
= 1/3 x 1/2 ∫dt/ t2 - 6t + 11/3
= 1/6 ∫dt/ t2 - 2t(3) + (3)2 - (3)2 + 11/3
= 1/6 ∫dt/ (t - 3)2 - (16/3) .....(ii)
Let t - 3 = u
dt = du
Now, put the above value in eq(ii)
= 1/6 ∫du/ (u)2 - (4/√3)2
On integrating the above equation then, we get
= 1/6 x 1/ 2(4/√3) log|u - (4/√3)/u + (4/√3)| + c
Since, ∫ 1/ x2 - a2dx = 1/2a log|x - a/x + a| + c
= √3/48 log|(t - 3 - 4/√3)/(t - 3 + 4/√3)| + c
Hence, I = √3/48 log|(x2 - 3 - 4/√3)/(x2 - 3 + 4/√3)| + c
Question 14. Evaluate ∫ ex/ (1 + ex)(2 + ex) dx
**Solution:
Let us assume I = ∫ ex/ (1 + ex)(2 + ex) dx .....(i)
Let ex = t
ex dx = dt
So, put the above value in eq(i)
= ∫ dt/ (1 + t)(2 + t)
= ∫dt/ (1 + t) - ∫dt/(2 + t)
On integrating the above equation then, we get
= log|1 + t| - log|2 + t| + c
= log|1 + t/2 + t| + c
Hence, I = log|1 + ex/2 + ex| + c
Practice Questions on Indefinite Integrals
**Question 1. ∫ sin^2(x) dx
**Question 2. ∫ sec^2(x) tan(x) dx
**Question 3. ∫ cos^3(x) dx
**Question 4. ∫ tan^2(x) dx
**Question 5. ∫ sin^3(x) cos^2(x) dx
**Read More:
Conclusion
1. Trigonometric Integrals: Exercise 19.16 likely focuses on integrating various combinations of trigonometric functions.
2. Key Trigonometric Identities:
- sin^2(x) + cos^2(x) = 1
- tan^2(x) + 1 = sec^2(x)
- sin(2x) = 2sin(x)cos(x)
- cos(2x) = cos^2(x) - sin^2(x) = 2cos^2(x) - 1 = 1 - 2sin^2(x)
3. Integration Strategies:
a) For even powers of sin or cos, use half-angle formulas
b) For odd powers, try substitution u = sin(x) or u = cos(x)
c) For products of sin and cos, try using trigonometric identities to simplify
4. Common Integrals to Remember:
- ∫ sin(x) dx = -cos(x) + C
- ∫ cos(x) dx = sin(x) + C
- ∫ tan(x) dx = -ln|cos(x)| + C
- ∫ sec(x) dx = ln|sec(x) + tan(x)| + C
5. Reduction Formulas: For higher powers, reduction formulas can be useful to reduce the power step by step.