NCERT Exemplar: Integrals

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 | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

## FAQs on NCERT Exemplar: Integrals - Mathematics (Maths) for JEE Main & Advanced

 1. What is the concept of integration?
Ans. Integration is a mathematical operation that involves finding the antiderivative of a function. It is the reverse process of differentiation and is used to calculate the area under a curve, the accumulation of quantities over time, and various other applications in mathematics and science.
 2. How can integrals be used to find the area under a curve?
Ans. Integrals can be used to find the area under a curve by evaluating the definite integral of the function over a specific interval. The integral represents the area between the curve and the x-axis within that interval. By evaluating this integral, we can determine the exact value of the area.
 3. What are the different methods of integration?
Ans. There are several methods of integration, including: - Integration by substitution: This method involves substituting a variable to simplify the integral. - Integration by parts: This method is based on the product rule of differentiation and involves splitting the integral into two parts. - Partial fraction decomposition: This method is used to integrate rational functions by decomposing them into simpler fractions. - Trigonometric substitution: This method involves substituting trigonometric functions to simplify the integral.
 4. How can integrals be applied in physics and engineering?
Ans. Integrals have numerous applications in physics and engineering. They can be used to calculate the work done by a force, the displacement of an object, or the velocity and acceleration of a particle. In engineering, integrals are commonly used for determining the area, volume, and moment of inertia of objects, as well as for solving differential equations that model physical systems.
 5. What is the fundamental theorem of calculus?
Ans. The fundamental theorem of calculus states that if a function is continuous over a closed interval and has an antiderivative, then the definite integral of the function over that interval can be evaluated by subtracting the antiderivative at the upper limit from the antiderivative at the lower limit. In other words, it relates differentiation and integration, providing a powerful tool for evaluating integrals.

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

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