Verify the following :
Q.1.
Ans.
L.H.S. =
[Dividing the numerator by the denominator]
[where C_{1} = C – log 2^{2}]
L.H.S. = R.H.S.
Hence proved.
Q.2.
Ans.
L.H.S. =
Put x^{2} + 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, ∴ sec^{2} 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 = t^{2} ∵ dx = 2t . dt
∴
Hence,
Q.11.
Ans.
Let
Let I = I_{1} + I_{2}
Now
and
Put a^{2} – x^{2} = t ⇒ – 2x dx = dt
∴
Since I = I_{1} + I_{2}
∴
Hence,[C = C_{1} + C_{2}]
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 = z^{4})
Ans.
Let
Put x = t^{4} ⇒ dx = 4t^{3} dt
I = I_{1} – I_{2}
Now
Put t^{3} + 1 = z ⇒ 3t^{2} dt = dz
∴
∴ I = I_{1} – I_{2}
Hence,[∵ C = C_{1}  C_{2}]
Q.13.
Ans.
Let
Put
∴
Hence,
Q.14.
Ans.
Let
Hence,
Q.15.
Ans.
Let
[Making perfect square]
Hence,
Q.16.
Ans.
Let
I = I_{1} – I_{2}
Now
Put x^{2} + 9 = t ⇒ 2x dx = dt
x dx = dt
∴ I = I_{1} – I_{2}
Hence,
Q.17.
Ans.
Let
(Making perfect square)
Hence,
Q.18.
Ans.
Let
Put x^{2} = t ⇒ 2x dx = dt ⇒ x dx =
Hence,
Q.19.
Ans.
Let
Put x^{2} = 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 = x^{2})
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 x^{2} = sec θ)
Ans.
Let
Put x^{2} = 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) = x^{2} + 3
f(0) = 0 + 3 = 3
f(0 + h) = (0 + h)^{2} + 3 = h^{2} + 3
f(0 + 2h) = (0 + 2h)^{2} + 3 = 4h^{2} + 3
..............................
..............................
Now
Hence,
Q.28.
Ans.
Let
Here, a = 0 and b = 2
Here f(x) = e^{x}
f(0) = e^{0} = 1
f(0 + h) = e^{0 + h} = e^{h}
f(0 + 2h) = e^{0 + 2h} = e^{2h}
................................
................................
Hence, I = e^{2 }–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 = e^{0} = 1
When x = 1 ∴ t = e^{1} = e
Hence,
Q.30.
Ans.
Let
Put sin^{2} x = t
2 sin x cos x dx = dt
sin x cos x dx =
Changing the limits we get,
When x = 0 ∴ t = sin^{2} 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 + x^{2} = 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 cos^{2} θ, we get
Put tan θ = t
∴ sec^{2} θ dθ = dt
Changing the limits, we get
When θ = 0 ∴ t = tan 0 = 0
Long Answer (L.A.)
Q.35.
Ans.
Let
Put x^{2} = 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 x^{2} = t for the purpose of partial fraction.
We get
⇒ t = At + Ab^{2} + Bt + Ba^{2}
Comparing the like terms, we get
A + B = 1 and Ab^{2} + Ba^{2} = 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) = sec^{2} t
= e^{t} . f(t) = e^{t} tan t =  + e ^{tan1 x} .x + C
Hence, I =  + e ^{tan1 x} .x + C.
Q.40.(Hint: Put x = a tan^{2}θ)
Ans.
Let
Put x = a tan^{2} θ
dx = 2a tan θ . sec^{2} 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 = t^{2})
Ans.
Let
Put tan x = t^{2}
sec^{2} x dx = 2t dt
∴
[Dividing the numerator and denominator by t^{2}]
Put
∴ I = I_{1} + I_{2} ...(i)
Now
Put
Now
Put
So I = I_{1} + I_{2}
⇒
Hence,
Q.44.
Ans.
Let
Dividing the numerator and denominator by cos^{4} x, we have
Put tan x = t ⇒ sec^{2} 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
a^{2}A + B = 1 and b^{2}A = 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 = tan^{2} θ ⇒ dx = 2 tan θ sec^{2} θ dθ
∴
Let us take
Put tan θ = t ⇒ sec^{2} θ dθ = dt
∴
∴
∴ I = tan ^{ 1} √x .x  √x+ tan1 √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(x^{2} + 1) + (x + 2) (Bx + C)
1 = Ax^{2} + A + Bx^{2} + Cx + 2Bx + 2C
1 = (A + B)x^{2} + (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 + x^{2} = t ⇒ 2x dx = dt ⇒ x dx =
But I = a(1 + x^{2} )^{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 ofsin^{3}x cos^{2}x dx is _______.
Ans.
Let I =
Let f(x) = sin^{3} x cos^{2} x f(– x)
= sin^{3}(– x).cos^{2} (– x) = – sin^{3} x cos^{2} x = – f(x)
209 videos443 docs143 tests

1. What is the concept of integration? 
2. How can integrals be used to find the area under a curve? 
3. What are the different methods of integration? 
4. How can integrals be applied in physics and engineering? 
5. What is the fundamental theorem of calculus? 
209 videos443 docs143 tests


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