Page No 18.104:
Answer:
Comparing the Coefficients of like powers of
x we get
Page No 18.104:
Question 2:
Answer:
Comparing Coefficients of like powers of
x
Page No 18.104:
Question 3:
Answer:
Comparing Coefficients of like powers of
x
Page No 18.104:
Question 4:
Answer:
Comparing Coefficients of like powers of
x
Page No 18.104:
Question 5:
Answer:
Comparing the Coefficients of like powers of
x
Page No 18.104:
Answer:
Comparing the Coefficients of like powers of
x
Page No 18.104:
Question 7:
Answer:
Comparing the Coefficients of like powers of
x
Page No 18.104:
Question 8:
Answer:
Comparing the Coefficients of like powers of
x
Page No 18.104:
Question 9:
Answer:
Page No 18.104:
Question 10:
Answer:
Comparing the Coefficients of like powers of
t
Page No 18.104:
Question 11:
Answer:
Comparing the Coefficients of like powers of
x
Page No 18.104:
Question 12:
Evaluate the following integrals:
Answer:
Page No 18.104:
Question 13:
Answer:
Page No 18.104:
Question 14:
Answer:
Using partial fraction, we get
Comparing coefficients, we get
A =
7 and
B = 10
So,
Page No 18.104:
Question 15:
Answer:
Page No 18.104:
Question 16:
Answer:
Page No 18.104:
Question 17:
Evaluate the following integrals:
Answer:
Page No 18.104:
Question 18:
Evaluate the following integrals:
Answer:
Let
, or,
Now,
Let
Now,
Page No 18.106:
Question 1:
Answer:
Page No 18.106:
Question 2:
Answer:
Page No 18.106:
Question 3:
Answer:
Comparing coefficients of like terms
Page No 18.106:
Question 4:
Answer:
Page No 18.106:
Question 5:
Answer:
Page No 18.106:
Question 6:
Answer:
Page No 18.106:
Question 7:
Answer:
Page No 18.106:
Question 8:
Answer:
Page No 18.106:
Question 9:
Answer:
Page No 18.106:
Question 10:
Answer:
Page No 18.110:
Question 1:
Answer:
Page No 18.110:
Question 2:
Answer:
Page No 18.110:
Question 3:
Answer:
Page No 18.110:
Question 4:
Answer:
Page No 18.110:
Question 5:
Answer:
Page No 18.110:
Answer:
Page No 18.110:
Question 7:
Answer:
â
Page No 18.110:
Answer:
â
Page No 18.110:
Answer:
â
Page No 18.110:
Answer:
Page No 18.110:
Answer:
â
Page No 18.110:
Question 12:
Answer:
Page No 18.110:
Question 13:
Answer:
Page No 18.110:
Answer:
Page No 18.111:
Question 15:
Answer:
Page No 18.111:
Question 16:
Answer:
â
Page No 18.111:
Question 17:
Answer:
Page No 18.111:
Question 18:
Evaluate the following integrals:
Answer:
Page No 18.114:
Question 1:
Answer:
Page No 18.114:
Question 2:
Answer:
Page No 18.114:
Question 3:
Answer:
Page No 18.114:
Question 4:
Answer:
Page No 18.114:
Question 5:
Answer:
Page No 18.114:
Question 6:
Answer:
Page No 18.114:
Question 7:
Answer:
Page No 18.114:
Question 8:
Answer:
Page No 18.114:
Question 9:
Answer:
Page No 18.114:
Question 10:
Answer:
Page No 18.114:
Question 11:
Answer:
Page No 18.117:
Question 1:
Answer:
Page No 18.117:
Question 2:
Answer:
Page No 18.117:
Question 3:
Answer:
Page No 18.117:
Question 4:
Answer:
Page No 18.117:
Question 5:
Answer:
Page No 18.117:
Question 6:
Answer:
Page No 18.117:
Question 7:
Answer:
Page No 18.117:
Question 8:
Answer:
Page No 18.117:
Question 9:
Answer:
Page No 18.117:
Question 10:
Answer:
Page No 18.117:
Question 11:
Answer:
Page No 18.117:
Question 12:
Answer:
Page No 18.117:
Question 13:
Answer:
Page No 18.117:
Question 14:
Answer:
Page No 18.117:
Question 15:
Answer:
Page No 18.122:
Answer:
Page No 18.122:
Answer:
Page No 18.122:
Question 3:
Answer:
Comparing the coefficients of like terms
Multiplying eq (1) by 2 and adding it to eq (2) we get ,
Putting value of
A = 2 in eq (1)
Page No 18.122:
Question 4:
Answer:
Comparing coefficients of like terms
Multiplying eq (1) by
p and eq (2) by
q and then adding
Putting value of
A in eq (1)
Page No 18.122:
Question 5:
Answer:
Comparing coefficients of like terms
Multiplying eq (3) by 2 and then adding to eq (2)
4
A + 2
B +
A – 2
B = 10
A = 2
Putting value of
A in eq (2) and eq (4) we get,
B = 1
& C = 0
Page No 18.122:
Question 6:
Answer:
By comparing the coefficients of like terms we get,
Multiplying eq (2) by 3 and eq (3) by 4 and then adding,
Thus, substituting the values of
A,B and
C in eq (1) we get ,
Page No 18.122:
Question 7:
Answer:
Multiplying eq (2) by 3 and equation (3) by 4 , then by adding them we get
Page No 18.122:
Question 8:
Answer:
Multiplying equation (2) by 3 and equation (3) by 4 ,then by adding them we get
Page No 18.122:
Question 9:
Answer:
Solving (1) and (2), we get
Page No 18.122:
Question 10:
Answer:
Solving eq (2) and eq (3) we get,
A = 2,
B = 1
Thus, by substituting the values of
A and
B in eq (1) we get ,
Page No 18.122:
Question 11:
Answer:
By solving eq (2) and eq (3) we get,
Page No 18.133:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Question 8:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Question 10:
Answer:
Page No 18.133:
Question 11:
Answer:
Page No 18.133:
Question 12:
Answer:
Page No 18.133:
Question 13:
Answer:
Page No 18.133:
Question 14:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Question 16:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Question 18:
Answer:
Page No 18.133:
Question 19:
Answer:
Page No 18.133:
Question 20:
Answer:
Page No 18.133:
Question 21:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Question 23:
Answer:
Page No 18.133:
Question 24:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Answer:
Page No 18.133:
Question 27:
Evaluate the following integrals:
Answer:
Page No 18.133:
Question 28:
Evaluate the following integrals:
Answer:
Page No 18.133:
Question 29:
Answer:
Page No 18.133:
Answer:
Page No 18.134:
Answer:
Page No 18.134:
Question 32:
Answer:
x tan
2x dx
= â∫
x (sec
2x – 1)
dx
Page No 18.134:
Question 33:
Answer:
Page No 18.134:
Question 34:
Answer:
Page No 18.134:
Question 35:
Answer:
sin
–1 (3
x – 4
x3)
dx
Let
x = sin
θ
⇒
dx = cosâ
θ.
dâ
θ
&
θ = sin
–1x sin
–1 (3
x – 4
x3)
dx =
sin
–1 (3 sin â
θ – 4 sin
3 â
θ) . cos
âθ dâθ
= ∫ sin
–1 (sin 3â
θ) . cos â
θ dâθ
Page No 18.134:
Question 36:
Answer:
Page No 18.134:
Question 37:
Answer:
Page No 18.134:
Question 38:
Answer:
Page No 18.134:
Question 39:
Answer:
Page No 18.134:
Question 40:
Answer:
Page No 18.134:
Question 41:
Answer:
Page No 18.134:
Question 42:
Answer:
Page No 18.134:
Question 43:
Answer:
Page No 18.134:
Question 44:
Answer:
Page No 18.134:
Question 45:
Answer:
Page No 18.134:
Question 46:
Answer:
Page No 18.134:
Question 47:
Answer:
Page No 18.134:
Answer:
Page No 18.134:
Question 49:
Answer:
Page No 18.134:
Question 50:
Answer:
Page No 18.134:
Question 51:
Answer:
Let I = (tan–1x2) x dx
Putting x2 = t
⇒â 2x dx = dt
Page No 18.134:
Question 52:
Answer:
Page No 18.134:
Question 53:
Answer:
Page No 18.134:
Answer:
Note: The answer in indefinite integration may vary depending on the integral constant.
Page No 18.134:
Question 55:
Answer:
Let I =
sin (3A) = 3 sin A – 4 sin3A
Page No 18.134:
Answer:
Note: The final answer in indefinite integration may vary based on the integration constant.
Page No 18.134:
Question 57:
Answer:
Let I=
Page No 18.134:
Question 58:
Answer:
Page No 18.134:
Question 59:
Answer:
Page No 18.134:
Question 60:
Answer:
Page No 18.134:
Question 61:
Answer:
Page No 18.14:
Answer:
Page No 18.14:
Question 2:
Answer:
Page No 18.14:
Question 3:
Answer:
Page No 18.14:
Question 4:
Answer:
Page No 18.14:
Question 5:
Answer:
Page No 18.14:
Answer:
Page No 18.14:
Answer:
Page No 18.14:
Question 8:
Answer:
Page No 18.14:
Answer:
Page No 18.14:
Answer:
Page No 18.14:
Answer:
Page No 18.14:
Question 12:
Answer:
Page No 18.14:
Question 13:
Answer:
Page No 18.14:
Answer:
Page No 18.143:
Question 1:
Answer:
Page No 18.143:
Question 2:
Answer:
Page No 18.143:
Question 3:
Answer:
Page No 18.143:
Question 4:
Answer:
Page No 18.143:
Question 5:
Answer:
Page No 18.143:
Question 6:
Answer:
Page No 18.143:
Question 7:
Answer:
Page No 18.143:
Question 8:
Answer:
Page No 18.143:
Question 9:
Answer:
Page No 18.143:
Question 10:
Answer:
Page No 18.143:
Question 11:
Answer:
Page No 18.143:
Question 12:
Answer:
Page No 18.143:
Question 13:
Answer:
Page No 18.143:
Question 14:
Answer:
Page No 18.143:
Question 15:
Answer:
Page No 18.143:
Question 16:
Answer:
Page No 18.143:
Question 17:
Answer:
Page No 18.143:
Question 18:
Answer:
Page No 18.143:
Question 19:
Answer:
Page No 18.143:
Question 20:
Answer:
Page No 18.143:
Question 21:
Answer:
Page No 18.143:
Question 22:
Answer:
Page No 18.143:
Question 23:
Answer:
Page No 18.143:
Question 24:
Evaluate the following integrals:
Answer:
Page No 18.149:
Question 1:
Answer:
Page No 18.149:
Question 2:
Answer:
Page No 18.149:
Question 3:
Answer:
Page No 18.149:
Question 4:
Answer:
Page No 18.149:
Question 5:
Answer:
Page No 18.149:
Question 6:
Answer:
Page No 18.149:
Question 7:
Evaluate the following integrals:
Answer:
Page No 18.149:
Question 8:
Answer:
Page No 18.149:
Question 9:
Answer:
Page No 18.149:
Question 10:
Answer:
Page No 18.149:
Question 11:
Answer:
Page No 18.149:
Question 12:
Answer:
Page No 18.15:
Answer:
Page No 18.15:
Answer:
Page No 18.15:
Question 17:
Answer:
Page No 18.15:
Answer:
Page No 18.15:
Question 19:
Answer:
Page No 18.15:
Question 20:
Answer:
Page No 18.15:
Question 21:
Answer:
Page No 18.15:
Question 22:
Answer:
Page No 18.15:
Question 23:
Answer:
Page No 18.15:
Question 24:
Answer:
Page No 18.15:
Question 25:
Answer:
Page No 18.15:
Question 26:
Answer:
Page No 18.15:
Question 27:
Answer:
Page No 18.15:
Question 28:
Answer:
Page No 18.15:
Question 29:
Answer:
Page No 18.15:
Question 30:
Answer:
Page No 18.15:
Question 31:
Answer:
Page No 18.15:
Question 32:
Answer:
Page No 18.15:
Question 33:
Answer:
Page No 18.15:
Question 34:
Answer:
Page No 18.15:
Question 35:
Answer:
Page No 18.15:
Question 36:
Answer:
Page No 18.15:
Question 37:
Answer:
Page No 18.15:
Question 38:
Answer:
Page No 18.15:
Question 39:
Answer:
Page No 18.15:
Question 40:
Answer:
Page No 18.15:
Question 41:
Answer:
Page No 18.15:
Question 42:
Answer:
Page No 18.15:
Question 43:
Answer:
Page No 18.15:
Question 45:
If f' (x) = x − and f(1) = find f(x)
Answer:
Page No 18.15:
Question 46:
If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)
Answer:
Page No 18.15:
Question 47:
If f' (x) = 8x3 − 2x, f(2) = 8, find f(x)
Answer:
Page No 18.15:
Question 48:
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f = 5, find f(x)
Answer:
Page No 18.15:
Question 49:
Write the primitive or anti-derivative of
Answer:
Integrating both sides
Page No 18.154:
Answer:
Page No 18.154:
Answer:
Page No 18.154:
Answer:
Page No 18.154:
Answer:
Page No 18.154:
Question 5:
Answer:
Page No 18.154:
Answer:
Page No 18.154:
Answer:
Page No 18.154:
Answer:
Page No 18.154:
Answer:
Page No 18.154:
Question 10:
Answer:
Page No 18.154:
Question 11:
Answer:
Page No 18.154:
Answer:
Page No 18.154:
Answer:
Page No 18.154:
Question 14:
Answer:
Page No 18.155:
Answer:
Page No 18.155:
Answer:
Page No 18.155:
Answer:
Page No 18.155:
Answer:
Let
Page No 18.158:
Question 1:
Answer:
Page No 18.158:
Question 2:
Answer:
Page No 18.159:
Question 3:
Answer:
Page No 18.159:
Question 4:
Answer:
Page No 18.159:
Question 5:
Answer:
Page No 18.159:
Question 6:
Answer:
Page No 18.159:
Question 7:
Answer:
Page No 18.159:
Question 8:
Answer:
Page No 18.159:
Question 9:
Answer:
Page No 18.159:
Answer:
Page No 18.159:
Question 11:
Evaluate the following integrals:
Answer:
Page No 18.159:
Question 12:
Evaluate the following integrals:
Answer:
Page No 18.159:
Question 13:
Answer:
Using (i), we get
Page No 18.159:
Question 14:
Answer:
Using (i), we get
Page No 18.176:
Question 1:
Answer:
Page No 18.176:
Question 2:
Answer:
Page No 18.176:
Question 3:
Answer:
Page No 18.176:
Question 4:
Answer:
Page No 18.176:
Answer:
Page No 18.176:
Question 6:
Answer:
Page No 18.176:
Question 7:
Answer:
Page No 18.176:
Question 8:
Answer:
Page No 18.176:
Question 9:
Answer:
Page No 18.176:
Question 10:
Answer:
Page No 18.176:
Question 11:
Answer:
Page No 18.176:
Question 12:
Answer:
Page No 18.176:
Question 13:
Answer:
Page No 18.176:
Question 14:
Evaluate the following integrals:
Answer:
Page No 18.176:
Question 15:
Answer:
Page No 18.176:
Question 16:
Evaluate the following integrals:
Answer:
Page No 18.176:
Question 17:
Answer:
Page No 18.176:
Question 18:
Evaluate the following integrals:
Answer:
Page No 18.176:
Question 19:
Answer:
Page No 18.176:
Question 20:
Answer:
Page No 18.176:
Question 21:
Answer:
Page No 18.177:
Question 22:
Answer:
Page No 18.177:
Answer:
Page No 18.177:
Question 24:
Answer:
Page No 18.177:
Question 25:
Evaluate the following integrals:
Answer:
A
Page No 18.177:
Question 26:
Evaluate the following integrals:
Answer:
Page No 18.177:
Question 27:
Evaluate the following integrals:
Answer:
Page No 18.177:
Question 28:
Answer:
Page No 18.177:
Question 29:
Answer:
Page No 18.177:
Question 30:
Answer:
Page No 18.177:
Question 31:
Answer:
Page No 18.177:
Question 32:
Answer:
Page No 18.177:
Question 33:
Answer:
Page No 18.177:
Question 34:
Answer:
Page No 18.177:
Question 35:
Answer:
Page No 18.177:
Question 36:
Answer:
Page No 18.177:
Question 37:
Answer:
Page No 18.177:
Question 38:
Answer:
Page No 18.177:
Question 39:
Answer:
Page No 18.177:
Answer:
Page No 18.177:
Question 41:
Answer:
Page No 18.177:
Question 42:
Answer:
Page No 18.177:
Question 43:
Answer:
Page No 18.177:
Question 44:
Answer:
Degree of numerator is equal to degree of denominator.
We divide numerator by denominator.
Equating coefficient of like terms
A + B = 0
C = 1
A = 1
B = –1
Page No 18.177:
Question 45:
Answer:
Page No 18.177:
Answer:
Page No 18.177:
Question 47:
Evaluate the following integrals:
Answer:
Page No 18.177:
Question 48:
Answer:
Page No 18.177:
Question 49:
Answer:
Page No 18.177:
Question 50:
Evaluate the following integrals:
Answer:
Page No 18.177:
Question 51:
Answer:
Page No 18.177:
Question 52:
Answer:
Page No 18.177:
Question 53:
Answer:
Page No 18.177:
Answer:
Page No 18.177:
Answer:
Page No 18.177:
Question 56:
Find
Answer:
Substituting the values of A, B and C in (1)Hence, =
Page No 18.177:
Question 57:
Evaluate the following integrals:
Answer:
I =
Comparing coefficients, we get
Page No 18.177:
Question 58:
Evaluate the following integrals:
Answer:
Page No 18.177:
Question 59:
Answer:
Page No 18.178:
Question 60:
Answer:
Page No 18.178:
Question 61:
Answer:
Page No 18.178:
Question 62:
Answer:
Page No 18.178:
Question 63:
Answer:
Degree of numerator is equal to degree of denominator.We divide numerator by denominator.
Page No 18.178:
Question 64:
Answer:
Page No 18.178:
Question 65:
Answer:
Page No 18.178:
Question 66:
Evaluate the following integrals:
Answer:
Page No 18.178:
Question 67:
Evaluate the following integrals:
Answer:
Page No 18.178:
Question 68:
Evaluate the following integrals:
â
Answer:
Page No 18.178:
Question 69:
Answer:
I =
Since,
...(i)
Let and
Comparing coefficients, we get
From (i), we get
Page No 18.178:
Question 70:
Answer:
Page No 18.190:
Question 1:
Answer:
Page No 18.190:
Answer:
Page No 18.190:
Question 3:
Answer:
Page No 18.190:
Answer:
Page No 18.190:
Question 5:
Answer:
Page No 18.190:
Question 6:
Answer:
Page No 18.190:
Answer:
Page No 18.190:
Question 8:
Answer:
Page No 18.190:
Question 9:
Answer:
Page No 18.190:
Question 10:
Answer:
Page No 18.190:
Question 11:
Evaluate the following integrals:
Answer:
Page No 18.196:
Answer:
Page No 18.196:
Question 2:
Answer:
Page No 18.196:
Question 3:
Answer:
Page No 18.196:
Question 4:
Answer:
Page No 18.196:
Answer:
Page No 18.196:
Answer:
Page No 18.196:
Question 7:
Answer:
Page No 18.196:
Question 8:
Answer:
Page No 18.196:
Question 9:
Answer:
Page No 18.196:
Question 10:
Answer:
Page No 18.196:
Question 11:
Answer:
Page No 18.196:
Question 12:
Answer:
Page No 18.196:
Question 13:
Answer:
â
Page No 18.196:
Question 14:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Question 4:
Answer:
Page No 18.197:
Question 5:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Question 7:
Answer:
Page No 18.197:
Question 8:
Answer:
Page No 18.197:
Question 9:
Answer:
Page No 18.197:
Question 10:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Question 13:
Answer:
Page No 18.197:
Question 14:
Answer:
Page No 18.197:
Question 15:
Answer:
Page No 18.197:
Answer:
Multiplying numerator and Denominator of eq (1) by
ex
Page No 18.197:
Question 17:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Question 19:
Answer:
Page No 18.197:
Question 20:
Answer:
Page No 18.197:
Question 21:
Answer:
Page No 18.197:
Question 22:
Answer:
Page No 18.197:
Question 23:
Answer:
Page No 18.197:
Question 24:
Answer:
Page No 18.197:
Question 25:
Answer:
Page No 18.197:
Question 26:
Answer:
Page No 18.197:
Question 27:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Question 36:
Answer:
Page No 18.197:
Question 37:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Answer:
Page No 18.197:
Question 40:
Answer:
Page No 18.197:
Question 41:
Answer:
Page No 18.197:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Question 44:
Answer:
Page No 18.198:
Question 45:
Answer:
Page No 18.198:
Question 46:
Answer:
Page No 18.198:
Question 47:
Answer:
Page No 18.198:
Question 48:
Answer:
Page No 18.198:
Question 49:
Answer:
Page No 18.198:
Question 50:
Answer:
Page No 18.198:
Question 51:
Answer:
Page No 18.198:
Question 52:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Answer:
Note: The answer in indefinite integration may vary depending on the integral constant.
Page No 18.198:
Question 56:
Answer:
Dividing numerator and denominator by cos
2x we get ,
Page No 18.198:
Question 57:
Answer:
Dividing numerator and denominator by cos
2x we get
Page No 18.198:
Question 58:
Answer:
Page No 18.198:
Question 59:
Answer:
Dividing numerator and denominator by cos
2x, we get
Page No 18.198:
Question 60:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Question 62:
Answer:
Dividing numerator and denominator by cos
2x, we get
Page No 18.198:
Question 63:
Answer:
Page No 18.198:
Question 64:
Answer:
Page No 18.198:
Question 65:
Answer:
Page No 18.198:
Question 66:
Answer:
Page No 18.198:
Question 67:
Answer:
Page No 18.198:
Question 68:
Answer:
Dividing numerator and denominator by cos
4x
Page No 18.198:
Question 69:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Question 71:
Answer:
Page No 18.198:
Question 72:
Answer:
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Question 73:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Question 75:
Answer:
Page No 18.198:
Question 76:
Answer:
Page No 18.198:
Question 77:
Answer:
Page No 18.198:
Question 78:
Answer:
Page No 18.198:
Question 79:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Question 81:
Answer:
Page No 18.198:
Question 82:
Answer:
Page No 18.198:
Question 83:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Question 87:
Answer:
Page No 18.198:
Question 88:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Question 90:
Answer:
Page No 18.198:
Question 91:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Question 93:
Answer:
Page No 18.198:
Question 94:
Answer:
Page No 18.198:
Question 95:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Question 97:
Answer:
Page No 18.198:
Answer:
Page No 18.198:
Question 99:
Answer:
Page No 18.198:
Question 100:
Answer:
Page No 18.198:
Question 101:
Answer:
Page No 18.198:
Answer:
Page No 18.199:
Answer:
Page No 18.199:
Question 104:
Answer:
Page No 18.199:
Question 105:
Answer:
Page No 18.199:
Answer:
Page No 18.199:
Question 107:
Answer:
Page No 18.199:
Question 108:
Answer:
Page No 18.199:
Answer:
Page No 18.199:
Answer:
Page No 18.199:
Answer:
Page No 18.199:
Question 112:
Answer:
Page No 18.199:
Question 113:
Answer:
Page No 18.199:
Question 114:
Answer:
Page No 18.199:
Question 115:
Answer:
Page No 18.199:
Question 116:
Answer:
Page No 18.199:
Question 117:
Answer:
Page No 18.199:
Question 118:
Answer:
Page No 18.199:
Question 119:
Answer:
Page No 18.199:
Question 120:
Answer:
Page No 18.199:
Question 121:
Answer:
Page No 18.199:
Question 122:
Answer:
Page No 18.199:
Answer:
Page No 18.199:
Question 124:
Answer:
Page No 18.199:
Question 125:
Answer:
Page No 18.199:
Question 126:
Answer:
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Answer:
Page No 18.199:
Question 128:
Answer:
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Question 129:
Answer:
Page No 18.199:
Question 130:
Answer:
Page No 18.203:
Question 1:
is equal to
(a)
(b)
(c)
(d) none of these
Answer:
(b)
Page No 18.203:
Question 2:
is equal to
(a)
(b)
(c)
(d) none of these
Answer:
(d) none of these
Page No 18.203:
Question 3:
(a)
log (sec
x2 + tan
x2) +
C
(b)
log (sec
x2 + tan
x2) +
C
(c) 2 log (sec
x2 + tan
x2) +
C
(d) none of these
Answer:
(a) log (sec x2 + tan x2) + C
Page No 18.203:
Question 4:
If then
(a) A = , B =
(b) A = , B =
(c) A = , B =
(d) A = , B =
Answer:
(a) A = , B =
Page No 18.203:
Question 5:
(a)
xsinx +
C
(b)
xsinx cos
x +
C
(c)
(d) none of these
Answer:
(a) xsinx + C
Page No 18.204:
Question 6:
Integration of with respect to logex is
(a)
(b)
(c)
(d) none of these
Answer:
(b)
Page No 18.204:
Question 7:
If = a cos 8x + C, then a =
(a)
(b)
(c)
(d)
Answer:
(c)
Page No 18.204:
Question 8:
If = a sin 2x + C, then a =
(a) −1/2
(b) 1/2
(c) −1
(d) 1
Answer:
(a) −1/2
Page No 18.204:
Question 9:
is equal to
(a) −
xex +
C
(b)
xex +
C
(c) −
xe−x +
C
(d)
xe−x +
C
Answer:
Page No 18.204:
Question 10:
If then k is equal to
(a)
(b) − loge 2
(c) − 1
(d)
Answer:
(a)
Page No 18.204:
Question 11:
(a) log
e (
x + sin
x) +
C
(b) log
e (sin
x + cos
x) +
C
(c)
(d)
[
x + log (sin
x + cos
x)] +
C
Answer:
Disclaimer : Generally here book is taking loge x as log x . So we are writing ln x or loge x instead log x only .
(d) [x + ln (sin x + cos x)] + C
Page No 18.204:
Question 12:
is equal to
(a)
(b)
(c)
(d) none of these
Answer:
(d) none of these
Page No 18.204:
Question 13:
The value of is
(a) 2 cos + C
(b)
(c) sin
(d) 2 sin
Answer:
(d) 2 sin
Page No 18.204:
Question 14:
(a)
ex cot
x +
C
(b) −
ex cot
x +
C
(c)
ex cosec
x +
C
(d) −
ex cosec
x +
C
Answer:
(b) −ex cot x + C
Page No 18.204:
Question 15:
(a) tan 7
x +
C
(b)
(c)
(d) sec
7x +
C
Answer:
(b)
Page No 18.204:
Question 16:
(a)
(b)
(c)
(d)
Answer:
(a)
Page No 18.205:
Question 17:
(a)
(b)
(c)
(d)
Answer:
(c)
Disclaimer : Here in answer refers to
Page No 18.205:
Question 18:
(a)
(b)
(c)
(d)
Answer:
(a)
Page No 18.205:
Question 19:
(a) log (3 + 4 cos
2x) + C
(b)
(c)
(d)
Answer:
(c)
Page No 18.205:
Question 20:
(a)
(c)
(d)
Answer:
(b)
Page No 18.205:
Question 21:
(a)
(b)
(c)
(d)
Answer:
(a)
Dividing numerator and denominator by ex
Page No 18.205:
Question 22:
(a) 2 log
e cos (
xex) +
C
(b) sec (
xex) + C
(c) tan (
xex) +
C
(d) tan (
x +
ex) +
C
Answer:
(c) tan (xex) + C
Page No 18.205:
Question 23:
(a)
(b)
(c)
(d) none of these
Answer:
(c)
Page No 18.205:
Question 24:
The primitive of the function
(a)
(b)
(c)
(d)
Answer:
(a)
Page No 18.205:
Question 25:
The value of is
(a) 1 + log x
(b) x + log x
(c) x log (1 + log x)
(d) log (1 + log x)
Answer:
(d) log (1 + log x)
Page No 18.206:
Question 26:
is equal to
(a)
(b)
(c)
(d)
Answer:
(d)
Page No 18.206:
Question 27:
(a)
ex f (
x) +
C
(b)
ex +
f (
x)
(c) 2
ex f (
x)
(d)
ex −
f (
x)
Answer:
(a) ex f (x) + C
Page No 18.206:
Question 28:
The value of is equal to
(a)
(b)
(c) ± (sin x − cos x) + C
(d) ± log (sin x − cos x) + C
Answer:
(d) ± log (sin x − cos x) + C
Page No 18.206:
Question 29:
If = −x cos x + α, then α is equal to
(a) sin x + C
(b) cos x + C
(c) C
(d) none of these
Answer:
(a) sin x + C
Page No 18.206:
Question 30:
(a) tan
x −
x +
C
(b)
x + tan
x +
C
(c)
x − tan
x +
C
(d) −
x − cot
x +
C
Answer:
(c) x − tan x + C
Page No 18.206:
Question 31:
is equal to
(a)
(b)
(c)
(d)
Answer:
Page No 18.206:
Question 32:
is equal to
Answer:
Page No 18.206:
Question 33:
, then
Answer:
Page No 18.206:
Question 34:
is equal to
Answer:
Page No 18.206:
Question 35:
If , then
Answer:
Page No 18.206:
Question 36:
is equal to
(a)
ex cos
x + C (b)
ex sin
x + C
(c) -
ex cos
x + C (d) -
ex sin
x + C
Answer:
Let
Hence, the correct answer is option (a).
Page No 18.207:
Question 37:
Answer:
Let
and
.....(1)
.....(2)
Solving (1) and (2), we get
and
Comparing with the given expression, we get
and
Hence, the correct answer is option (a).
Page No 18.207:
Question 38:
2)2+C (d) -ex(1+x2)2+C
Answer:
Let
Hence, the correct answer is option (c).
Page No 18.207:
Question 39:
Answer:
Hence, the correct answer is option (d).
Page No 18.207:
Question 40:
Answer:
Put
Hence, the correct answer is option (a).
Page No 18.207:
Question 41:
Answer:
Hence, the correct answer is option (c).
Page No 18.207:
Question 1:
If 0 < x < , then dx is equal to ______________.
Answer:
When
If 0 <
x <
, then
is equal to
.
Page No 18.207:
Question 2:
If then dx is equal to ______________
Answer:
When
If
then
is equal to
.
Page No 18.207:
Question 3:
If dx = k cos 2x + C, then k = ___________________.
Answer:
Comparing with the given expression, we get
If
=
k cos 2
x +
C, then
k =
.
Page No 18.207:
Question 4:
The value of the integral dx is ________________.
Answer:
The value of the integral
is
.
Page No 18.208:
Question 5:
If the value of integral then k = _________________.
Answer:
Let
Comparing with the given expression, we get
k =
If the value of integral
then
k =
.
Page No 18.208:
Question 6:
Answer:
Let
.
Page No 18.208:
Question 7:
Answer:
Let
Page No 18.208:
Question 8:
Answer:
Let
.
Page No 18.208:
Question 9:
Answer:
Let
Page No 18.208:
Question 10:
Answer:
Page No 18.208:
Question 11:
Answer:
Page No 18.208:
Question 12:
Answer:
Page No 18.208:
Question 13:
Answer:
Let
Page No 18.208:
Question 14:
ex sin
ex dx = ___________________.
Answer:
Let
ex sin
ex dx =
.
Page No 18.208:
Question 15:
Answer:
Let
Page No 18.208:
Question 16:
xx(1 + log
x)
dx = ________________.
Answer:
Let
xx(1 + log
x)
dx =
____xx + C____.
Page No 18.208:
Question 17:
Answer:
Let
Page No 18.208:
Question 18:
(
x + 3) (
x2 + 6
x + 10)
9dx = ______________________.
Answer:
Let
(
x + 3)(
x2 + 6
x + 10)
9dx =
.
Page No 18.208:
Question 19:
Answer:
Let
Page No 18.208:
Question 20:
ex(1 - cot
x + cosec
2x)
dx = ______________.
Answer:
Let
ex(1 − cot
x + cosec
2x)
dx =
.
Page No 18.208:
Question 1:
Write a value of .
Answer:
Page No 18.208:
Question 2:
Write a value of .
Answer:
Page No 18.208:
Question 3:
Write a value of .
Answer:
Let I=x2.sin x3dx
Let x3 = t
⇒ â3x2dx = dt
Page No 18.208:
Question 4:
Write a value of .
Answer:
Let I= tan3x . sec2x dx
Let tan x = t
⇒ sec2x dx = dt
I= t3 . dt
Page No 18.208:
Question 5:
Write a value of .
Answer:
Let I=ex (sin x + cos x) dx
Let ex sin x = t
⇒ (ex . sin x + ex cos x) dx = dt
Page No 18.208:
Question 6:
Write a value of .
Answer:
Let I= tan6x . sec2x dx
Let tan x = t
sec2x dx = dt
t6 . dt
Page No 18.209:
Question 7:
Write a value of .
Answer:
Page No 18.209:
Question 8:
Write a value of .
Answer:
Let I=ex sec x(1 + tan x) dx
=ex (sec x + sec x tan x) dx
Let ex sec x = t
⇒ (ex sec x + ex sec x tan x)dx = dt
⇒â ex sec x (1 + tan x) dx = dt
dt
= t + C
= ex sec x + C
Page No 18.209:
Question 9:
Write a value of .
Answer:
Page No 18.209:
Question 10:
Write a value of .
Answer:
Page No 18.209:
Question 11:
Write a value of .
Answer:
Let I=elog sin x . cos x dx
= sin x × cos x dx
Let sin x = t
⇒â cos x dx = dt
= t . dt
Page No 18.209:
Question 12:
Write a value of .
Answer:
Let I= sin3x . cos x dx
Let sin x = t
⇒â cos x dx = dt
â t3 . dt
Page No 18.209:
Question 13:
Write a value of .
Answer:
Let I=cos4x .sin x dx
Let cos x = t
⇒â –sin x dx = dt
⇒ sin x dx = –dt
–âât4dt
Page No 18.209:
Question 14:
Write a value of .
Answer:
Let I=tan x . sec3x dx
=sec2x . sec x tan x dx
Let sec x = t
⇒ sec x tan x dx = dt
∫ t2dt
Page No 18.209:
Question 15:
Write a value of .
Answer:
Page No 18.209:
Question 16:
Write a value of .
Answer:
Page No 18.209:
Question 17:
Write a value of .
Answer:
Page No 18.209:
Question 18:
Write a value of .
Answer:
Page No 18.209:
Question 19:
Write a value of .
Answer:
Page No 18.209:
Question 20:
Write a value of .
Answer:
log
ex dx
=
dx
=
= log
ex 1 .
dx – â
.
dx
= log
e x ×
x – â
dx
=â
x log
e x –
x +
C
=â
x log
e x –
x +
C
=
x (log
ex – 1) +
C
Page No 18.209:
Question 21:
Write a value of .
Answer:
∫ ax . ex dx
= â∫ (ae)x dx
Page No 18.209:
Question 22:
Write a value of .
Answer:
Page No 18.209:
Question 23:
Write a value of .
Answer:
Page No 18.209:
Question 24:
Write a value of .
Answer:
Page No 18.209:
Question 25:
Write a value of .
Answer:
Page No 18.209:
Question 26:
Write a value of .
Answer:
Page No 18.209:
Question 27:
Write a value of .
Answer:
Page No 18.209:
Question 28:
Write a value of .
Answer:
Page No 18.209:
Question 29:
Write a value of .
Answer:
Page No 18.209:
Question 30:
Write a value of .
Answer:
Page No 18.209:
Question 31:
Write a value of .
Answer:
Page No 18.209:
Question 32:
Write a value of .
Answer:
Page No 18.209:
Question 33:
Write a value of .
Answer:
Page No 18.209:
Question 34:
Write a value of .
Answer:
Page No 18.209:
Question 35:
Write a value of .
Answer:
Page No 18.209:
Question 36:
Write a value of .
Answer:
Page No 18.209:
Question 37:
Write a value of .
Answer:
Page No 18.209:
Question 38:
Evaluate:
Answer:
Page No 18.209:
Question 39:
Evaluate:
Answer:
Page No 18.209:
Question 40:
Evaluate:
Answer:
Page No 18.209:
Question 41:
Evaluate:
Answer:
Page No 18.209:
Question 42:
Evaluate:
Answer:
Page No 18.209:
Question 43:
Evaluate:
Answer:
Page No 18.209:
Question 44:
Evaluate:
Answer:
Page No 18.209:
Question 45:
Evaluate:
Answer:
Page No 18.209:
Question 46:
Evaluate:
Answer:
Page No 18.209:
Question 47:
Write a value of
Answer:
Page No 18.209:
Question 48:
Evaluate:
Answer:
Page No 18.209:
Question 49:
Evaluate:
Answer:
Page No 18.209:
Question 50:
Evaluate:
Answer:
Page No 18.209:
Question 51:
Evaluate:
Answer:
Page No 18.209:
Question 52:
Write the value of
Answer:
Page No 18.21:
Question 53:
Evaluate:
Answer:
Page No 18.21:
Question 54:
Evaluate:
Answer:
Page No 18.21:
Question 55:
Evaluate:
Answer:
Page No 18.21:
Question 56:
Answer:
Page No 18.21:
Question 57:
Answer:
Page No 18.21:
Question 58:
Evaluate:
Answer:
Page No 18.21:
Question 59:
Write the anti-derivative of
Answer:
Page No 18.21:
Question 60:
Evaluate:
Answer:
Page No 18.21:
Question 61:
Evaluate:
Answer:
Page No 18.21:
Question 62:
Evaluate :
Answer:
I =
Let (1 + log
x) =
t
Page No 18.23:
Question 1:
Answer:
Page No 18.23:
Question 2:
Answer:
Page No 18.23:
Question 3:
Answer:
Page No 18.23:
Answer:
Page No 18.23:
Answer:
Rationalise the denominator
Page No 18.23:
Question 6:
Answer:
Rationalise the denominator
Page No 18.23:
Answer:
Page No 18.23:
Answer:
Page No 18.23:
Question 9:
Answer:
Page No 18.23:
Question 10:
Answer:
Page No 18.23:
Question 11:
Answer:
Page No 18.23:
Question 12:
Answer:
Page No 18.23:
Question 13:
Answer:
Page No 18.23:
Question 14:
Answer:
Page No 18.23:
Answer:
Page No 18.23:
Question 16:
Answer:
Page No 18.23:
Question 17:
Answer:
Rationalising the denominator
Page No 18.23:
Question 18:
Answer:
Page No 18.24:
Question 19:
Answer:
Let 1
tan
x =
t
Page No 18.30:
Question 1:
Answer:
Page No 18.30:
Answer:
Page No 18.30:
Question 3:
Answer:
Page No 18.30:
Answer:
Page No 18.30:
Question 5:
Answer:
Putting
x + 1 =
t
⇒
x =
t – 1
& dx =
dt
Let C + 1 = C′
Page No 18.30:
Answer:
Page No 18.33:
Answer:
Page No 18.33:
Answer:
Putting
x + 2 =
t
Then
, x =
t – 2
Difference both sides
dx =
dt
Now, integral becomes
Page No 18.33:
Answer:
Putting
x + 4 =
t
Then,
x =
t – 4
Difference both sides
dx =
dt
Now integral becomes,
Page No 18.33:
Answer:
Page No 18.33:
Answer:
Page No 18.33:
Answer:
Page No 18.33:
Answer:
Page No 18.33:
Answer:
Putting 1 + 3
x =
t
⇒ 3
x =
t – 1
Page No 18.33:
Question 9:
Answer:
Page No 18.33:
Answer:
Page No 18.36:
Question 1:
Answer:
Page No 18.36:
Question 2:
Answer:
Page No 18.36:
Answer:
Page No 18.36:
Answer:
Page No 18.36:
Answer:
Page No 18.36:
Answer:
Page No 18.36:
Answer:
Page No 18.36:
Question 8:
Answer:
Page No 18.38:
Question 1:
Answer:
Page No 18.38:
Question 2:
Answer:
Page No 18.38:
Question 3:
Answer:
Page No 18.38:
Question 4:
Answer:
Page No 18.38:
Question 5:
Integrate the following integrals:
Answer:
â
Hence, .
Page No 18.38:
Question 6:
Integrate the following integrals:
Answer:
â
Hence, .
Page No 18.4:
Question 1:
Evaluate each of the following integrals:(i) (ii) (iii) (iv) (v) (vi) (vii) (viii)
Answer:
(i)(ii)(iii)(iv) (v)(vi)(vii)(viii)
Page No 18.4:
Question 2:
Evaluate: (i)
(ii)
Answer:
(i)
(ii)
Page No 18.4:
Question 3:
Evaluate:
Answer:
Page No 18.4:
Question 4:
Evaluate:
Answer:
Page No 18.4:
Question 5:
Evaluate: (i)
(ii)
Answer:
(i)
(ii)
Page No 18.4:
Question 6:
Evaluate:
Answer:
Page No 18.47:
Question 1:
Answer:
Page No 18.47:
Answer:
Page No 18.47:
Question 3:
Answer:
Page No 18.47:
Question 4:
Answer:
Page No 18.47:
Question 5:
Evaluate the following integrals:
Answer:
Page No 18.47:
Question 6:
Answer:
Page No 18.47:
Question 7:
Answer:
Page No 18.47:
Question 8:
Answer:
Page No 18.47:
Question 9:
Answer:
Page No 18.47:
Question 10:
Answer:
Page No 18.47:
Question 11:
Answer:
Page No 18.47:
Question 12:
Answer:
Page No 18.47:
Question 13:
Answer:
Page No 18.47:
Question 14:
Answer:
Page No 18.47:
Question 15:
Answer:
Page No 18.47:
Question 16:
Answer:
Page No 18.47:
Question 17:
Answer:
Page No 18.47:
Question 18:
Answer:
Page No 18.47:
Question 19:
Answer:
Page No 18.47:
Question 20:
Answer:
Page No 18.47:
Question 21:
Answer:
Page No 18.47:
Answer:
Page No 18.47:
Answer:
Page No 18.47:
Question 24:
Answer:
Page No 18.47:
Question 25:
Answer:
Page No 18.47:
Question 26:
Answer:
Page No 18.48:
Question 27:
Answer:
Page No 18.48:
Question 28:
Answer:
â
Hence, .
Page No 18.48:
Question 29:
Answer:
Page No 18.48:
Question 30:
Answer:
Page No 18.48:
Question 31:
Answer:
Page No 18.48:
Question 32:
Answer:
Page No 18.48:
Question 33:
Answer:
Page No 18.48:
Question 34:
Answer:
Page No 18.48:
Question 35:
Answer:
Page No 18.48:
Question 36:
Answer:
Page No 18.48:
Question 37:
Answer:
Page No 18.48:
Question 38:
Answer:
Page No 18.48:
Question 39:
Answer:
Page No 18.48:
Question 40:
Answer:
Page No 18.48:
Question 41:
Answer:
Page No 18.48:
Question 42:
Answer:
Page No 18.48:
Question 43:
Answer:
Page No 18.48:
Question 44:
Answer:
Page No 18.48:
Question 45:
Answer:
Page No 18.48:
Answer:
Page No 18.48:
Question 47:
Answer:
Page No 18.48:
Question 48:
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Page No 18.69:
Question 1:
Answer:
∫ tan3x sec2x dx
Let tan x = t
⇒ sec2x dx = dt
Now, ∫ tan3x sec2x dx
= â∫ t3.dt
Page No 18.69:
Question 2:
Answer:
∫ tan x. sec4x dx
= â∫ tan x. sec2x . sec2 x dx
= ∫ tan x (1 + tan2x) sec2x dx
Let tan x = t
⇒ sec2x dx = dt
Now, ∫ tan x (1 + tan2x) sec2x dx
= ∫ t (1 + t2) dt
= ∫ (t + t3) dt
Page No 18.69:
Question 3:
Answer:
∫ tan5x sec4x dx
= ∫ tan5x. sec2x . sec2x dx
= ∫ tan5x (1 + tan2x) sec2x dx
Let tan x = t
⇒ sec2x dx = dt
Now, ∫tan5x (1+tan2x) sec2x dx
= ∫ t5 (1 + t2) dt
= ∫ (t5 + t7) dt
Page No 18.69:
Question 4:
Answer:
∫ sec6 x tan x dx
=∫ sec6x.sec x tan x dx
âLet sec x = t
⇒ sec x tan x dx = dt
Now, ∫ sec6x.sec x tan x dx
= ∫ t6. dt
â
Page No 18.69:
Answer:
∫ tan5x dx
= ∫ tan4x. tan x dx
= ∫(sec2x – 1)2 . tan x dx
= ââ∫ (sec4x – 2 sec2x + 1) tan x dx
= ∫ tan x . sec4x dx – 2 â∫ sec2x . tan x dx+ â∫ tan x dx
= ∫ sec2x. sec2x . tan x dx – 2 â∫ tan x sec2x dx + â∫ tan x dx
= ∫ (1 + tan2x) . tan x . sec2x dx – 2 â∫ tan x . sec2x dx + â∫ tan x dx
Let I1=∫ (1 + tan2x) . tan x . sec2x dx – 2 â∫ tan x . sec2x dx
And I2=∫ tan x dx
∫ tan5x dx=I1 + I2
Now, I1=∫ (1 + tan2x) . tan x . sec2x dx – 2 â∫ tan x . sec2x dx
Let tan x = t
⇒ sec2x dx = dt
I1=∫ (1 + tan2x) . tan x . sec2x dx – 2 â∫ tan x . sec2x dx
∫ (1 + t2) . t. dt – 2 â∫ t. dt
∫ (t + t3) dt – 2 â∫ t dt
And I2=∫ tan x dx
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Question 6:
Answer:
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Answer:
∫ sec4 2x dx
=â ∫ sec2 2x . sec2 2x dx
= ∫ (1 + tan2 2x) . sec2 2x dx
Let tan 2x = t
⇒sec2 2x . 2 dx = dt
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Question 8:
Answer:
â∫ cosec4 3x dx
= ∫ cosec23x . cosec2 3x dx
= ∫ (1 + cot2 3x) cosec2 3x dx
Let cot (3x) = t
⇒–cosec2 (3x) × 3 dx = dt
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Question 9:
Answer:
∫ cotn x cosec2x dx
Let cot x = t
⇒ –cosec2x dx = dt
⇒ cosec2x dx = –dt
Page No 18.69:
Question 10:
Answer:
∫ cot5x . cosec4x dx
= ∫ cot5 x . cosec2x . cosec2x dx
= ââ∫ cot5x . (1 + cot2x) . âcosec2x dx
Let cot x = t
⇒ – cosec2x dx = dt
⇒ cosec2x dx = –dt
Now, ∫ cot5x . cosec4x dx
= ∫ t5 (1 + t2) dt
= ∫(t5 + t7) dt
Page No 18.69:
Answer:
∫ cot5x dx
= ∫ cot4x . cot x dx
= ∫ (cosec2x – 1)2 cot x dx
= ∫ (cosec4x – 2 cosec2x + 1) cot x dx
= ∫ cosec4x . cot x dx – 2 â∫ cot x . cosec2x dx + â∫ cot x dx
= ∫ cosec2x . cosec2x . cot x . dx – 2 â∫ cot x cosec2x dx + ∫â cot x dx
=∫ (1 + cot 2x) . cot x . cosec2x dx – 2 â∫ cot x cosec2x dx + â∫ cot x dx
= ∫ (cot x + cot3x) cosec2x dx – 2 â∫ cot x cosec2x dx + â∫ cot x dx
Now, let I1= ∫ (cot x + cot3x) cosec2x dx – 2 â∫ cot x cosec2x dx
And I2= ∫ cot x dx
First we integrate I1
I1= ∫ (cot x + cot3x) cosec2x dx – 2 â∫ cot x cosec2x dx
Let cot x = t
⇒ – cosec2x dx = dt
⇒ cosec2x dx = – dt
I1= ∫ (t + t3) (– dt) – 2â∫ t (–dt)
= –∫(t + t3) + 2â∫t dt
Now we integrate I2
I2= ∫ cot x dx
=
Now, ∫ cot5x dx=I1 + I2
=
=
Page No 18.69:
Answer:
∫ cot6x dx
= ∫ cot4x . (cosec2x – 1) dx
= ∫ cot4x × cosec2x dx – â∫ cot4x dx
= ∫ cot4 x . cosec2x dx – â∫ cot2x . cot2x dx
= ∫ cot4x – cosec2x dx – â∫ (cosec2x – 1) cot2x dx
= ∫ cot4x . cosec2x dx – â∫ cot2x . cosec2x dx + â∫ cot2x dx
= ∫ cot4x . cosec2x dx – â∫ cot2x . cosec2x dx + â∫ (cosec2x – 1) dx
Now, let I1= ∫ cot4x . cosec2x dx – â∫ cot2x . cosec2x dx
And I2= ∫ (cosec2x – 1) dx
First we integrate I1
I1= ∫ cot4x . cosec2x dx – â∫ cot2x . cosec2x dx
Let cot x = t
⇒ –cosec2x dx = dt
⇒ cosec2dx = – dt
I1=– ∫ ât4dt + â∫ t2dt
Now we integrate I2
I2= ∫ (cosec2x – 1) dx
= – cot x – x + C1
Now, ∫ cot6x dx=I1 + I2
=
= −14cot4x+12cot2x+log|sin x|+C [∴C=C1+C2]
Page No 18.73:
Question 1:
Answer:
∫ sin4x cos3x dx
=â ∫ sin4x . cos2x cos x dx
= ∫ sin4x . (1 – sin2x ) cos x dx
Let sin x = t
⇒ââ cos x dx = dt
Now, ∫ sin4x . (1 – sin2x ) cos x dx
= ââ∫ t4 (1 – t2) dt
= ∫ (t4 – t6) dt
Page No 18.73:
Answer:
∫ sin5x dx
= â∫ sin4 x . sin x dx
= ∫ (1 – cos2x)2 sin x dx
= ∫ (1 – cos4 x – 2 cos2x) sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos4 x – 2 cos2x) sin x dx
=–â∫ (1 + t4 – 2t2) dt
Page No 18.73:
Answer:
∫ cos5x dx
= ∫ cos4x . cos x dx
= ∫ (1 – sin2x)2 cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ (1 – sin2x)2 cos x dx
= ââ∫ (1 – t2)2 . dt
= ∫ (1 + t4 – 2t2) dt
= ∫ dt + â∫ t4dt – 2 â∫t2dt
Page No 18.73:
Question 4:
Answer:
∫ sin5x cos x dx
Let sin x = t
cos x dx = dt
Now, ∫ sin5x cos x dx
â= ∫ t5 . dt
Page No 18.73:
Question 5:
Answer:
∫ sin3x . cos6x dx
=â ∫ sin2x . cos6x . sin x dx
= ââ∫ (1 – cos2x) . cos6x . sin x dx
Let cos x = t
⇒ –sin x dx = dt
Now, ââ∫ (1 – cos2x) . cos6x . sin x dx
= –∫ (1 – t2) . t6dt
= ∫ (t2 – 1) t6dt
= ∫ (t8 – t6) dt
Page No 18.73:
Answer:
∫â cos7x dx
= â∫ cos6x . cos x dx
= ∫ (cos2x)3 cos x dx
= ∫ (1 – sin2x)3 . cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ (1 – sin2x)3.cos x dx
= ∫ (1 – t2)3dt
= ∫ (1 – t6 – 3t2 + 3t4) dt
Page No 18.73:
Question 7:
Answer:
∫ x . cos3x2 sin x2dx
Let x2 = t
⇒â 2x dx = dt
Page No 18.73:
Answer:
∫ sin7x dx
= ∫ sin6x . sin x dx
= ∫ (sin2x)3 sin x dx
= ââ∫ (1 – cos2x)3 sin x dx
Let cos x = t
⇒ –sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos2x)3 sin x dx
= ∫ (1 – t2)3 . (–dt)
= –â∫ (1 – t6 – 3t2 + 3t4) dt
Page No 18.73:
Question 9:
Answer:
∫ sin3x . cos5x dx
â= ∫ sin2x . cos5x . sin x dx
= ∫ (1 – cos2x) . cos5x sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos2x) . cos5x sin x dx
= –â∫ (1 – t2) t5 dt
= –∫ (t5 – t7) dt
= ∫(t7 – t5) dt
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Question 13:
Answer:
Page No 18.79:
Question 1:
Evaluate the following integrals:
Answer:
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Question 3:
Evaluate the following integrals:
Answer:
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Question 4:
Evaluate the following integrals:
Answer:
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Question 5:
Evaluate the following integrals:
Answer:
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Question 12:
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Question 14:
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Question 15:
Answer:
Let
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Question 16:
Evaluate the following integrals:
Answer:
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Question 3:
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Question 16:
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Question 17:
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Question 18:
Answer:
Page No 19.79:
Question 2:
Evaluate the following integrals:
Answer:
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