NCERT Solutions for Class 12 Maths Chapter 7 – Integrals Miscellaneous Exercise
Page No 352:
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
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
Question 7:
Answer:
Let x − a = t ⇒ dx = dt
Question 8:
Answer:
Question 9:
Answer:
Let sin x = t ⇒ cos x dx = dt
Question 10:
Answer:
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:
Question 17:
Answer:
Question 18:
Answer:
Question 19:
Answer:
Let I=∫sin-1x-cos-1xsin-1x+cos-1xdx
It is known that, sin-1x+cos-1x=π2
⇒I=∫π2-cos-1x-cos-1xπ2dx
=2π∫π2-2cos-1xdx
=2π.π2∫1.dx-4π∫cos-1xdx
=x-4π∫cos-1xdx …(1)
Let I1=∫cos-1x dx
Also, let x=t⇒dx=2 t dt
⇒I1=2∫cos-1t.t dt
=2cos-1t.t22-∫-11-t2.t22dt
=t2cos-1t+∫t21-t2dt
=t2cos-1t-∫1-t2-11-t2dt
=t2cos-1t-∫1-t2dt+∫11-t2dt
=t2cos-1t-t21-t2-12sin-1t+sin-1t
=t2cos-1t-t21-t2+12sin-1t
From equation (1), we obtain
I=x-4πt2cos-1t-t21-t2+12sin-1t =x-4πxcos-1x-x21-x+12sin-1x
=x-4πxπ2-sin-1x-x-x22+12sin-1x 
Question 20:
Answer:
Question 21:
Answer:
Question 22:
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
Page No 353:
Question 23:
Answer:
Question 24:
Answer:
Integrating by parts, we obtain
Question 25:
Answer:
Question 26:
Answer:
When x = 0, t = 0 and 
Question 27:
Answer:
When 
and when
and when
Question 28:
Answer:
When
and when 
and when 
As 
, therefore, 
is an even function.
, therefore, is an even function.
It is known that if f(x) is an even function, then 
Question 29:
Answer:
Question 30:
Answer:
Question 31:
Answer:
From equation (1), we obtain
Question 32:
Answer:
Adding (1) and (2), we obtain
Question 33:
Answer:
From equations (1), (2), (3), and (4), we obtain
Question 34:
Answer:
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:
Integrating by parts, we obtain
Hence, the given result is proved.
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:
Hence, the given result is proved.
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.
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.
Page No 354:
Question 43:
If 
then 
is equal to
then is equal to
A. 
B. 
C. 
D. 
Answer:
Hence, the correct answer is D.
Question 44:
The value of 
is
is
A. 1
B. 0
C. − 1
D. 
Answer:
Adding (1) and (2), we obtain
Hence, the correct answer is B.
