1.08i Integration by parts

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

Solutions relying entirely on calculator technology are not acceptable. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} Figure 2 shows the curve with equation $$y = 10 x \mathrm { e } ^ { - \frac { 1 } { 2 } x } \quad 0 \leqslant x \leqslant 10$$ The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the line with equation \(x = 10\) The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.

1. Show that the volume, \(V\), of this solid is given by $$V = k \int _ { 0 } ^ { 10 } x ^ { 2 } \mathrm { e } ^ { - x } \mathrm {~d} x$$ where \(k\) is a constant to be found.
2. Find \(\int x ^ { 2 } e ^ { - x } d x\) Figure 3 represents an exercise weight formed by joining two of these solids together.
   The exercise weight has mass 5 kg and is 20 cm long.
   Given that $$\text { density } = \frac { \text { mass } } { \text { volume } }$$ and using your answers to part (a) and part (b),
3. find the density of this exercise weight. Give your answer in grams per \(\mathrm { cm } ^ { 3 }\) to 3 significant figures.

1. (i) Find

$$\int x ^ { 2 } \mathrm { e } ^ { x } \mathrm {~d} x$$ (4)
(ii) Use the substitution \(u = \sqrt { 1 - 3 x }\) to show that $$\int \frac { 27 x } { \sqrt { 1 - 3 x } } \mathrm {~d} x = - 2 ( 1 - 3 x ) ^ { \frac { 1 } { 2 } } ( A x + B ) + k$$ where \(A\) and \(B\) are integers to be found and \(k\) is an arbitrary constant.

5. \begin{figure}[h] \end{figure}

1. Find \(\int \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x\) Figure 3 shows a sketch of part of the curve with equation $$y = \frac { 3 + 2 x + \ln x } { x ^ { 2 } } \quad x > 0.5$$ The finite region \(R\), shown shaded in Figure 3, is bounded by the curve, the line with equation \(x = 2\), the \(x\)-axis and the line with equation \(x = 4\)
2. Use the answer to part (a) to find the exact area of \(R\), writing your answer in simplest form.

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

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

1. Use integration by parts to find the exact value of $$\int _ { 0 } ^ { 4 } x ^ { 2 } \mathrm { e } ^ { 2 x } \mathrm {~d} x$$ giving your answer in simplest form.
2. Use integration by substitution to show that $$\int _ { 3 } ^ { \frac { 21 } { 2 } } \frac { 4 x } { ( 2 x - 1 ) ^ { 2 } } \mathrm {~d} x = a + \ln b$$ where \(a\) and \(b\) are constants to be found.

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0\) The curve meets the \(x\)-axis at the origin \(O\) and cuts the \(x\)-axis at the point \(A\) .

1. Find,in terms of \(\ln 2\) ,the \(x\) coordinate of the point \(A\) .
2. Find \(\int x \mathrm { e } ^ { \frac { 1 } { 2 } x } \mathrm {~d} x\) The finite region \(R\) ,shown shaded in Figure 2,is bounded by the \(x\)-axis and the curve with equation \(y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0\)
3. Find,by integration,the exact value for the area of \(R\) . Give your answer in terms of \(\ln 2\) \includegraphics[max width=\textwidth, alt={}, center]{4de08317-5fb9-4789-8d57-ccf463224c78-18_2655_1943_114_118}

7. (a) Use de Moivre's theorem to show that $$\sin 5 \theta \equiv 16 \sin ^ { 5 } \theta - 20 \sin ^ { 3 } \theta + 5 \sin \theta$$ (b) Hence find the five distinct solutions of the equation $$16 x ^ { 5 } - 20 x ^ { 3 } + 5 x + \frac { 1 } { 2 } = 0$$ giving your answers to 3 decimal places where necessary.
(c) Use the identity given in (a) to find $$\int _ { 0 } ^ { \frac { \pi } { 4 } } \left( 4 \sin ^ { 5 } \theta - 5 \sin ^ { 3 } \theta \right) \mathrm { d } \theta$$ expressing your answer in the form \(a \sqrt { } 2 + b\), where \(a\) and \(b\) are rational numbers.

6. $$I _ { n } = \int \mathrm { e } ^ { x } \sin ^ { n } x \mathrm {~d} x \quad n \in \mathbb { Z } \quad n \geqslant 0$$

1. Show that $$I _ { n } = \frac { \mathrm { e } ^ { x } \sin ^ { n - 1 } x } { n ^ { 2 } + 1 } ( \sin x - n \cos x ) + \frac { n ( n - 1 ) } { n ^ { 2 } + 1 } I _ { n - 2 } \quad n \geqslant 2$$
2. Hence find the exact value of $$\int _ { 0 } ^ { \frac { \pi } { 2 } } e ^ { x } \sin ^ { 4 } x d x$$ giving your answer in the form \(A \mathrm { e } ^ { \frac { \pi } { 2 } } + B\) where \(A\) and \(B\) are rational numbers to be determined.

8. $$I _ { n } = \int \cos ^ { n } x \mathrm {~d} x \quad n \geqslant 0$$

1. Prove that for \(n \geqslant 2\) $$I _ { n } = \frac { 1 } { n } \cos ^ { n - 1 } x \sin x + \frac { n - 1 } { n } I _ { n - 2 }$$
2. Show that for positive even integers \(n\) $$\int _ { 0 } ^ { \overline { 2 } } \cos ^ { n } x d x = \frac { ( n - 1 ) ( n - 3 ) \ldots 5 \times 3 \times 1 } { n ( n - 2 ) ( n - 4 ) \ldots 6 \times 4 \times 2 } \times \overline { 2 }$$
3. Hence determine the exact value of $$\int _ { 0 } ^ { \overline { 2 } } \cos ^ { 6 } x \sin ^ { 2 } x d x$$

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5. Given that \(y = \operatorname { artanh } ( \cos x )\)

1. show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } x$$
2. Hence find the exact value of $$\int _ { 0 } ^ { \frac { \pi } { 6 } } \cos x \operatorname { artanh } ( \cos x ) d x$$ giving your answer in the form \(a \ln ( b + c \sqrt { 3 } ) + d \pi\), where \(a , b , c\) and \(d\) are rational numbers to be found.
   (5)
   

5. $$I _ { n } = \int \operatorname { cosec } ^ { n } x \mathrm {~d} x , \quad 0 < x < \frac { \pi } { 2 } , \quad n \geqslant 0$$

1. Show that, for \(n \geqslant 2\) $$I _ { n } = \frac { n - 2 } { n - 1 } I _ { n - 2 } - \frac { 1 } { n - 1 } \cot x \operatorname { cosec } ^ { n - 2 } x$$
2. Hence, or otherwise, find $$\int \operatorname { cosec } ^ { 4 } x \mathrm {~d} x$$ giving your answer in terms of \(\cot x\).

5. $$\mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { \mathrm { n } } x \mathrm { dx } , \mathrm { n } \geqslant 0$$

1. Show that \(I _ { n } = \frac { n - 1 } { n } I _ { n - 2 }\), for \(n \geqslant 2\)
2. Using the result in part (a), find the exact value of $$\int _ { 0 } ^ { \frac { \pi } { 2 } } x \sin ^ { 5 } x \cos x d x$$

4.

1. Show that, for \(n \geqslant 2\)
2. Hence find the functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) such that $$\int x ^ { 4 } \cos x \mathrm {~d} x = \mathrm { f } ( x ) \sin x + \mathrm { g } ( x ) \cos x + c$$ where \(c\) is an arbitrary constant. $$I _ { n } = \int x ^ { n } \cos x \mathrm {~d} x$$
   1. Show that, for \(n \geqslant 2\) $$I _ { n } = x ^ { n } \sin x + n x ^ { n - 1 } \cos x - n ( n - 1 ) I _ { n - 2 }$$
   2. Hence find the functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) such that

1. (a) Differentiate \(x \operatorname { arcosh } 5 x\) with respect to \(x\) (b) Hence, or otherwise, show that

$$\int _ { \frac { 1 } { 4 } } ^ { \frac { 3 } { 5 } } \operatorname { arcosh } 5 x \mathrm {~d} x = \frac { 3 } { 20 } - \frac { 2 \sqrt { 2 } } { 5 } + \ln ( p + q \sqrt { 2 } ) ^ { k } - \frac { 1 } { 4 } \ln r$$ where \(p , q , r\) and \(k\) are rational numbers to be determined.

8. $$y = \arccos ( 2 \sqrt { x } )$$

1. Determine \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
2. Show that $$\int y \mathrm {~d} x = x \arccos ( 2 \sqrt { x } ) + \int \frac { \sqrt { x } } { \sqrt { 1 - 4 x } } \mathrm {~d} x$$
3. Use the substitution \(\sqrt { x } = \frac { 1 } { 2 } \cos \theta\) to show that $$\int _ { 0 } ^ { \frac { 1 } { 8 } } \frac { \sqrt { x } } { \sqrt { 1 - 4 x } } \mathrm {~d} x = \frac { 1 } { 4 } \int _ { a } ^ { b } \cos ^ { 2 } \theta \mathrm {~d} \theta$$ where \(a\) and \(b\) are limits to be determined.
4. Hence, determine the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 8 } } \arccos ( 2 \sqrt { x } ) d x$$

4. \(\quad I _ { n } = \int _ { 0 } ^ { a } ( a - x ) ^ { n } \cos x \mathrm {~d} x , \quad a > 0 , \quad n \geqslant 0\)

1. Show that, for \(n \geqslant 2\), $$I _ { n } = n \tilde { a } ^ { - 1 } - n ( n - 1 ) I _ { n - 2 }$$
2. Hence evaluate \(\int _ { 0 } ^ { \frac { \pi } { 2 } } \left( \frac { \pi } { 2 } - x \right) ^ { 2 } \cos x \mathrm {~d} x\).

4. $$I _ { n } = \int _ { 0 } ^ { \frac { \pi } { 4 } } x ^ { n } \sin 2 x \mathrm {~d} x , \quad n \geqslant 0$$

1. Prove that, for \(n \geqslant 2\), $$I _ { n } = \frac { 1 } { 4 } n \left( \frac { \pi } { 4 } \right) ^ { n - 1 } - \frac { 1 } { 4 } n ( n - 1 ) I _ { n - 2 }$$
2. Find the exact value of \(I _ { 2 }\)
3. Show that \(I _ { 4 } = \frac { 1 } { 64 } \left( \pi ^ { 3 } - 24 \pi + 48 \right)\)

1. (a) Differentiate \(x \operatorname { arsinh } 2 x\) with respect to \(x\).
   (b) Hence, or otherwise, find the exact value of

$$\int _ { 0 } ^ { \sqrt { 2 } } \operatorname { arsinh } 2 x \mathrm {~d} x$$ giving your answer in the form \(A \ln B + C\), where \(A , B\) and \(C\) are real.

5. Given that $$I _ { n } = \int x ^ { n } \sqrt { ( x + 8 ) } \mathrm { d } x , \quad n \geqslant 0 , x \geqslant 0$$

1. show that, for \(n \geqslant 1\) $$I _ { n } = \frac { p x ^ { n } ( x + 8 ) ^ { \frac { 3 } { 2 } } } { 2 n + 3 } - \frac { q n } { 2 n + 3 } I _ { n - 1 }$$ where \(p\) and \(q\) are constants to be found.
2. Use part (a) to find the exact value of $$\int _ { 0 } ^ { 10 } x ^ { 2 } \sqrt { ( x + 8 ) } d x$$ giving your answer in the form \(k \sqrt { 2 }\), where \(k\) is rational.

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 \(\int _ { 0 } ^ { 1 } 2 x \ln ( 1 + x ) \mathrm { d } x\).

8 Fig. 8 shows part of the curve \(y = x \sin 3 x\). It crosses the \(x\)-axis at P . The point on the curve with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\) is Q . \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of P .
2. Show that Q lies on the line \(y = x\).
3. Differentiate \(x \sin 3 x\). Hence prove that the line \(y = x\) touches the curve at Q .
4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).

4. (i) Use Simpson's rule with four intervals, each of width 0.25 , to estimate the value of the integral $$\int _ { 0 } ^ { 1 } x \mathrm { e } ^ { 2 x } \mathrm {~d} x$$ (ii) Find the exact value of the integral $$\int _ { \frac { 1 } { 2 } } ^ { 1 } e ^ { 1 - 2 x } d x$$

7 Fig. 7 shows the curve $$y = 2 x - x \ln x , \text { where } x > 0 .$$ The curve crosses the \(x\)-axis at A , and has a turning point at B . The point C on the curve has \(x\)-coordinate 1 . Lines CD and BE are drawn parallel to the \(y\)-axis. \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of A , giving your answer in terms of e .
2. Find the exact coordinates of B .
3. Show that the tangents at A and C are perpendicular to each other.
4. Using integration by parts, show that $$\int x \ln x \mathrm {~d} x = \frac { 1 } { 2 } x ^ { 2 } \ln x - \frac { 1 } { 4 } x ^ { 2 } + c$$ Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines CD and BE . \section*{[Question 8 is printed overleaf.]}

2 Evaluate \(\int _ { 1 } ^ { 2 } x ^ { 2 } \ln x \mathrm {~d} x\), giving your answer in an exact form.

2 Show that \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } x \sin 2 x \mathrm {~d} x = \frac { 3 \sqrt { 3 } - \pi } { 24 }\).