Multiple substitutions or transformations

10
\includegraphics[max width=\textwidth, alt={}, center]{0f081749-4fe0-46e3-96c2-466e69cf49d3-4_620_894_338_687} The function f is defined by \(\mathrm { f } ( x ) = ( \ln x ) ^ { 2 }\) for \(x > 0\). The diagram shows a sketch of the graph of \(y = \mathrm { f } ( x )\). The minimum point of the graph is \(A\). The point \(B\) has \(x\)-coordinate e .

1. State the \(x\)-coordinate of \(A\).
2. Show that \(\mathrm { f } ^ { \prime \prime } ( x ) = 0\) at \(B\).
3. Use the substitution \(x = \mathrm { e } ^ { u }\) to show that the area of the region bounded by the \(x\)-axis, the line \(x = \mathrm { e }\), and the part of the curve between \(A\) and \(B\) is given by $$\int _ { 0 } ^ { 1 } u ^ { 2 } \mathrm { e } ^ { u } \mathrm {~d} u .$$
4. Hence, or otherwise, find the exact value of this area.

10

1. Use the substitution \(u = \tan x\) to show that, for \(n \neq - 1\), $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \left( \tan ^ { n + 2 } x + \tan ^ { n } x \right) \mathrm { d } x = \frac { 1 } { n + 1 }$$
2. Hence find the exact value of
   (a) \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \left( \sec ^ { 4 } x - \sec ^ { 2 } x \right) \mathrm { d } x\),
   (b) \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \left( \tan ^ { 9 } x + 5 \tan ^ { 7 } x + 5 \tan ^ { 5 } x + \tan ^ { 3 } x \right) \mathrm { d } x\).

7 A curve is defined by the equation \(y = 2 x \ln ( 1 + x )\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence verify that the origin is a stationary point of the curve.
2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that the origin is a minimum point.
3. Using the substitution \(u = 1 + x\), show that \(\int \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x = \int \left( u - 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x\), giving your answer in an exact form.
4. Using integration by parts and your answer to part (iii), evaluate \(\int _ { 0 } ^ { 1 } 2 x \ln ( 1 + x ) \mathrm { d } x\).

2 Fig. 8 shows the curve \(y = \frac { x } { \sqrt { x - 2 } }\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \(( 11,11 )\). \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 3 .
2. Show that, for the curve, \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 4 } { 2 ( x - 2 ) ^ { \frac { 3 } { 2 } } }\). Hence find the gradient of the curve at P . Use the result to show that the curve is not symmetrical about \(y = x\).
3. Using the substitution \(u = x - 2\), show that \(\int _ { 3 } ^ { 11 } \frac { x } { \sqrt { x - 2 } } \mathrm {~d} x = 25 \frac { 1 } { 3 }\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\).

3

1. Use the substitution \(u = 1 + x\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$ where \(a\) and \(b\) are to be found.
   Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x\), giving your answer in exact form. Fig. 8 shows the curve \(y = x ^ { 2 } \ln ( 1 + x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{d1206ce8-7716-4205-b98e-664e7ead8a25-3_830_806_907_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
3. Using integration by parts, and the result of part (i), find the exact area enclosed by the curve \(y = x ^ { 2 } \ln ( 1 + x )\), the \(x\)-axis and the line \(x = 1\).

2 Fig. 7 shows part of the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = x \sqrt { 1 + x }\). The curve meets the \(x\)-axis at the origin and at the point P . \begin{figure}[h] \end{figure}

1. Verify that the point P has coordinates \(( - 1,0 )\). Hence state the domain of the function \(\mathrm { f } ( x )\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + 3 x } { 2 \sqrt { 1 + x } }\).
3. Find the exact coordinates of the turning point of the curve. Hence write down the range of the function.
4. Use the substitution \(u = 1 + x\) to show that $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( \begin{array} { l l } u ^ { \frac { 3 } { 2 } } & u ^ { \frac { 1 } { 2 } } \end{array} \right) \mathrm { d } u .$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.

5 Fig. 8 shows the curve \(y = \frac { x } { \sqrt { x - 2 } }\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \(( 11,11 )\). \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 3 .
2. Show that, for the curve, \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 4 } { 2 ( x - 2 ) ^ { \frac { 3 } { 2 } } }\). Hence find the gradient of the curve at P . Use the result to show that the curve is not symmetrical about \(y = x\).
3. Using the substitution \(u = x - 2\), show that \(\int _ { 3 } ^ { 11 } \frac { x } { \sqrt { x - 2 } } \mathrm {~d} x = 25 \frac { 1 } { 3 }\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\).

2

1. Use the substitution \(u = 1 + x\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$ where \(a\) and \(b\) are to be found.
   Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x\), giving your answer in exact form. Fig. 8 shows the curve \(y = x ^ { 2 } \ln ( 1 + x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{35646966-3747-4f1d-bf94-60e9e3130afe-2_829_806_944_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
3. Using integration by parts, and the result of part (i), find the exact area enclosed by the curve \(y = x ^ { 2 } \ln ( 1 + x )\), the \(x\)-axis and the line \(x = 1\).

1. (a) Use the substitution \(x = \sec \theta\) to show that

$$\int \sqrt { } \left( x ^ { 2 } - 1 \right) d x$$ can be written as $$\int \sec \theta \tan ^ { 2 } \theta \mathrm {~d} \theta$$ (3)
(b) Use integration by parts to show that $$\int \sec \theta \tan ^ { 2 } \theta \mathrm {~d} \theta = \frac { 1 } { 2 } [ \sec \theta \tan \theta - \ln | \sec \theta + \tan \theta | ] + \text { constant. }$$ (c) Evaluate \(\int _ { 0 } ^ { \frac { \pi } { 4 } } \sin x \sqrt { } ( \cos 2 x ) \mathrm { d } x\).

8

1. Use the substitution \(u = 1 + x\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$ where \(a\) and \(b\) are to be found.
   Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x\), giving your answer in exact form. Fig. 8 shows the curve \(y = x ^ { 2 } \ln ( 1 + x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{9857ad89-b315-4cd6-8d8d-26a509ca52a8-4_830_809_902_667} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
3. Using integration by parts, and the result of part (i), find the exact area enclosed by the curve \(y = x ^ { 2 } \ln ( 1 + x )\), the \(x\)-axis and the line \(x = 1\).

8 Fig. 8 shows the curve \(y = \frac { x } { \sqrt { x - 2 } }\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \(( 11,11 )\). \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 3 .
2. Show that, for the curve, \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 4 } { 2 ( x - 2 ) ^ { \frac { 3 } { 2 } } }\). Hence find the gradient of the curve at P . Use the result to show that the curve is not symmetrical about \(y = x\).
3. Using the substitution \(u = x - 2\), show that \(\int _ { 3 } ^ { 11 } \frac { x } { \sqrt { x - 2 } } \mathrm {~d} x = 25 \frac { 1 } { 3 }\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\).

5 It is given that \(I = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \cos \theta } { 1 + \cos \theta } \mathrm { d } \theta\).

1. By using the substitution \(t = \tan \frac { 1 } { 2 } \theta\), show that \(I = \int _ { 0 } ^ { 1 } \left( \frac { 2 } { 1 + t ^ { 2 } } - 1 \right) \mathrm { d } t\).
2. Hence find \(I\) in terms of \(\pi\).

5 Show that \(\int _ { 0 } ^ { 1 } x \mathrm { e } ^ { - x ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 2 } - \frac { 1 } { 2 \mathrm { e } }\). Let \(I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \mathrm { e } ^ { - x ^ { 2 } } \mathrm {~d} x\). Show that \(I _ { 2 n + 1 } = n I _ { 2 n - 1 } - \frac { 1 } { 2 \mathrm { e } }\) for \(n \geqslant 1\). Find the exact value of \(I _ { 7 }\).

8

1. The curve \(C _ { 1 }\) has equation \(y = - \ln ( \cos x )\). Show that the length of the arc of \(C _ { 1 }\) from the point where \(x = 0\) to the point where \(x = \frac { 1 } { 3 } \pi\) is \(\ln ( 2 + \sqrt { 3 } )\).
2. The curve \(C _ { 2 }\) has equation \(y = 2 \sqrt { } ( x + 3 )\). The arc of \(C _ { 2 }\) joining the point where \(x = 0\) to the point where \(x = 1\) is rotated through one complete revolution about the \(x\)-axis. Show that the area of the surface generated is $$\frac { 8 } { 3 } \pi ( 5 \sqrt { } 5 - 8 )$$

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 \(\int _ { 1 } ^ { 2 } \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x = 8 \sin \frac { 1 } { 2 } - 4 \cos \frac { 1 } { 2 }\).

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 $$\int _ { 3 } ^ { 27 } \frac { 1 } { x + 3 \sqrt { 2 x - 5 } } \mathrm {~d} x$$

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. (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.

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.

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.

1. In this question you must show all stages of your working.

\section*{Solutions relying on calculator technology are not acceptable.}

1. Use the substitution \(t = \tan \left( \frac { \theta } { 2 } \right)\) to show that $$\int \frac { 1 } { 2 \sin \theta + \cos \theta + 2 } d \theta = \int \frac { a } { ( t + b ) ^ { 2 } + c } d t$$ where \(a\), \(b\) and \(c\) are constants to be determined.
2. Hence show that $$\int _ { \frac { \pi } { 2 } } ^ { \frac { 2 \pi } { 3 } } \frac { 1 } { 2 \sin \theta + \cos \theta + 2 } d \theta = \ln \left( \frac { 2 \sqrt { 3 } } { 3 } \right)$$

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}