1.08h Integration by substitution

1. A particle \(P\) of mass 0.5 kg is moving along the positive \(x\)-axis in the direction of \(x\) increasing. At time \(t\) seconds \(( t \geqslant 0 ) , P\) is \(x\) metres from the origin \(O\) and the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The resultant force acting on \(P\) is directed towards \(O\) and has magnitude \(k v ^ { 2 } \mathrm {~N}\), where \(k\) is a positive constant.

When \(x = 1 , v = 4\) and when \(x = 2 , v = 2\)

1. Show that \(v = a b ^ { x }\), where \(a\) and \(b\) are constants to be found. The time taken for the speed of \(P\) to decrease from \(4 \mathrm {~ms} ^ { - 1 }\) to \(2 \mathrm {~ms} ^ { - 1 }\) is \(T\) seconds.
2. Show that \(T = \frac { 1 } { 4 \ln 2 }\)

1. A cyclist and her cycle have a combined mass of 60 kg . The cyclist is moving along a straight horizontal road and is working at a constant rate of 200 W .

When she has travelled a distance \(x\) metres, her speed is \(v \mathrm {~ms} ^ { - 1 }\) and the magnitude of the resistance to motion is \(3 v ^ { 2 } \mathrm {~N}\).

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} x } = \frac { 200 - 3 v ^ { 3 } } { 60 v ^ { 2 } }\) The distance travelled by the cyclist as her speed increases from \(2 \mathrm {~ms} ^ { - 1 }\) to \(4 \mathrm {~ms} ^ { - 1 }\) is \(D\) metres.
2. Find the exact value of \(D\)

1. A particle of mass 2 kg is moving in a straight line on a smooth horizontal surface under the action of a horizontal force of magnitude \(F\) newtons.

At time \(t\) seconds \(( t > 0 )\),

• the particle is moving with speed \(v \mathrm {~ms} ^ { - 1 }\)
• \(F = 2 + v\)

The time taken for the speed of the particle to increase from \(5 \mathrm {~ms} ^ { - 1 }\) to \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is \(T\) seconds.

1. Show that \(T = 2 \ln \frac { 12 } { 7 }\) The distance moved by the particle as its speed increases from \(5 \mathrm {~ms} ^ { - 1 }\) to \(10 \mathrm {~ms} ^ { - 1 }\) is \(D\) metres.
2. Find the exact value of \(D\).

1. A car of mass 500 kg moves along a straight horizontal road.

The engine of the car produces a constant driving force of 1800 N .
The car accelerates from rest from the fixed point \(O\) at time \(t = 0\) and at time \(t\) seconds the car is \(x\) metres from \(O\), moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When the speed of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the resistance to the motion of the car has magnitude \(2 v ^ { 2 } \mathrm {~N}\). At time \(T\) seconds, the car is at the point \(A\), moving with speed \(10 \mathrm {~ms} ^ { - 1 }\).

1. Show that \(T = \frac { 25 } { 6 } \ln 2\)
2. Show that the distance from \(O\) to \(A\) is \(125 \ln \frac { 9 } { 8 } \mathrm {~m}\).

5 In this question you must show detailed reasoning. Using the substitution \(\mathrm { u } = \mathrm { x } + 1\), find the value of the positive integer \(c\) such that \(\int _ { \mathrm { c } } ^ { \mathrm { c } + 4 } \frac { \mathrm { x } } { ( \mathrm { x } + 1 ) ^ { 2 } } \mathrm { dx } = \ln 3 - \frac { 1 } { 3 }\).

6 \includegraphics[max width=\textwidth, alt={}, center]{28beb431-45d5-4300-88fe-00d05d78790b-06_463_702_264_685} The diagram shows the curve \(C\) with parametric equations $$x = \frac { 1 } { 4 } \sin t , \quad y = t \cos t$$ where \(0 \leqslant t \leqslant k\).

1. Find the value of \(k\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) in terms of \(t\). The maximum point on \(C\) is denoted by \(P\).
3. Using your answer to part (ii) and the standard small angle approximations, find an approximation for the \(x\)-coordinate of \(P\).
4. (a) Show that the area of the finite region bounded by \(C\) and the \(x\)-axis is given by $$b \int _ { 0 } ^ { a } t ( 1 + \cos 2 t ) \mathrm { d } t$$ where \(a\) and \(b\) are constants to be determined.
   (b) In this question you must show detailed reasoning. Hence find the exact area of the finite region bounded by \(C\) and the \(x\)-axis.

4. Given that \(y = \operatorname { arsinh } ( \sqrt { } x ) , x > 0\),

1. find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer as a simplified fraction.
2. Hence, or otherwise, find $$\int _ { \frac { 1 } { 4 } } ^ { 4 } \frac { 1 } { \sqrt { [ x ( x + 1 ) ] } } \mathrm { d } x$$ giving your answer in the form \(\ln \left( \frac { a + b \sqrt { } 5 } { 2 } \right)\), where \(a\) and \(b\) are integers.

5. $$I _ { n } = \int _ { 0 } ^ { 5 } \frac { x ^ { n } } { \sqrt { } \left( 25 - x ^ { 2 } \right) } \mathrm { d } x , \quad n \geq 0$$

1. Find an expression for \(\int \frac { x } { \sqrt { \left( 25 - x ^ { 2 } \right) } } \mathrm { d } x , \quad 0 \leq x \leq 5\).
2. Using your answer to part (a), or otherwise, show that $$I _ { n } = \frac { 25 ( n - 1 ) } { n } I _ { n - 2 } , \quad n \geq 2$$
3. Find \(I _ { 4 }\) in the form \(k \pi\), where \(k\) is a fraction.

8. A curve, which is part of an ellipse, has parametric equations $$x = 3 \cos \theta , \quad y = 5 \sin \theta , \quad 0 \leq \theta \leq \frac { \pi } { 2 }$$ The curve is rotated through \(2 \pi\) radians about the \(x\)-axis.

1. Show that the area of the surface generated is given by the integral $$k \pi \int _ { 0 } ^ { a } \sqrt { } \left( 16 c ^ { 2 } + 9 \right) \mathrm { d } c , \text { where } c = \cos \theta$$ and where \(k\) and \(\alpha\) are constants to be found.
2. Using the substitution \(c = \frac { 3 } { 4 } \sinh u\), or otherwise, evaluate the integral, showing all of your working and giving the final answer to 3 significant figures.

10

1. 1. By writing \(\ln x\) as \(( \ln x ) \times 1\), use integration by parts to find \(\int \ln x \mathrm {~d} x\).
   2. Find \(\int ( \ln x ) ^ { 2 } \mathrm {~d} x\).
2. Use the substitution \(u = \sqrt { x }\) to find the exact value of $$\int _ { 1 } ^ { 4 } \frac { 1 } { x + \sqrt { x } } \mathrm {~d} x$$ (7 marks)

3

1. 1. Given that \(\mathrm { f } ( x ) = x ^ { 4 } + 2 x\), find \(\mathrm { f } ^ { \prime } ( x )\).
   2. Hence, or otherwise, find \(\int \frac { 2 x ^ { 3 } + 1 } { x ^ { 4 } + 2 x } \mathrm {~d} x\).
2. 1. Use the substitution \(u = 2 x + 1\) to show that $$\int x \sqrt { 2 x + 1 } \mathrm {~d} x = \frac { 1 } { 4 } \int \left( u ^ { \frac { 3 } { 2 } } - u ^ { \frac { 1 } { 2 } } \right) \mathrm { d } u$$
   2. Hence show that \(\int _ { 0 } ^ { 4 } x \sqrt { 2 x + 1 } \mathrm {~d} x = 19.9\) correct to three significant figures.

9

1. Given that \(y = \frac { 4 x } { 4 x - 3 }\), use the quotient rule to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { ( 4 x - 3 ) ^ { 2 } }\), where \(k\) is an integer.
2. 1. Given that \(y = x \ln ( 4 x - 3 )\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Find an equation of the tangent to the curve \(y = x \ln ( 4 x - 3 )\) at the point where \(x = 1\).
3. 1. Use the substitution \(u = 4 x - 3\) to find \(\int \frac { 4 x } { 4 x - 3 } \mathrm {~d} x\), giving your answer in terms of \(x\).
   2. By using integration by parts, or otherwise, find \(\int \ln ( 4 x - 3 ) \mathrm { d } x\).

8

1. Using integration by parts, find \(\int x \sin ( 2 x - 1 ) \mathrm { d } x\).
2. Use the substitution \(u = 2 x - 1\) to find \(\int \frac { x ^ { 2 } } { 2 x - 1 } \mathrm {~d} x\), giving your answer in terms of \(x\).
   (6 marks)

5 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{aligned} & \mathrm { f } ( x ) = \sqrt { x - 2 } \text { for } x \geqslant 2 \\ & \mathrm {~g} ( x ) = \frac { 1 } { x } \quad \text { for real values of } x , x \neq 0 \end{aligned}$$

1. State the range of f .
2. 1. Find fg(x).
   2. Solve the equation \(\operatorname { fg } ( x ) = 1\).
3. The inverse of f is \(\mathrm { f } ^ { - 1 }\). Find \(\mathrm { f } ^ { - 1 } ( x )\).

6

1. Use integration by parts to find \(\int x \mathrm { e } ^ { 5 x } \mathrm {~d} x\).
2. 1. Use the substitution \(u = \sqrt { x }\) to show that $$\int \frac { 1 } { \sqrt { x } ( 1 + \sqrt { x } ) } \mathrm { d } x = \int \frac { 2 } { 1 + u } \mathrm {~d} u$$
   2. Find the exact value of \(\int _ { 1 } ^ { 9 } \frac { 1 } { \sqrt { x } ( 1 + \sqrt { x } ) } \mathrm { d } x\).

7 Use the substitution \(u = 6 - x ^ { 2 }\) to find the value of \(\int _ { 1 } ^ { 2 } \frac { x ^ { 3 } } { \sqrt { 6 - x ^ { 2 } } } \mathrm {~d} x\), giving your answer in the form \(p \sqrt { 5 } + q \sqrt { 2 }\), where \(p\) and \(q\) are rational numbers.
[0pt] [7 marks]

7

1. Given that \(y = \ln \tanh \frac { x } { 2 }\), where \(x > 0\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosech } x$$
2. A curve has equation \(y = \ln \tanh \frac { x } { 2 }\), where \(x > 0\). The length of the arc of the curve between the points where \(x = 1\) and \(x = 2\) is denoted by \(s\).
   1. Show that $$s = \int _ { 1 } ^ { 2 } \operatorname { coth } x \mathrm {~d} x$$
   2. Hence show that \(s = \ln ( 2 \cosh 1 )\).

5

1. Given that \(u = \cosh ^ { 2 } x\), show that \(\frac { \mathrm { d } u } { \mathrm {~d} x } = \sinh 2 x\).
2. Hence show that $$\int _ { 0 } ^ { 1 } \frac { \sinh 2 x } { 1 + \cosh ^ { 4 } x } \mathrm {~d} x = \tan ^ { - 1 } \left( \cosh ^ { 2 } 1 \right) - \frac { \pi } { 4 }$$

6

1. Given that \(y = ( x - 2 ) \sqrt { 5 + 4 x - x ^ { 2 } } + 9 \sin ^ { - 1 } \left( \frac { x - 2 } { 3 } \right)\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = k \sqrt { 5 + 4 x - x ^ { 2 } }$$ where \(k\) is an integer.
2. Hence show that $$\int _ { 2 } ^ { \frac { 7 } { 2 } } \sqrt { 5 + 4 x - x ^ { 2 } } \mathrm {~d} x = p \sqrt { 3 } + q \pi$$ where \(p\) and \(q\) are rational numbers.
   [0pt] [3 marks]

4.(a)Use the substitution \(x = \sqrt { 3 } \tan u\) to show that $$\int \frac { 1 } { 3 + x ^ { 2 } } \mathrm {~d} x = p \arctan ( p x ) + c$$ where \(p\) is a real constant to be determined and \(c\) is an arbitrary constant.
(b)Use the substitution \(x = \frac { 3 u + 3 } { u - 3 }\) to determine the exact value of \(I\) where $$I = \int _ { - 3 } ^ { 1 } \frac { \ln ( 3 - x ) } { 3 + x ^ { 2 } } \mathrm {~d} x$$ giving your answer in simplest form. \includegraphics[max width=\textwidth, alt={}, center]{a8e9db6b-dfad-4278-82d8-a8fa5ba61008-10_2264_47_314_1984}

6. (a) Use the substitution \(u = \sqrt { t }\) to show that $$\int _ { 1 } ^ { x } \frac { \ln t } { \sqrt { t } } \mathrm {~d} t = 4 - 4 \sqrt { x } + 2 \sqrt { x } \ln x \quad x \geqslant 1$$ (b) The function g is such that $$\int _ { 1 } ^ { x } \mathrm {~g} ( t ) \mathrm { d } t = x - \sqrt { x } \ln x - 1 \quad x \geqslant 1$$

1. Use differentiation to find the function g .
2. Evaluate \(\int _ { 4 } ^ { 16 } \mathrm {~g} ( t ) \mathrm { d } t\) and simplify your answer.
   (c) Find the value of \(x\) (where \(x > 1\) ) that gives the maximum value of $$\int _ { x } ^ { x + 1 } \frac { \ln t } { 2 ^ { t } } \mathrm {~d} t$$

7

1. Find \(\int 5 x ^ { 3 } \sqrt { x ^ { 2 } + 1 } \mathrm {~d} x\).
2. Find \(\int \theta \tan ^ { 2 } \theta \mathrm {~d} \theta\). You may use the result \(\int \tan \theta \mathrm { d } \theta = \ln | \sec \theta | + c\).