# NCERT Exemplar: Integrals Notes | Study Mathematics (Maths) Class 12 - JEE

## JEE: NCERT Exemplar: Integrals Notes | Study Mathematics (Maths) Class 12 - JEE

The document NCERT Exemplar: Integrals Notes | Study Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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Verify the following :
Q.1.
Ans.
L.H.S. =
[Dividing the numerator by the denominator]

[where C1 = C – log 22]
L.H.S. = R.H.S.
Hence proved.

Q.2.
Ans.
L.H.S. =
Put x2 + 3x = t
∴ (2x + 3) dx = dt

L.H.S. = R.H.S.
Hence verified.

Evaluate the following:
Q.3.
Ans.

Hence, the required solution is

Q.4.
Ans.
Let

Hence, the required solution is

Q.5.
Ans.
Let
Put x + sin x = t ⇒ (1 + cos x) dx = dt

Hence, the required solution is

Q.6.
Ans.
Let

Hence, the required solution is

Q.7.
Ans.
Let

Put tan x = t, ∴ sec2 x dx = dt

Hence, the required solution is

Q.8.
Ans.
Let

= x + C
Hence, the required solution is x + C.

Q.9.
Ans.
Let

Hence, the required solution is

Q.10.
Ans.

Put √x = t  x = t2 ∵ dx = 2t . dt

Hence,

Q.11.
Ans.
Let

Let I = I1 + I2
Now
and
Put a2 – x2 = t ⇒ – 2x dx = dt

Since I = I1 + I2

Hence,[C = C1 + C2]
Alternate method:

Put x = a cos 2θ
∴ dx = a (– 2 sin 2θ) dθ =  – 2a sin 2θ dθ

Now x = a cos 2θ

Hence,

Q.12. (Hint : Put x = z4)
Ans.
Let

Put x = t4 ⇒ dx = 4t3 dt

I = I1 – I2

Now

Put t3 + 1 = z ⇒ 3t2 dt = dz

∴ I = I1 – I2

Hence,[∵ C = C1 - C2]

Q.13.
Ans.
Let
Put

Hence,

Q.14.
Ans.
Let

Hence,

Q.15.
Ans.
Let
[Making perfect square]

Hence,

Q.16.
Ans.
Let

I = I1 – I2

Now
Put x2 + 9 = t ⇒ 2x dx = dt
x dx = dt

∴ I = I1 – I2

Hence,

Q.17.
Ans.
Let
(Making perfect square)

Hence,

Q.18.
Ans.
Let
Put x2 = t ⇒ 2x dx = dt ⇒ x dx =

Hence,

Q.19.
Ans.
Let

Put x2 = t for the purpose of partial fractions.

We get
Resolving into partial fractions we put

[where A and B are arbitrary constants]

⇒ t = A + At + B – Bt
Comparing the like terms, we get A – B = 1 and A + B = 0
Solving the above equations, we have
(Putting t = x2)

Hence,

Q.20.
Ans.
Let

Hence,

Q.21.
Ans.
Let
Put x = sin θ ⇒ dx  = cos θ dθ

Hence,

Q.22.
Ans.
Let I =

[∵ 2 cos A cos B = cos (A + B) + cos (A - B)]

Hence,

Q.23.
Ans.
Let

= tan x – cot x – 3x + C
Hence, I = tan x – cot x – 3x + C.

Q.24.
Ans.
Let I =
Put

Hence,

Q.25.
Ans.
Let

= x + 2 sin x + C
Hence, I = x + 2 sin x + C.

Q.26.(Hint : Put x2 = sec θ)
Ans.
Let
Put x2 = sec θ
∴ 2x dx = sec θ tan θ dθ

So
Hence,

Evaluate the following as limit of sums:
Q.27.
Ans.
Let
Using the formula,

where h =
Here, a  = 0 and b = 2

Here, f(x) = x2 + 3
f(0) = 0 + 3 = 3
f(0 + h) = (0 + h)2 + 3 = h2 + 3
f(0 + 2h) = (0 + 2h)2 + 3 = 4h2 + 3

..............................

..............................

Now

Hence,

Q.28.
Ans.
Let
Here, a  = 0 and b = 2

Here f(x) = ex
f(0) = e0 = 1
f(0 + h) = e0 + h = eh
f(0 + 2h) = e0 + 2h = e2h
................................

................................

Hence, I = e–1.

Evaluate the following:
Q.29.
Ans.
Let

Put ex = t ⇒ ex dx = dt
Changing the limit, we have
When x  = 0 ∴ t = e0 = 1
When x  = 1 ∴ t = e1 = e

Hence,

Q.30.
Ans.
Let

Put sin2 x = t
2 sin x cos x dx = dt
sin x cos x dx =
Changing the limits we get,
When x  = 0 ∴ t = sin2 0 = 0;  When x =

Hence,

Q.31.
Ans.
Let

[Making perfect square]

= sin– 1 (4 – 3) – sin– 1 (2 – 3)
= sin– 1 (1) – sin– 1 (– 1) = sin– 1 (1) + sin– 1 (1)

Hence, I = π.

Q.32.
Ans.
Let
Put 1 + x2 = t ⇒ 2x dx = dt ⇒ x dx =
Changing the limits, we have
When x  = 0 ∴ t = 1
When x  = 1 ∴ t = 2

Hence, I = √2 -1 .

Q.33.
Ans.
Let...(i)

...(ii)
Adding (i) and (ii) we get,

Put cos x = t ⇒ – sin x dx = dt ⇒ sin x dx = – dt
Changing the limits, we have
When x  = 0, t = cos 0 = 1; When x = p, t = cos p = – 1

Q.34.(Hint: let x = sinθ)
Ans.
Let
Put x = sin θ
∴ dx = cos θ dθ
Changing the limits, we get
When x  = 0 ∴ sin θ = 0 ∴ θ = 0

Now, dividing the numerator and denominator by cos2 θ, we get

Put tan θ = t
∴ sec2 θ dθ = dt
Changing the limits, we get
When θ = 0 ∴ t = tan 0 = 0

Q.35.
Ans.
Let

Put x2 = t for the purpose of partial fraction.

We get
Let
[where A and B are arbitrary constants]

⇒ t = At + 3A + Bt – 4B
Comparing the like terms, we get
A + B = 1 and 3A – 4B = 0
⇒ 3A = 4B

Now

So,

Hence,

Q.36.
Ans.
Let

Put x2 = t for the purpose of partial fraction.

We get

⇒ t = At + Ab2 + Bt + Ba2
Comparing the like terms, we get
A + B = 1 and Ab2 + Ba2 = 0

So

Hence,

Q.37.
Ans.
Let...(i)

...(ii)
Adding (i) and (ii), we get

2I = π [0 - (- 1 - 1)] = π(2)
∴ I = π
Hence, I = π.

Q.38.
Ans.
Let
Resolving into partial fraction, we put

2x – 1 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2)
put x  = 1, 1 = A(3)(– 2)
put x  = – 2, – 5 = B(– 3)(– 5)
put x  = 3, 5 = C(2)(5)

Hence,

Q.39.
Ans.
Let
Put tan– 1x = t

Here f(t) = tan t
∴ f ′(t) = sec2 t
= et . f(t) = et tan t = - + e tan-1 x .x + C

Hence, I = - + e tan-1 x .x + C.

Q.40.(Hint: Put x = a tan2θ)
Ans.
Let
Put x = a tan2 θ
dx = 2a tan θ . sec2 θ . dθ

Hence,

Q.41.
Ans.
Let

Put
Changing the limits, we have

Hence,

Q.42.
Ans.
Let

Now, put

Hence,

Q.43.(Hint: Put tanx = t2)
Ans.
Let
Put tan x = t2
sec2 x dx = 2t dt

[Dividing the numerator and denominator by t2]

Put
∴ I = I1 + I2 ...(i)
Now
Put

Now
Put

So I = I1 + I2

Hence,

Q.44.
Ans.
Let
Dividing the numerator and denominator by cos4 x, we have

Put tan x = t ⇒ sec2 x dx = dt
Changing the limits, we get
When x  = 0, t = tan 0 = 0

Put t 2 = u only for the purpose of partial fraction

Comparing the coefficients of like terms, we get
a2A + B = 1 and b2A = 1
Now

Hence, I =

Q.45.
Ans.
Let I =

Hence,

Q.46.
Ans.
Let I =
log sin (π - x) dx

...(ii)
Adding (i) and (ii), we get

...(iii)

...(iv)
On adding (iii) and (iv), we get

Put 2x = t ⇒ 2 dx = dt
dx [Changing the limit]
2I = I -π . log 2 [ x]0π/2 [from eqn. (iii)]

So

Q.47.
Ans.
Let I = ...(i)

...(ii)
Adding (i) and (ii), we get

cos 2x dx
Put 2x = t
Changing the limits we get
When x = 0 ∴ t = 0; When x =
...(iii)

...(iv)

On adding (iii) and (iv), we get,

Put 2t = u ⇒ 2 dt = du

[From eq. (ii)]

Hence,

Objective Type Questions
Q.48.is equal to
(a) 2(sinx + xcosθ) + C
(b) 2(sinx – xcosθ) + C
(c) 2(sinx + 2xcosθ) + C
(d) 2(sinx – 2x cosθ) + C
Ans. (a)
Solution.

Let I =

∴ I = 2(sin x +cos θ .x)+ C.

Hence, correct option is (a).

Q.49.is equal to
(a) sin (b – a) log
(b) cosec (b – a) log
(c) cosec (b – a) log
(d) sin (b – a) log
Ans. (c)
Solution.

Let I =
Multiplying and dividing by sin (b – a) we get,

Hence, the correct option is (c).

Q.50. is equal to
(a) (x + 1) tan –1 √x – √x + C
(b) x tan –1 √x – √x + C
(c) √x – x tan –1 √x + C
(d) √x – ( x + 1) tan –1 √x + C
Ans. (a)
Solution.

Let I =
Put √x = tan θ ⇒  x = tan2 θ ⇒ dx = 2 tan θ sec2 θ dθ

Let us take

Put tan θ = t ⇒ sec2 θ dθ = dt

∴ I = tan - 1 √x .x - √x+ tan-1 √x + C =

( x + 1) tan - 1 √x -√x+ C
Hence, the correct option is (a).

Q.51.is equal to
(a)
(b)
(c)
(d)
Ans. (a)
Solution.

Let I =

Here f(x) =
Using

Hence, the correct option is (a).

Q.52.is equal to
(a)
(b)
(c)
(d)
Ans. (d)
Solution.

Let
Put

Hence, the correct option is (d).

Q.53. Ifthen
(a)
(b)
(c)
(d)
Ans. (c)
Solution.

Let I =
Put
1 = A(x2 + 1) + (x + 2) (Bx + C)
1 = Ax2 + A + Bx2 + Cx + 2Bx + 2C
1 = (A + B)x2 + (C + 2B)x + (A + 2C)
Comparing the like terms, we have
A + B = 0 ...(i)
2B + C = 0 ...(ii)
A + 2C = 1 ...(iii)
Subtracting (i) from (iii) we get
2C  B = 1 ∴ B = 2C – 1

Putting the value of B in eqn. (ii) we have
2(2C – 1) + C = 0 ⇒ 4C – 2 + C = 0

5C = 2

Putting the given value of I

Hence, the correct option is (c).

Q.54.is equal to
(a)
(b)
(c)
(d)
Ans. (d)
Solution.

Let I =

Hence, the correct option is (d).

Q.55.is equal to
(a) log 1 + cos x + C
(b) log x + sin x + C
(c)
(d)
Ans. (d)
Solution.

Let I =

Hence, the correct option is (d).

Q.56. Ifthen
(a)
(b)
(c)
(d)
Ans. (d)
Solution.

Let I =
Put 1 + x2 = t ⇒ 2x dx = dt ⇒ x dx =

But I = a(1 + x2 )3/2 +
Hence, the correct option is (d).

Q.57.is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (a)
Solution.

Let I =

Hence, the correct option is (a).

Q.58.is equal to
(a) 2√2
(b) 2( √2 + 1)
(c) 2
(d) 2( √2 - 1)
Ans. (d)
Solution.

Let I =

Hence, the correct option is (d).

Q.59.is equal to _______.
Ans.
Let I =
Put sin x = t ⇒ cos x dx = dt
When x  = 0 then t = sin 0 = 0;  When x =

Hence, I = e – 1.

Q.60.= ________.

Ans.
Let I =

Put
Let f(x) =
Using

Hence,

Fill in the blanks
Q.61. Ifthen a = ________.
Ans.
Given that:

Q.62.= ________.
Ans.
Let I =
Put cos x = t
∴ – sin x dx = dt ⇒ sin x dx = – dt

Hence,

Q.63. The value ofsin3x cos2x dx is _______.
Ans.
Let I =
Let f(x) = sin3 x cos2 x f(– x)
= sin3(– x).cos2 (– x) = – sin3 x cos2 x = – f(x)

The document NCERT Exemplar: Integrals Notes | Study Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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