1.08h Integration by substitution

474 questions

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AQA M2 2006 January Q5
8 marks Moderate -0.3
5 A particle moves such that at time \(t\) seconds its acceleration is given by $$( 2 \cos t \mathbf { i } - 5 \sin t \mathbf { j } ) \mathrm { m } \mathrm {~s} ^ { - 2 }$$
  1. The mass of the particle is 6 kg . Find the magnitude of the resultant force on the particle when \(t = 0\).
  2. When \(t = 0\), the velocity of the particle is \(( 2 \mathbf { i } + 10 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). Find an expression for the velocity of the particle at time \(t\).
AQA M2 2011 January Q1
10 marks Moderate -0.3
1 The velocity of a particle at time \(t\) seconds is \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\), where $$\mathbf { v } = \left( 4 + 3 t ^ { 2 } \right) \mathbf { i } + ( 12 - 8 t ) \mathbf { j }$$
  1. When \(t = 0\), the particle is at the point with position vector \(( 5 \mathbf { i } - 7 \mathbf { j } ) \mathrm { m }\). Find the position vector, \(\mathbf { r }\) metres, of the particle at time \(t\).
  2. Find the acceleration of the particle at time \(t\).
  3. The particle has mass 2 kg . Find the magnitude of the force acting on the particle when \(t = 1\).
AQA M2 2012 January Q2
10 marks Standard +0.3
2 A particle, of mass 50 kg , moves on a smooth horizontal plane. A single horizontal force $$\left[ \left( 300 t - 60 t ^ { 2 } \right) \mathbf { i } + 100 \mathrm { e } ^ { - 2 t } \mathbf { j } \right] \text { newtons }$$ acts on the particle at time \(t\) seconds.
The vectors \(\mathbf { i }\) and \(\mathbf { j }\) are perpendicular unit vectors.
  1. Find the acceleration of the particle at time \(t\).
  2. When \(t = 0\), the velocity of the particle is \(( 7 \mathbf { i } - 4 \mathbf { j } ) \mathrm { ms } ^ { - 1 }\). Find the velocity of the particle at time \(t\).
  3. Calculate the speed of the particle when \(t = 1\).
AQA M3 2007 June Q3
9 marks Moderate -0.3
3 A particle \(P\), of mass 2 kg , is initially at rest at a point \(O\) on a smooth horizontal surface. The particle moves along a straight line, \(O A\), under the action of a horizontal force. When the force has been acting for \(t\) seconds, it has magnitude \(( 4 t + 5 ) \mathrm { N }\).
  1. Find the magnitude of the impulse exerted by the force on \(P\) between the times \(t = 0\) and \(t = 3\).
  2. Find the speed of \(P\) when \(t = 3\).
  3. The speed of \(P\) at \(A\) is \(37.5 \mathrm {~ms} ^ { - 1 }\). Find the time taken for the particle to reach \(A\).
AQA M3 2008 June Q3
4 marks Standard +0.3
3 A particle of mass 0.2 kg lies at rest on a smooth horizontal table. A horizontal force of magnitude \(F\) newtons acts on the particle in a constant direction for 0.1 seconds. At time \(t\) seconds, $$F = 5 \times 10 ^ { 3 } t ^ { 2 } , \quad 0 \leqslant t \leqslant 0.1$$ Find the value of \(t\) when the speed of the particle is \(2 \mathrm {~ms} ^ { - 1 }\).
(4 marks)
AQA M3 2011 June Q1
6 marks Standard +0.3
1 A ball of mass 0.2 kg is hit directly by a bat. Just before the impact, the ball is travelling horizontally with speed \(18 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Just after the impact, the ball is travelling horizontally with speed \(32 \mathrm {~ms} ^ { - 1 }\) in the opposite direction.
  1. Find the magnitude of the impulse exerted on the ball.
  2. At time \(t\) seconds after the ball first comes into contact with the bat, the force exerted by the bat on the ball is \(k \left( 0.9 t - 10 t ^ { 2 } \right)\) newtons, where \(k\) is a constant and \(0 \leqslant t \leqslant 0.09\). The bat stays in contact with the ball for 0.09 seconds. Find the value of \(k\).
Edexcel M4 2015 June Q4
14 marks Standard +0.8
4. A car of mass 900 kg is moving along a straight horizontal road with the engine of the car working at a constant rate of 22.5 kW . At time \(t\) seconds, the speed of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 } ( 0 < v < 30 )\) and the total resistance to the motion of the car has magnitude \(25 v\) newtons.
  1. Show that when the speed of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the acceleration of the car is $$\frac { 900 - v ^ { 2 } } { 36 v } \mathrm {~m} \mathrm {~s} ^ { - 2 }$$ The time taken for the car to accelerate from \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is \(T\) seconds.
  2. Show that $$T = 18 \ln \frac { 8 } { 5 }$$
  3. Find the distance travelled by the car as it accelerates from \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)
OCR M4 2004 June Q3
6 marks Challenging +1.2
3 The region between the curve \(y = x \sqrt { } ( 3 - x )\) and the \(x\)-axis for \(0 \leqslant x \leqslant 3\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid of revolution. Find the \(x\)-coordinate of the centre of mass of this solid.
OCR M4 2005 June Q5
8 marks Challenging +1.2
5 In this question, \(a\) and \(k\) are positive constants.
The region enclosed by the curve \(y = a \mathrm { e } ^ { - \frac { x } { a } }\) for \(0 \leqslant x \leqslant k a\), the \(x\)-axis, the \(y\)-axis and the line \(x = k a\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid of mass \(m\). Show that the moment of inertia of this solid about the \(x\)-axis is \(\frac { 1 } { 4 } m a ^ { 2 } \left( 1 + \mathrm { e } ^ { - 2 k } \right)\).
OCR M4 2007 June Q2
7 marks Challenging +1.2
2 The region \(R\) is bounded by the curve \(y = \sqrt { 4 a ^ { 2 } - x ^ { 2 } }\) for \(0 \leqslant x \leqslant a\), the \(x\)-axis, the \(y\)-axis and the line \(x = a\), where \(a\) is a positive constant. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid of revolution. Find the \(x\)-coordinate of the centre of mass of this solid.
AQA FP2 2008 June Q6
5 marks Standard +0.8
6 By using the substitution \(u = x - 2\), or otherwise, find the exact value of $$\int _ { - 1 } ^ { 5 } \frac { \mathrm {~d} x } { \sqrt { 32 + 4 x - x ^ { 2 } } }$$
AQA FP2 2010 June Q5
18 marks Standard +0.8
5
  1. Using the identities $$\cosh ^ { 2 } t - \sinh ^ { 2 } t = 1 , \quad \tanh t = \frac { \sinh t } { \cosh t } \quad \text { and } \quad \operatorname { sech } t = \frac { 1 } { \cosh t }$$ show that:
    1. \(\tanh ^ { 2 } t + \operatorname { sech } ^ { 2 } t = 1\);
    2. \(\frac { \mathrm { d } } { \mathrm { d } t } ( \tanh t ) = \operatorname { sech } ^ { 2 } t\);
    3. \(\frac { \mathrm { d } } { \mathrm { d } t } ( \operatorname { sech } t ) = - \operatorname { sech } t \tanh t\).
  2. A curve \(C\) is given parametrically by $$x = \operatorname { sech } t , y = 4 - \tanh t$$
    1. Show that the arc length, \(s\), of \(C\) between the points where \(t = 0\) and \(t = \frac { 1 } { 2 } \ln 3\) is given by $$s = \int _ { 0 } ^ { \frac { 1 } { 2 } \ln 3 } \operatorname { sech } t \mathrm {~d} t$$
    2. Using the substitution \(u = \mathrm { e } ^ { t }\), find the exact value of \(s\).
      REFERENCE
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AQA FP2 2013 June Q6
8 marks Challenging +1.2
6
  1. Show that \(\frac { 1 } { 5 \cosh x - 3 \sinh x } = \frac { \mathrm { e } ^ { x } } { m + \mathrm { e } ^ { 2 x } }\), where \(m\) is an integer.
  2. Use the substitution \(u = \mathrm { e } ^ { x }\) to show that $$\int _ { 0 } ^ { \ln 2 } \frac { 1 } { 5 \cosh x - 3 \sinh x } \mathrm {~d} x = \frac { \pi } { 8 } - \frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$$
AQA FP2 2013 June Q7
12 marks Challenging +1.8
7
    1. Show that $$\frac { \mathrm { d } } { \mathrm {~d} u } \left( 2 u \sqrt { 1 + 4 u ^ { 2 } } + \sinh ^ { - 1 } 2 u \right) = k \sqrt { 1 + 4 u ^ { 2 } }$$ where \(k\) is an integer.
    2. Hence show that $$\int _ { 0 } ^ { 1 } \sqrt { 1 + 4 u ^ { 2 } } \mathrm {~d} u = p \sqrt { 5 } + q \sinh ^ { - 1 } 2$$ where \(p\) and \(q\) are rational numbers.
  1. The arc of the curve with equation \(y = \frac { 1 } { 2 } \cos 4 x\) between the points where \(x = 0\) and \(x = \frac { \pi } { 8 }\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
    1. Show that the area \(S\) of the curved surface formed is given by $$S = \pi \int _ { 0 } ^ { \frac { \pi } { 8 } } \cos 4 x \sqrt { 1 + 4 \sin ^ { 2 } 4 x } \mathrm {~d} x$$
    2. Use the substitution \(u = \sin 4 x\) to find the exact value of \(S\).
AQA FP2 2014 June Q6
12 marks Challenging +1.2
6
    1. Use De Moivre's Theorem to show that if \(z = \cos \theta + \mathrm { i } \sin \theta\), then $$z ^ { n } - \frac { 1 } { z ^ { n } } = 2 \mathrm { i } \sin n \theta$$
    2. Write down a similar expression for \(z ^ { n } + \frac { 1 } { z ^ { n } }\).
    1. Expand \(\left( z - \frac { 1 } { z } \right) ^ { 2 } \left( z + \frac { 1 } { z } \right) ^ { 2 }\) in terms of \(z\).
    2. Hence show that $$8 \sin ^ { 2 } \theta \cos ^ { 2 } \theta = A + B \cos 4 \theta$$ where \(A\) and \(B\) are integers.
  1. Hence, by means of the substitution \(x = 2 \sin \theta\), find the exact value of $$\int _ { 1 } ^ { 2 } x ^ { 2 } \sqrt { 4 - x ^ { 2 } } \mathrm {~d} x$$ \includegraphics[max width=\textwidth, alt={}, center]{5287255f-5ac4-401a-b850-758257412ff7-14_1180_1707_1525_153}
AQA FP2 2014 June Q8
11 marks Challenging +1.8
8 A curve has equation \(y = 2 \sqrt { x - 1 }\), where \(x > 1\). The length of the arc of the curve between the points on the curve where \(x = 2\) and \(x = 9\) is denoted by \(s\).
  1. Show that \(s = \int _ { 2 } ^ { 9 } \sqrt { \frac { x } { x - 1 } } \mathrm {~d} x\).
    1. Show that \(\cosh ^ { - 1 } 3 = 2 \ln ( 1 + \sqrt { 2 } )\).
    2. Use the substitution \(x = \cosh ^ { 2 } \theta\) to show that $$s = m \sqrt { 2 } + \ln ( 1 + \sqrt { 2 } )$$ where \(m\) is an integer.
      [0pt] [6 marks]
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Edexcel CP1 2023 June Q2
6 marks Standard +0.3
  1. (a) Write \(x ^ { 2 } + 4 x - 5\) in the form \(( x + p ) ^ { 2 } + q\) where \(p\) and \(q\) are integers.
    (b) Hence use a standard integral from the formula book to find
$$\int \frac { 1 } { \sqrt { x ^ { 2 } + 4 x - 5 } } \mathrm {~d} x$$ (c) Determine the mean value of the function $$\mathrm { f } ( x ) = \frac { 1 } { \sqrt { x ^ { 2 } + 4 x - 5 } } \quad 3 \leqslant x \leqslant 13$$ giving your answer in the form \(A \ln B\) where \(A\) and \(B\) are constants in simplest form.
Edexcel FP1 2019 June Q5
8 marks Challenging +1.2
5. $$I = \int \frac { 1 } { 4 \cos x - 3 \sin x } \mathrm {~d} x \quad 0 < x < \frac { \pi } { 4 }$$ Use the substitution \(t = \tan \left( \frac { x } { 2 } \right)\) to show that $$I = \frac { 1 } { 5 } \ln \left( \frac { 2 + \tan \left( \frac { x } { 2 } \right) } { 1 - 2 \tan \left( \frac { x } { 2 } \right) } \right) + k$$ where \(k\) is an arbitrary constant.
Edexcel FP1 2020 June Q8
16 marks Challenging +1.8
8. $$f ( x ) = \frac { 3 } { 13 + 6 \sin x - 5 \cos x }$$ Using the substitution \(t = \tan \left( \frac { x } { 2 } \right)\)
  1. show that \(\mathrm { f } ( x )\) can be written in the form $$\frac { 3 \left( 1 + t ^ { 2 } \right) } { 2 ( 3 t + 1 ) ^ { 2 } + 6 }$$
  2. Hence solve, for \(0 < x < 2 \pi\), the equation $$\mathrm { f } ( x ) = \frac { 3 } { 7 }$$ giving your answers to 2 decimal places where appropriate.
  3. Use the result of part (a) to show that $$\int _ { \frac { \pi } { 3 } } ^ { \frac { 4 \pi } { 3 } } f ( x ) d x = K \left( \arctan \left( \frac { \sqrt { 3 } - 9 } { 3 } \right) - \arctan \left( \frac { \sqrt { 3 } + 3 } { 3 } \right) + \pi \right)$$ where \(K\) is a constant to be determined.
Edexcel FP1 2021 June Q2
10 marks Challenging +1.2
  1. (i) Use the substitution \(t = \tan \frac { X } { 2 }\) to prove the identity
$$\frac { \sin x - \cos x + 1 } { \sin x + \cos x - 1 } \equiv \sec x + \tan x \quad x \neq \frac { n \pi } { 2 } \quad n \in \mathbb { Z }$$ (ii) Use the substitution \(t = \tan \frac { \theta } { 2 }\) to determine the exact value of $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \frac { 5 } { 4 + 2 \cos \theta } d \theta$$ giving your answer in simplest form.
Edexcel FP1 2022 June Q2
7 marks Standard +0.8
  1. During 2029, the number of hours of daylight per day in London, H, is modelled by the equation
$$H = 0.3 \sin \left( \frac { x } { 60 } \right) - 4 \cos \left( \frac { x } { 60 } \right) + 11.5 \quad 0 \leqslant x < 365$$ where \(x\) is the number of days after 1st January 2029 and the angle is in radians.
  1. Show that, according to the model, the number of hours of daylight in London on the 31st January 2029 will be 8.13 to 3 significant figures.
  2. Use the substitution \(t = \tan \left( \frac { x } { 120 } \right)\) to show that \(H\) can be written as $$H = \frac { a t ^ { 2 } + b t + c } { 1 + t ^ { 2 } }$$ where \(a\), \(b\) and \(c\) are constants to be determined.
  3. Hence determine, according to the model, the date of the first day of 2029 when there will be at least 12 hours of daylight in London.
Edexcel FP2 2023 June Q10
12 marks Challenging +1.8
10. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{78543314-72b7-4366-98a1-dbb6b852632f-32_385_679_280_694} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} A solid playing piece for a board game is modelled by rotating the curve \(C\), shown in Figure 2, through \(2 \pi\) radians about the \(x\)-axis. The curve \(C\) has equation $$y = \sqrt { 1 + \frac { x ^ { 2 } } { 9 } } \quad - 4 \leqslant x \leqslant 4$$ with units as centimetres.
  1. Show that the total surface area, \(S \mathrm {~cm} ^ { 2 }\), of the playing piece is given by $$S = p \pi \int _ { - 4 } ^ { 4 } \sqrt { 81 + 10 x ^ { 2 } } \mathrm {~d} x + q \pi$$ where \(p\) and \(q\) are constants to be determined. Using the substitution \(x = \frac { 9 } { \sqrt { 10 } } \sinh u\), or another algebraic integration method, and showing all your working,
  2. determine the total surface area of the playing piece, giving your answer to the nearest \(\mathrm { cm } ^ { 2 }\)
Edexcel FP2 2024 June Q6
11 marks Challenging +1.8
  1. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
$$I _ { n } = \int \frac { \cos ( n x ) } { \sin x } \mathrm {~d} x \quad n \geqslant 1$$
  1. Show that, for \(n \geqslant 1\) $$I _ { n + 2 } = \frac { 2 \cos ( n + 1 ) x } { n + 1 } + I _ { n }$$
  2. Hence determine the exact value of $$\int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 3 } } \frac { \cos ( 5 x ) } { \sin x } d x$$ giving the answer in the form \(a + b \ln c\) where \(a , b\) and \(c\) are rational numbers to be found.
Edexcel FP2 2024 June Q8
13 marks Challenging +1.8
8. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c20a4592-74c6-4f58-b63b-984b171b1bfd-28_552_380_264_468} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c20a4592-74c6-4f58-b63b-984b171b1bfd-28_524_446_274_1151} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 1 shows a French horn with a detachable bell section.
The shape of the bell section can be modelled by rotating an exponential curve through \(360 ^ { \circ }\) about the \(x\)-axis, where units are centimetres. The model uses the curve shown in Figure 2, with equation $$y = \frac { 9 } { 2 } e ^ { \frac { 1 } { 9 } x } \quad 0 \leqslant x \leqslant 9$$
  1. Show that, according to this model, the external surface area of the bell section is given by $$K \int _ { 0 } ^ { 9 } \mathrm { e } ^ { \frac { 1 } { 9 } x } \sqrt { 4 + \mathrm { e } ^ { \frac { 2 } { 9 } x } } \mathrm {~d} x$$ where \(K\) is a real constant to be determined.
  2. Use the substitution \(u = e ^ { \frac { 1 } { 9 } x }\) to show that $$\int _ { 0 } ^ { 9 } \mathrm { e } ^ { \frac { 1 } { 9 } x } \sqrt { 4 + \mathrm { e } ^ { \frac { 2 } { 9 } x } } \mathrm {~d} x = 9 \int _ { a } ^ { b } \frac { 2 u + u ^ { 3 } } { \sqrt { 4 u ^ { 2 } + u ^ { 4 } } } \mathrm {~d} u + 18 \int _ { a } ^ { b } \frac { 1 } { \sqrt { 4 + u ^ { 2 } } } \mathrm {~d} u$$ where \(a\) and \(b\) are constants to be determined. Hence, using algebraic integration,
  3. determine, according to the model, the external surface area of the bell section of the horn, giving your answer to 3 significant figures.
Edexcel FM2 2019 June Q2
10 marks Challenging +1.2
  1. A particle, \(P\), of mass 0.4 kg is moving along the positive \(x\)-axis, in the positive \(x\) direction under the action of a single force. At time \(t\) seconds, \(t > 0 , P\) is \(x\) metres from the origin \(O\) and the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The force is acting in the direction of \(x\) increasing and has magnitude \(\frac { k } { v }\) newtons, where \(k\) is a constant.
At \(x = 3 , v = 2\) and at \(x = 6 , v = 2.5\)
  1. Show that \(v ^ { 3 } = \frac { 61 x + 9 } { 24 }\) The time taken for the speed of \(P\) to increase from \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is \(T\) seconds.
  2. Use algebraic integration to show that \(T = \frac { 81 } { 61 }\)