1.08h Integration by substitution

4 It is given that, for \(n \geqslant 0\), $$I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \mathrm { e } ^ { x ^ { 3 } } \mathrm {~d} x$$

1. Show that \(I _ { 2 } = \frac { 1 } { 3 } ( \mathrm { e } - 1 )\).
2. Show that, for \(n \geqslant 3\), $$3 I _ { n } = \mathrm { e } - ( n - 2 ) I _ { n - 3 }$$
3. Hence find the exact value of \(I _ { 8 }\).

5 A curve \(C\) is defined parametrically by $$x = \frac { 2 } { \mathrm { e } ^ { t } + \mathrm { e } ^ { - t } } \quad \text { and } \quad y = \frac { \mathrm { e } ^ { t } - \mathrm { e } ^ { - t } } { \mathrm { e } ^ { t } + \mathrm { e } ^ { - t } }$$ for \(0 \leqslant t \leqslant 1\). The area of the surface generated when \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis is denoted by \(S\).

1. Show that \(S = 4 \pi \int _ { 0 } ^ { 1 } \frac { \mathrm { e } ^ { t } - \mathrm { e } ^ { - t } } { \left( \mathrm { e } ^ { t } + \mathrm { e } ^ { - t } \right) ^ { 2 } } \mathrm {~d} t\).
2. Using the substitution \(u = \mathrm { e } ^ { t } + \mathrm { e } ^ { - t }\), or otherwise, find \(S\) in terms of \(\pi\) and e .

2 The curve \(C\) has polar equation \(r ^ { 2 } = \ln ( 1 + \theta )\), for \(0 \leqslant \theta \leqslant 2 \pi\).

1. Sketch \(C\).
2. Using the substitution \(u = 1 + \theta\), or otherwise, find the area of the region bounded by \(C\) and the initial line, leaving your answer in an exact form.

6 Show that $$\frac { \mathrm { d } } { \mathrm {~d} x } \left[ x ^ { n - 1 } \sqrt { } \left( 4 - x ^ { 2 } \right) \right] = \frac { 4 ( n - 1 ) x ^ { n - 2 } } { \sqrt { } \left( 4 - x ^ { 2 } \right) } - \frac { n x ^ { n } } { \sqrt { } \left( 4 - x ^ { 2 } \right) }$$ Let $$I _ { n } = \int _ { 0 } ^ { 1 } \frac { x ^ { n } } { \sqrt { } \left( 4 - x ^ { 2 } \right) } \mathrm { d } x$$ where \(n \geqslant 0\). Prove that $$n I _ { n } = 4 ( n - 1 ) I _ { n - 2 } - \sqrt { } 3$$ for \(n \geq 2\). Given that \(I _ { 0 } = \frac { 1 } { 6 } \pi\), find \(I _ { 4 }\), leaving your answer in an exact form.

9 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x \sin x \mathrm {~d} x\). Given that \(I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x ^ { n } \sin x \mathrm {~d} x\), prove that, for \(n > 1\), $$I _ { n } = n \left( \frac { 1 } { 2 } \pi \right) ^ { n - 1 } - n ( n - 1 ) I _ { n - 2 }$$ By first using the substitution \(x = \cos ^ { - 1 } u\), find the value of $$\int _ { 0 } ^ { 1 } \left( \cos ^ { - 1 } u \right) ^ { 3 } \mathrm {~d} u$$ giving your answer in an exact form.

11 Show that \(\int x \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x = - \frac { 1 } { 3 } \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } + c\), where \(c\) is a constant. Given that \(I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x\), prove that, for \(n \geqslant 2\), $$( n + 2 ) I _ { n } = ( n - 1 ) I _ { n - 2 }$$ Use the substitution \(x = \sin u\) to show that $$\int _ { 0 } ^ { 1 } \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x = \frac { 1 } { 4 } \pi$$ Find \(I _ { 4 }\).

7 In a chemical process, the mass \(M\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac { \mathrm { d } M } { \mathrm {~d} t } = \frac { M } { t \left( 1 + t ^ { 2 } \right) }$$

1. Find \(\int \frac { t } { 1 + t ^ { 2 } } \mathrm {~d} t\).
2. Find constants \(A , B\) and \(C\) such that $$\frac { 1 } { t \left( 1 + t ^ { 2 } \right) } = \frac { A } { t } + \frac { B t + C } { 1 + t ^ { 2 } } .$$
3. Use integration, together with your results in parts (i) and (ii), to show that $$M = \frac { K t } { \sqrt { 1 + t ^ { 2 } } } ,$$ where \(K\) is a constant.
4. When \(t = 1 , M = 25\). Calculate \(K\). What is the mass of the chemical in the long term?

5

1. Explain why \(\int _ { \frac { 1 } { 2 } } ^ { \infty } \frac { x ( 1 - 2 x ) } { x ^ { 2 } + 3 \mathrm { e } ^ { 4 x } } \mathrm {~d} x\) is an improper integral.
   (1 mark)
2. By using the substitution \(u = x ^ { 2 } \mathrm { e } ^ { - 4 x } + 3\), find $$\int \frac { x ( 1 - 2 x ) } { x ^ { 2 } + 3 \mathrm { e } ^ { 4 x } } \mathrm {~d} x$$
3. Hence evaluate \(\int _ { \frac { 1 } { 2 } } ^ { \infty } \frac { x ( 1 - 2 x ) } { x ^ { 2 } + 3 \mathrm { e } ^ { 4 x } } \mathrm {~d} x\), showing the limiting process used.

8 The diagram shows a sketch of a curve. \includegraphics[max width=\textwidth, alt={}, center]{f05737eb-adb1-4228-aebf-6b5c7f26a434-5_464_574_402_726} The polar equation of the curve is $$r = \sin 2 \theta \sqrt { \left( 2 + \frac { 1 } { 2 } \cos \theta \right) } , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$ The point \(P\) is the point of the curve at which \(\theta = \frac { \pi } { 3 }\). The perpendicular from \(P\) to the initial line meets the initial line at the point \(N\).

1. 1. Find the exact value of \(r\) when \(\theta = \frac { \pi } { 3 }\).
   2. Show that the polar equation of the line \(P N\) is \(r = \frac { 3 \sqrt { 3 } } { 8 } \sec \theta\).
   3. Find the area of triangle \(O N P\) in the form \(\frac { k \sqrt { 3 } } { 128 }\), where \(k\) is an integer.
2. 1. Using the substitution \(u = \sin \theta\), or otherwise, find \(\int \sin ^ { n } \theta \cos \theta \mathrm {~d} \theta\), where \(n \geqslant 2\).
   2. Find the area of the shaded region bounded by the line \(O P\) and the arc \(O P\) of the curve. Give your answer in the form \(a \pi + b \sqrt { 3 } + c\), where \(a , b\) and \(c\) are constants.
      (8 marks)

10 \includegraphics[max width=\textwidth, alt={}, center]{febe231d-200a-4957-b41b-de5b9be98b0a-7_352_545_258_239} The diagram shows the curve \(y = \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right)\), for \(1 \leqslant x \leqslant 2\).

1. Use rectangles of width 0.25 to find upper and lower bounds for \(\int _ { 1 } ^ { 2 } \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x\). Give your answers correct to 3 significant figures.
2. 1. Use the substitution \(t = \sqrt { x - 1 }\) to show that \(\int \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x = \int 2 t \sin \left( \frac { 1 } { 2 } t \right) \mathrm { d } t\).
   2. Hence show that \(\int _ { 1 } ^ { 2 } \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x = 8 \sin \frac { 1 } { 2 } - 4 \cos \frac { 1 } { 2 }\).

11

1. Use the substitution \(u ^ { 2 } = x ^ { 2 } + 3\) to show that \(\int \frac { 4 x ^ { 3 } } { \sqrt { x ^ { 2 } + 3 } } \mathrm {~d} x = \frac { 4 } { 3 } \left( x ^ { 2 } - 6 \right) \sqrt { x ^ { 2 } + 3 } + c\).
2. In this question you must show detailed reasoning. \includegraphics[max width=\textwidth, alt={}, center]{6b766f5c-8533-4e0c-bb10-0d9949dc777b-7_620_951_1836_317} The graph shows part of the curve \(y = \frac { 4 x ^ { 3 } } { \sqrt { x ^ { 2 } + 2 } }\).
   Find the exact area enclosed by the curve \(y = \frac { 4 x ^ { 3 } } { \sqrt { x ^ { 2 } + 3 } }\), the normal to this curve at the point \(( 1,2 )\) and the \(x\)-axis.

9 Use the substitution \(x = 2 \sin \theta\) to show that \(\int _ { 1 } ^ { \sqrt { 3 } } \sqrt { 4 - x ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 3 } \pi\).

12

1. Use the substitution \(u = \mathrm { e } ^ { x } - 2\) to show that $$\int \frac { 7 \mathrm { e } ^ { x } - 8 } { \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } } \mathrm {~d} x = \int \frac { 7 u + 6 } { u ^ { 2 } ( u + 2 ) } \mathrm { d } u$$
2. Hence show that $$\int _ { \ln 4 } ^ { \ln 6 } \frac { 7 \mathrm { e } ^ { x } - 8 } { \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } } \mathrm {~d} x = a + \ln b$$ where \(a\) and \(b\) are rational numbers to be determined. \section*{END OF QUESTION PAPER}

1

1. Differentiate the following.
   1. \(\frac { x ^ { 2 } } { 2 x + 1 }\)
   2. \(\tan \left( x ^ { 2 } - 3 x \right)\)
2. Use the substitution \(u = \sqrt { x } - 1\) to integrate \(\frac { 1 } { \sqrt { x } - 1 }\).
3. Integrate \(\frac { x - 2 } { 2 x ^ { 2 } - 8 x - 1 }\).

5

1. Use the trapezium rule, with two strips of equal width, to show that $$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { x } } \mathrm {~d} x \approx \frac { 11 } { 4 } - \sqrt { 2 }$$
2. Use the substitution \(x = u ^ { 2 }\) to find the exact value of $$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { x } } \mathrm {~d} x$$
3. Using your answers to parts (i) and (ii), show that $$\ln 2 \approx k + \frac { \sqrt { 2 } } { 4 }$$ where \(k\) is a rational number to be determined.

9. a. Use the substitution \(t ^ { 2 } = 2 x - 5\) to show that $$\int \frac { 1 } { x + 3 \sqrt { 2 x - 5 } } \mathrm {~d} x = \int \frac { 2 t } { t ^ { 2 } + 6 t + 5 } \mathrm {~d} t$$ b. Hence find the exact value of $$\int _ { 3 } ^ { 27 } \frac { 1 } { x + 3 \sqrt { 2 x - 5 } } \mathrm {~d} x$$

13. \includegraphics[max width=\textwidth, alt={}, center]{63d85737-99d4-4916-a479-fe44f77b1505-25_679_1043_413_607} Figure 5 shows a sketch of part of the curve with equation \(y = \frac { 6 x } { \sqrt { 3 x + 1 } } , \quad x \geq 0\) The finite region \(\mathbf { R }\), shown shaded in figure 5 is bounded by the curve, the \(x\)-axis and the lines \(x = 2\) and \(x = 5\). Use the substitution \(u = 3 x + 1\) to find the exact area of \(\mathbf { R }\).
(Total for Question 13 is 7 marks)

1. Show that

$$\int _ { 0 } ^ { 2 } 2 x \sqrt { x + 2 } \mathrm {~d} x = \frac { 32 } { 15 } ( 2 + \sqrt { 2 } )$$

1. (a) Given that \(a\) is a positive constant, use the substitution \(x = a \sin ^ { 2 } \theta\) to show that

$$\int _ { 0 } ^ { a } x ^ { \frac { 1 } { 2 } } \sqrt { a - x } \mathrm {~d} x = \frac { 1 } { 2 } a ^ { 2 } \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { 2 } 2 \theta \mathrm {~d} \theta$$ (b) Hence use algebraic integration to show that $$\int _ { 0 } ^ { a } x ^ { \frac { 1 } { 2 } } \sqrt { a - x } d x = k \pi a ^ { 2 }$$ where \(k\) is a constant to be found.

1. (a) Use the substitution \(x = u ^ { 2 } + 1\) to show that

$$\int _ { 5 } ^ { 10 } \frac { 3 \mathrm {~d} x } { ( x - 1 ) ( 3 + 2 \sqrt { x - 1 } ) } = \int _ { p } ^ { q } \frac { 6 \mathrm {~d} u } { u ( 3 + 2 u ) }$$ where \(p\) and \(q\) are positive constants to be found.
(b) Hence, using algebraic integration, show that $$\int _ { 5 } ^ { 10 } \frac { 3 \mathrm {~d} x } { ( x - 1 ) ( 3 + 2 \sqrt { x - 1 } ) } = \ln a$$ where \(a\) is a rational constant to be found.

1. Show that

$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \frac { \sin 2 \theta } { 1 + \cos \theta } d \theta = 2 - 2 \ln 2$$

1. (a) Use the substitution \(u = 4 - \sqrt { h }\) to show that

$$\int \frac { \mathrm { d } h } { 4 - \sqrt { h } } = - 8 \ln | 4 - \sqrt { h } | - 2 \sqrt { h } + k$$ where \(k\) is a constant A team of scientists is studying a species of slow growing tree.
The rate of change in height of a tree in this species is modelled by the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { t ^ { 0.25 } ( 4 - \sqrt { h } ) } { 20 }$$ where \(h\) is the height in metres and \(t\) is the time, measured in years, after the tree is planted.
(b) Find, according to the model, the range in heights of trees in this species. One of these trees is one metre high when it is first planted.
According to the model,
(c) calculate the time this tree would take to reach a height of 12 metres, giving your answer to 3 significant figures.

1. (a) Use the substitution \(u = 1 + \sqrt { x }\) to show that

$$\int _ { 0 } ^ { 16 } \frac { \mathrm { x } } { 1 + \sqrt { \mathrm { x } } } \mathrm {~d} x = \int _ { p } ^ { q } \frac { 2 ( u - 1 ) ^ { 3 } } { u } \mathrm {~d} u$$ where \(p\) and \(q\) are constants to be found.
(b) Hence show that $$\int _ { 0 } ^ { 16 } \frac { \mathrm { x } } { 1 + \sqrt { \mathrm { x } } } \mathrm {~d} x = A - B \ln 5$$ where \(A\) and \(B\) are constants to be found.

10. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of the curve \(C\) with parametric equations $$x = \ln ( t + 2 ) , \quad y = \frac { 1 } { t + 1 } , \quad t > - \frac { 2 } { 3 }$$

1. State the domain of values of \(x\) for the curve \(C\). The finite region \(R\), shown shaded in Figure 4, is bounded by the curve \(C\), the line with equation \(x = \ln 2\), the \(x\)-axis and the line with equation \(x = \ln 4\)
2. Use calculus to show that the area of \(R\) is \(\ln \left( \frac { 3 } { 2 } \right)\).