NCERT Exemplar: Integrals | Mathematics (Maths) for JEE Main & Advanced PDF Download

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) = e^x 
f(0) = e⁰ = 1
f(0 + h) = e^{0 + h} = e^h 
f(0 + 2h) = e^{0 + 2h} = e^2h 
................................

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





Hence, I = e² –1.

Evaluate the following:
Q.29.
Ans.
Let 

Put e^x = t ⇒ e^x 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




Long Answer (L.A.)
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^{– 1}x = t

Here f(t) = tan t
∴ f ′(t) = sec² t
= e^t . f(t) = e^t 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]₀^π/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)