Integrals
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Class XII Chapter 7 – Integrals Maths
Exercise 7.1
Question 1: sin 2x Answer
The anti derivative of sin 2x is a function of x whose derivative is sin 2x.It is known that,
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Therefore, the anti derivative of
Question 2:
Cos 3x Answer
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The anti derivative of cos 3x is a function of x whose derivative is cos 3x. It is known that,
Therefore, the anti derivative of .
Question 3:
e2x
Answer
The anti derivative of e2x is the function of x whose derivative is e2x. It is known that,
Therefore, the anti derivative of .
Question 4:
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Answer
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The anti derivative of is the function of x whose
derivative is . It is known
that,
Therefore, the anti derivative of .
Question 5:
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Answer
The anti derivative of is the function of x whose
derivative is
.
It is known that,
Therefore, the anti derivative of is .
Question 6:
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Answer
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Question 7:
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Answer
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Question 8:
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Answer
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Question 9:
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Answer
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Question 10:
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Answer
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Question 11:
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Answer
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Question 12:
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Answer
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Question 13:
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Answer
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On dividing, we obtain
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Question 14:
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Answer
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Question 15:
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Answer
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Question 16:
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Answer
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Question 17:
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Answer
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Question 18:
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Answer
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Question 19:
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Answer
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Question 20:
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Answer
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Question 21:
The anti derivative of equals
(A) (B)
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(C) (D)
Answer
Hence, the correct Answer is C.
Question 22:
If such that f(2) = 0, then f(x) is
(A) (B)
(C) (D)
Answer
It is given that,
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∴Anti derivative of
∴
Also,
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Hence, the correct Answer is A.
Exercise 7.2
Question 1:
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Answer
Let = t
∴2x dx = dt
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Question 2:
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Answer
Let
log |x| = t
∴
Question 3:
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Answer
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Let 1 + log x = t
∴
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Question 4:
sin x ⋅ sin (cos x)
Answer
sin x ⋅ sin (cos x)
Let cos x = t
∴ −sin x dx = dt
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Question 5:
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Answer
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Let
∴ 2adx = dt
Question 6:
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Answer
Let ax + b = t
⇒ adx = dt
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Question 7:
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Answer Let
∴ dx = dt
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Question 8:
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Answer
Let 1 + 2x2 = t
∴ 4xdx = dt
Question 9:
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Answer Let
∴ (2x + 1)dx = dt
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Question 10:
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Answer
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Let
∴
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Question 11:
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Answer Let
∴ dx = dt
Question 12:
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Answer Let
∴
Question 13:
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Answer Let
∴ 9x2 dx = dt
Question 14:
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Answer
Let log x
= t
∴
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Question 15:
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Answer
Let
∴ −8x dx = dt
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Question 16:
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Answer Let
∴ 2dx = dt
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Question 17:
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Answer Let
∴ 2xdx = dt
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Question 18:
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Answer Let
∴
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Question 19:
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Answer
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Dividing numerator and denominator by ex, we obtain
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Let
∴
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Question 20:
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Answer Let
∴
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Question 21:
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Answer
Let 2x − 3 = t
∴ 2dx = dt
Question 22:
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Answer
Let 7 − 4x = t
∴ −4dx = dt
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Question 23:
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Answer Let
∴
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Question 24:
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Answer
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Let
∴
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Question 25:
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Answer
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Let
∴
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Question 26:
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Answer Let
∴
Question 27:
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Answer
Let sin 2x = t
∴
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Question 28:
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Answer Let
∴ cos x dx = dt
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Question 29: cot x log sin x Answer
Let log sin x = t
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Question 30:
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Answer
Let 1 + cos x = t
∴ −sin x dx = dt
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Question 31:
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Answer
Let 1 + cos x = t
∴ −sin x dx = dt
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Question 32:
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Answer
Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt
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Question 33:
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Answer
Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt
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Question 34:
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Answer
Question 35:
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Answer
Let 1 + log x = t
∴
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Question 36:
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Answer
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Let
∴
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Question 37:
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Answer Let x4 = t
∴ 4x3 dx = dt
Let
∴
From (1), we obtain
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Question 38:
equals
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Answer Let
∴
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Hence, the correct Answer is D.
Question 39:
equals
A.
B.
C.
D.
Answer
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Hence, the correct Answer is B.
Exercise 7.3
Question 1:
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Answer
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Question 2:
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Answer
It is known that,
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Question 3:
cos 2x cos
4x cos 6x
Answer
It is known that,
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Question 4: sin3 (2x + 1) Answer
Let
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Question 5: sin3 x cos3 x Answer
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Question 6:
sin x sin 2x sin 3x Answer
It is known that,
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Question 7: sin
4x sin 8x Answer
It is known that,
Question 8:
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Answer
Question 9:
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Answer
Question 10:
sin4 x
Answer
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Question 11: cos4 2x Answer
Question 12:
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Answer
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Question 13:
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Answer
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Question 14:
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Answer
Question 15:
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Answer
Question 16:
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tan4x Answer
From equation (1), we obtain
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Question 17:
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Answer
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Question 18:
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Answer
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Question 19:
Answer
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Question 20:
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Answer
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Question 21: sin−1 (cos x) Answer
It is known that,
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Substituting in equation (1), we obtain
Question 22:
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Answer
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Question 23:
is equal to
A. tan x + cot x + C
B. tan x + cosec x + C
C. − tan x + cot x + C
D.
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tan x + sec x + C Answer
Hence, the correct Answer is A.
Question 24:
equals
A. − cot (exx) + C
B. tan (xex) + C
C. tan (ex) + C
D.
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cot (ex) + C Answer
Let exx = t
Hence, the correct Answer is B.
Exercise 7.4
Question 1:
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Answer Let x3 = t
∴ 3x2 dx = dt
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Question 2:
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Answer Let 2x = t
∴ 2dx = dt
Question 3:
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Answer
Let 2 − x = t
⇒ −dx = dt
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Question 4:
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Answer Let 5x = t
∴ 5dx = dt
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Question 5:
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Answer
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Question 6:
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Answer Let x3 = t
∴ 3x2 dx = dt
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Question 7:
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Answer
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From (1), we obtain
Question 8:
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Answer Let x3 = t
⇒ 3x2 dx = dt
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Question 9:
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Answer
Let tan x = t
∴ sec2x dx = dt
Question 10:
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Answer
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Question 11:
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Answer
Question 12:
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Answer
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Question 13:
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Answer
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Question 14:
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Answer
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Question 15:
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Answer
Question 16:
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Answer
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Equating the coefficients of x and constant term on both sides, we obtain
4A = 4 ⇒ A = 1
A + B = 1 ⇒ B = 0
Let 2x2 + x − 3 = t
∴ (4x + 1) dx = dt
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Question 17:
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Answer
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Equating the coefficients of x and constant term on both sides, we obtain
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From (1), we obtain
From equation (2), we obtain
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Question 18:
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Answer
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Equating the coefficient of x and constant term on both sides, we obtain
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Substituting equations (2) and (3) in equation (1), we obtain
Question 19:
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Answer
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Equating the coefficients of x and constant term, we obtain
2A = 6 ⇒ A = 3
−9A + B = 7 ⇒ B = 34
∴ 6x + 7 = 3 (2x − 9) + 34
Substituting equations (2) and (3) in (1), we obtain
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Question 20:
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Answer
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Equating the coefficients of x and constant term on both sides, we obtain
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Using equations (2) and (3) in (1), we obtain
Question 21:
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Answer
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Let x2 + 2x +3 = t
⇒ (2x + 2) dx =dt
Using equations (2) and (3) in (1), we obtain
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Question 22:
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Answer
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Equating the coefficients of x and constant term on both sides, we obtain
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Substituting (2) and (3) in (1), we obtain
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Question 23:
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Answer
Equating the coefficients of x and constant term, we obtain
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Using equations (2) and (3) in (1), we obtain
Question 24:
equals
A. x tan−1 (x + 1) + C
B. tan− 1 (x + 1) + C
C. (x + 1) tan−1 x + C
D.
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tan−1 x + C Answer
Hence, the correct Answer is B.
Question 25:
equals
A.
B.
C.
D.
Answer
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Hence, the correct Answer is B.
Exercise 7.5
Question 1:
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Answer
Let
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Equating the coefficients of x and constant term, we obtain
A + B = 1 2A + B = 0
On solving, we obtain
A = −1 and B = 2
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Question 2:
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Answer
Let
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Equating the coefficients of x and constant term, we obtain
A + B = 0
−3A + 3B = 1
On solving, we obtain
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Question 3:
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Answer
Let
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Substituting x = 1, 2, and 3 respectively in equation (1), we obtain
A = 1, B = −5, and C = 4
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Question 4:
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Answer
Let
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Substituting x = 1, 2, and 3 respectively in equation (1), we obtain
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Question 5:
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Answer
Let
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Substituting x = −1 and −2 in equation (1), we obtain
A = −2 and B = 4
Question 6:
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Answer
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It can be seen that the given integrand is not a proper fraction. Therefore, on dividing (1 − x2) by x(1 − 2x), we obtain
Let
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Substituting x = 0 and in equation
(1), we obtain
A = 2 and B = 3
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Substituting in equation (1), we obtain
Question 7:
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Answer
Let
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Equating the coefficients of x2, x, and constant term, we obtain
A + C = 0
−A + B = 1
−B + C = 0
On solving these equations, we obtain
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From equation (1), we obtain
Question 8:
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Answer
Let
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Substituting x = 1, we obtain
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Equating the coefficients of x2 and constant term, we obtain
A + C = 0
−2A + 2B + C = 0
On solving, we obtain
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Question 9:
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Answer
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Let
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Substituting x = 1 in equation (1), we obtain
B = 4
Equating the coefficients of x2 and x, we obtain
A + C = 0
B − 2C = 3
On solving, we obtain
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Question 10:
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Answer
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Let
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Equating the coefficients of x2 and x, we obtain
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Question 11:
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Answer
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Let
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Substituting x = −1, −2, and 2 respectively in equation (1), we obtain
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Question 12:
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Answer
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It can be seen that the given integrand is not a proper fraction. Therefore, on dividing (x3 + x + 1) by x2 − 1, we obtain
Let
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Substituting x = 1 and −1 in equation (1), we obtain
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Question 13:
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Answer
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Equating the coefficient of x2, x, and constant term, we obtain
A − B = 0 B − C = 0 A + C = 2
On solving these equations, we obtain
A = 1, B = 1, and C = 1
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Question 14:
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Answer
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Equating the coefficient of x and constant term, we obtain
A = 3
2A + B = −1 ⇒ B = −7
Question 15:
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Answer
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Equating the coefficient of x3, x2, x, and constant term, we obtain
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On solving these equations, we obtain
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Question 16:
[Hint: multiply
numerator and denominator by xn − 1 and put xn = t]
Answer
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Multiplying numerator and denominator by xn − 1, we obtain
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Substituting t = 0, −1 in equation (1), we obtain
A = 1 and B = −1
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Question 17:
[Hint: Put sin x = t]
Answer
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Substituting t = 2 and then t = 1 in equation (1), we obtain
A = 1 and B = −1
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Question 18:
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Answer
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Equating the coefficients of x3, x2, x, and constant term, we obtain
A + C = 0
B + D = 4 4A + 3C = 0
4B + 3D = 10
On solving these equations, we obtain
A = 0, B = −2, C = 0, and D = 6
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Question 19:
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Answer
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Let x2 = t ⇒ 2x dx = dt
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Substituting t = −3 and t = −1 in equation (1), we obtain
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Question 20:
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Answer
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Multiplying numerator and denominator by x3, we obtain
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Let x4 = t ⇒ 4x3dx = dt
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Substituting t = 0 and 1 in (1), we obtain
A = −1 and B = 1
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Question 21:
[Hint: Put ex = t]
Answer
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Let ex = t ⇒ ex dx = dt
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Substituting t = 1 and t = 0 in equation (1), we obtain
A = −1 and B = 1
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Question 22:
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A. 
B. 
C. 
D.
Answer
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Substituting x = 1 and 2 in (1), we obtain
A = −1 and B = 2
Hence, the correct Answer is B.
Question 23:
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A.
B.
C.
D.
Answer
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Equating the coefficients of x2, x, and constant term, we obtain
A + B = 0
C = 0
A = 1
On solving these equations, we obtain
A = 1, B = −1, and C = 0
Hence, the correct Answer is A.
Exercise 7.6
Question 1:
x sin x
Answer
Let I =
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Taking x as first function and sin x as second function and integrating by parts, we obtain
Question 2:
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Answer
Let I = 
Taking x as first function and sin 3x as second function and integrating by parts, we obtain
Question 3:
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Answer
Let 
Taking x2 as first function and ex as second function and integrating by parts, we obtain
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Again integrating by parts, we obtain
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Question 4:
x logx Answer
Let 
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Taking log x as first function and x as second function and integrating by parts, we obtain
Question 5:
x log 2x Answer
Let 
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Taking log 2x as first function and x as second function and integrating by parts, we obtain
Question 6:
x2 log x
Answer
Let 
Taking log x as first function and x2 as second function and integrating by parts, we obtain
Question 7:
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Answer
Let 
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Taking as first function and x as
second function and integrating by parts, we obtain
Question 8:
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Answer Let
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Taking as first function and x as
second function and integrating by parts, we obtain
Question 9:
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Answer
Let 
Taking cos−1 x as first function and x as second function and integrating by parts, we obtain
Page 106 of 216
Question 10:
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Answer
Let 
Taking as first function and 1 as second function and integrating by parts, we obtain
Question 11:
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Answer
Let
Taking as first function and as second
function and integrating by parts, we obtain
Question 12:
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Answer
Let 
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Taking x as first function and sec2x as second function and integrating by parts, we obtain
Question 13:
|
Answer
Let
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Taking as first function and 1 as second function
and integrating by parts, we obtain
Question 14:
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Answer
 |
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Taking as first function and 1 as second function
and integrating by parts, we obtain
Again integrating by parts, we obtain
Question 15:
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Answer
Let 
Let
I = I1 + I2 … (1)
Where, 
and 
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Taking log x as first function and x2 as second function and integrating by parts, we obtain
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Taking log x as first function and 1 as second function and integrating by parts, we obtain
Using equations (2) and (3) in (1), we obtain
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Question 16:
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Answer Let
Let
⇒
⇒
It is known that,
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Question 17:
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Answer
Let
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Let ⇒
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It is known that, 
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Question 18:
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Answer
Let ⇒
It is known that,
From equation (1), we obtain
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Question 19:
Answer
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Also, let ⇒
It is known that,
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Question 20:
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Answer
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Let ⇒
It is known that,
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Question 21:
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Answer
Let
Integrating by parts, we obtain
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Again integrating by parts, we obtain
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Question 22:
 |
Answer
Let ⇒
= 2θ
⇒
Integrating by parts, we obtain
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Question 23:
equals
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Answer
Let 
Also, let ⇒
Hence, the correct Answer is A.
Question 24:
equals
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Answer
 |
Let 
Also, let ⇒
It is known that, Hence, the correct Answer is B.
Exercise 7.7
Question 1:
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Answer
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Question 2:
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Answer
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Question 3:
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Answer
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Question 4:
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Answer
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Question 5:
Answer
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Question 6:
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Answer
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Question 7:
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Answer
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Question 8:
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Answer
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Question 9:
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Answer
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Question 10:
is equal to
A.
B.
C.
D.
Answer
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Hence, the correct Answer is A.
Question 11:
is equal to
A.
B.
C.
D.
Answer
 |
Hence, the correct Answer is D.
Exercise 7.8
Question 1:
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Answer
It is known that,
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Question 2:
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Answer
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It is known that,
 |
Question 3:
Answer
It is known that,
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Question 4:
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Answer
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It is known that,
 |
 |
From equations (2) and (3), we obtain
Question 5:
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Answer
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It is known that,
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Question 6:
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Answer
It is known that,
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Exercise 7.9
Question 1:
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Answer
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By second fundamental theorem of calculus, we obtain
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Question 2:
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Answer
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By second fundamental theorem of calculus, we obtain
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Question 3:
Answer
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By second fundamental theorem of calculus, we obtain
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Question 4:
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Answer
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By second fundamental theorem of calculus, we obtain
Question 5:
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Answer
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By second fundamental theorem of calculus, we obtain
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Question 6:
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Answer
By second fundamental theorem of calculus, we obtain
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Question 7:
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Answer
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By second fundamental theorem of calculus, we obtain
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Question 8:
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Answer
By second fundamental theorem of calculus, we obtain
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Question 9:
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Answer
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By second fundamental theorem of calculus, we obtain
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Question 10:
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Answer
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By second fundamental theorem of calculus, we obtain
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Question 11:
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Answer
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By second fundamental theorem of calculus, we obtain
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Question 12:
Answer
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By second fundamental theorem of calculus, we obtain
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Question 13:
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Answer
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By second fundamental theorem of calculus, we obtain
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Question 14:
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Answer
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By second fundamental theorem of calculus, we obtain
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Question 15:
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Answer
By second fundamental theorem of calculus, we obtain
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Question 16:
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Answer
Let 
 |
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Equating the coefficients of x and constant term, we obtain A = 10 and B = −25
Substituting the value of I1 in (1), we obtain
Question 17:
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Answer
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By second fundamental theorem of calculus, we obtain
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Question 18:
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Answer
By second fundamental theorem of calculus, we obtain
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Question 19:
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Answer
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By second fundamental theorem of calculus, we obtain
Question 20:
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Answer
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By second fundamental theorem of calculus, we obtain
Question 21:
equals
A.
B.
C.
D.
Answer
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By second fundamental theorem of calculus, we obtain
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Hence, the correct Answer is D.
Question 22:
equals
A.
B.
C.
D.
Answer
 |
By second fundamental theorem of calculus, we obtain
Hence, the correct Answer is C.
Exercise 7.10
Question 1:
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Answer
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When x = 0, t = 1 and when x = 1, t = 2
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Question 2:
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Answer
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Also, let
Question 3:
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Answer
 |
Also, let x = tanθ ⇒ dx = sec2θ dθ
When x = 0, θ = 0 and when x = 1,
Takingθas first function and sec2θ as second function and integrating by parts, we obtain
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Question 4:
 |
Answer
 |
Let x + 2 = t2 ⇒ dx = 2tdt
When x = 0, and when x = 2,
t = 2
Question 5:
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Answer
|
Let cos x = t ⇒ −sinx dx = dt
When x = 0, t = 1 and when
Question 6:
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Answer
 |
Let ⇒ dx = dt
Page 153 of 216
Question 7:
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Answer
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Let x + 1 = t ⇒ dx = dt
When x = −1, t = 0 and when x = 1, t = 2
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Question 8:
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Answer
|
Let 2x = t ⇒ 2dx = dt
When x = 1, t = 2 and when x = 2, t = 4
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Question 9:
The value of the integral is
A. 6
B. 0
C. 3
D. 4 Answer
 |
Let cotθ = t ⇒ −cosec2θ dθ= dt
Hence, the correct Answer is A.
Question 10:
If
A. cos x + x sin x
B. x sin x
C. x cos x
D. sin x + x cos x
Answer
 |
Integrating by parts, we obtain
Hence, the correct Answer is B.
Exercise 7.11
Question 1:
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Answer
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Adding (1) and (2), we obtain
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Question 2:
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Answer
Adding (1) and (2), we obtain
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Question 3:
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Answer
Adding (1) and (2), we obtain
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Question 4:
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Answer
Adding (1) and (2), we obtain
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Question 5:
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Answer
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It can be seen that (x + 2) ≤ 0 on [−5, −2] and (x + 2) ≥ 0 on [−2, 5].
Question 6:
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Answer
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It can be seen that (x − 5) ≤ 0 on [2, 5] and (x − 5) ≥ 0 on [5, 8].
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Question 7:
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Answer
Question 8:
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Answer
Question 9:
 |
Answer
Question 10:
 |
Answer
Adding (1) and (2), we obtain
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Question 11:
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Answer
 |
As sin2 (−x) = (sin (−x))2 = (−sin x)2 = sin2x, therefore, sin2x is an even function.
It is known that if f(x) is an even function, then
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Question 12:
 |
Answer
 |
Adding (1) and (2), we obtain
Question 13:
 |
Answer
As sin7 (−x) = (sin (−x))7 = (−sin x)7 = −sin7x, therefore, sin2x is an odd function. It is known that, if f(x) is an odd function, then
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Question 14:
 |
Answer
 |
It is known that,
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Question 15:
 |
Answer
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Adding (1) and (2), we obtain
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Question 16:
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Answer
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Adding (1) and (2), we obtain
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sin (π − x) = sin x
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Adding (4) and (5), we obtain
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Let 2x = t ⇒ 2dx = dt
When x = 0, t = 0 and when
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Question 17:
 |
Answer
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It is known that,
 |
Adding (1) and (2), we obtain
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Question 18:
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Answer
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It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4
Question 19:
Show that if f and g are defined
as and
 |
Answer
 |
Adding (1) and (2), we obtain
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Question 20:
The value of is
A. 0
B. 2
C. π
D.
 |
1 Answer
It is known that if f(x) is an even function, then and if f(x) is an odd function, then
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Hence, the correct Answer is C.
Question 21:
The value of is
A. 2
B.
C. 0
D. 
Answer
 |
Adding (1) and (2), we obtain
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Hence, the correct Answer is C.
Miscellaneous Solutions
Question 1:
 |
Answer
 |
Equating the coefficients of x2, x, and constant term, we obtain
−A + B − C = 0
B + C = 0
A = 1
On solving these equations, we obtain
 |
From equation (1), we obtain
Question 2:
 |
Answer
Question 3:
[Hint: Put ]
Answer
Question 4:
 |
Answer
Question 5:
 |
 |
Answer
 |
On dividing, we obtain
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Question 6:
 |
Answer
 |
Equating the coefficients of x2, x, and constant term, we obtain
A + B = 0 B + C = 5 9A + C = 0
On solving these equations, we obtain
 |
From equation (1), we obtain
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Question 7:
 |
Answer
Let x − a = t ⇒ dx = dt
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Question 8:
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Answer
 |
Question 9:
 |
Answer
Let sin x = t ⇒ cos x dx = dt
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Question 10:
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Answer
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Question 11:
 |
Answer
Question 12:
 |
Answer
|
Let x4 = t ⇒ 4x3 dx = dt
Question 13:
 |
Answer
|
Let ex = t ⇒ ex dx = dt
 |
Question 14:
 |
Answer
Equating the coefficients of x3, x2, x, and constant term, we obtain
A + C = 0 B + D = 0 4A + C = 0
4B + D = 1
On solving these equations, we obtain
 |
From equation (1), we obtain
 |
Question 15:
 |
Answer
= cos3 x × sin x
Let cos x = t ⇒ −sin x dx = dt
Question 16:
 |
Answer
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Question 17:
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Answer
Question 18:
 |
Answer
Question 19:
 |
Answer
From equation (1), we obtain
Question 20:
 |
Answer
 |
 |
Question 21:
 |
Answer
Question 22:
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Answer
 |
Equating the coefficients of x2, x,and constant term, we obtain
A + C = 1
3A + B + 2C = 1
2A + 2B + C = 1
On solving these equations, we obtain
A = −2, B = 1, and C = 3
From equation (1), we obtain
Question 23:
 |
Answer
 |
Question 24:
 |
Answer
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Integrating by parts, we obtain
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Question 25:
 |
Answer
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Question 26:
 |
Answer
When x = 0, t = 0 and
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Question 27:
 |
Answer
When and when
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Question 28:
 |
Answer
 |
When and when
 |
As , therefore, 
is an even function.
It is known that if f(x) is an even function, then
Question 29:
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Answer
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Question 30:
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Answer
 |
Question 31:
 |
Answer
From equation (1), we obtain
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Question 32:
 |
Answer
 |
Adding (1) and (2), we obtain
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Question 33:
Answer
 |
From equations (1), (2), (3), and (4), we obtain
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Question 34:
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Answer
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Equating the coefficients of x2, x, and constant term, we obtain
A + C = 0 A + B = 0 B = 1
On solving these equations, we obtain
A = −1, C = 1, and B = 1
 |
Hence, the given result is proved.
Question 35:
 |
Answer
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Integrating by parts, we obtain
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Question 36:
 |
Answer
 |
Therefore, f (x) is an odd function.
It is known that if f(x) is an odd function, then
 |
Hence, the given result is proved.
Question 37:
 |
Answer
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Question 38:
 |
Answer
 |
Hence, the given result is proved.
Question 39:
 |
Answer
 |
Integrating by parts, we obtain
Let 1 − x2 = t ⇒ −2x dx = dt
 |
Hence, the given result is proved.
Question 40:
 |
Evaluate as a limit of a sum. Answer
It is known that,
 |
 |
Question 41:
is equal to
A.
B.
C.
D.
Answer
 |
Hence, the correct Answer is A.
Question 42:
is equal to
A.
B.
C.
D.
Answer
Hence, the correct Answer is B.
Question 43:
If then is equal to
A.
B.
C.
D.
Answer
 |
Hence, the correct Answer is D.
Question 44:
The value of is
A. 1
B. 0
C. − 1
D.
Answer
 |
Adding (1) and (2), we obtain
 |
Hence, the correct Answer is B.