1.08h Integration by substitution

8. (a) Using the substitution \(x = 2 \cos u\), or otherwise, find the exact value of $$\int _ { 1 } ^ { \sqrt { 2 } } \frac { 1 } { x ^ { 2 } \sqrt { } \left( 4 - x ^ { 2 } \right) } d x$$ \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation \(y = \frac { 4 } { x \left( 4 - x ^ { 2 } \right) ^ { \frac { 1 } { 4 } } } , \quad 0 < x < 2\). The shaded region \(S\), shown in Figure 3, is bounded by the curve, the \(x\)-axis and the lines with equations \(x = 1\) and \(x = \sqrt { } 2\). The shaded region \(S\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
(b) Using your answer to part (a), find the exact volume of the solid of revolution formed.

7. $$I = \int _ { 2 } ^ { 5 } \frac { 1 } { 4 + \sqrt { } ( x - 1 ) } \mathrm { d } x$$

1. Given that \(y = \frac { 1 } { 4 + \sqrt { } ( x - 1 ) }\), complete the table below with values of \(y\) corresponding to \(x = 3\) and \(x = 5\). Give your values to 4 decimal places.

   \(x\)	2	3	4	5
   \(y\)	0.2		0.1745

2. Use the trapezium rule, with all of the values of \(y\) in the completed table, to obtain an estimate of \(I\), giving your answer to 3 decimal places.
3. Using the substitution \(x = ( u - 4 ) ^ { 2 } + 1\), or otherwise, and integrating, find the exact value of \(I\).

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve with equation \(y = \frac { 2 \sin 2 x } { ( 1 + \cos x ) } , 0 \leqslant x \leqslant \frac { \pi } { 2 }\).
The finite region \(R\), shown shaded in Figure 3, is bounded by the curve and the \(x\)-axis. The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 2 \sin 2 x } { ( 1 + \cos x ) }\).

1. Complete the table above giving the missing value of \(y\) to 5 decimal places.
2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of \(R\), giving your answer to 4 decimal places.
3. Using the substitution \(u = 1 + \cos x\), or otherwise, show that $$\int \frac { 2 \sin 2 x } { ( 1 + \cos x ) } d x = 4 \ln ( 1 + \cos x ) - 4 \cos x + k$$ where \(k\) is a constant.
4. Hence calculate the error of the estimate in part (b), giving your answer to 2 significant figures.

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \frac { x } { 1 + \sqrt { } x }\). The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis, the line with equation \(x = 1\) and the line with equation \(x = 4\).

1. Complete the table with the value of \(y\) corresponding to \(x = 3\), giving your answer to 4 decimal places.
   (1)

   \(x\)	1	2	3	4
   \(y\)	0.5	0.8284		1.3333

2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate of the area of the region \(R\), giving your answer to 3 decimal places.
3. Use the substitution \(u = 1 + \sqrt { } x\), to find, by integrating, the exact area of \(R\).
   \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{a98d4a7f-1e6d-4294-9b5c-c945e8fbe83e-07_743_1568_219_182} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Figure 2 shows a sketch of part of the curve \(C\) with parametric equations $$x = 1 - \frac { 1 } { 2 } t , \quad y = 2 ^ { t } - 1$$ The curve crosses the \(y\)-axis at the point \(A\) and crosses the \(x\)-axis at the point \(B\).

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \frac { 4 \mathrm { e } ^ { - x } } { 3 \sqrt { } \left( 1 + 3 \mathrm { e } ^ { - x } \right) }\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis, the line \(x = - 3 \ln 2\) and the \(y\)-axis. The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 4 \mathrm { e } ^ { - x } } { 3 \sqrt { } \left( 1 + 3 \mathrm { e } ^ { - x } \right) }\)

1. Complete the table above by giving the missing value of \(y\) to 4 decimal places.
2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of \(R\), giving your answer to 2 decimal places.
3. 1. Using the substitution \(u = 1 + 3 \mathrm { e } ^ { - x }\), or otherwise, find $$\int \frac { 4 \mathrm { e } ^ { - x } } { 3 \sqrt { } \left( 1 + 3 \mathrm { e } ^ { - x } \right) } \mathrm { d } x$$
   2. Hence find the value of the area of \(R\).

4. Use the substitution \(x = \sin \theta\) to find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x$$

2. Use the substitution \(u = 2 ^ { x }\) to find the exact value of $$\int _ { 0 } ^ { 1 } \frac { 2 ^ { x } } { \left( 2 ^ { x } + 1 \right) ^ { 2 } } d x$$

2. Using the substitution \(u = \cos x + 1\), or otherwise, show that $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { e } ^ { \cos x + 1 } \sin x \mathrm {~d} x = \mathrm { e } ( \mathrm { e } - 1 )$$ (6)

3. Using the substitution \(u = 2 + \sqrt { } ( 2 x + 1 )\), or other suitable substitutions, find the exact value of $$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { } ( 2 x + 1 ) } d x$$ giving your answer in the form \(A + 2 \ln B\), where \(A\) is an integer and \(B\) is a positive constant.

1. (a) Use the substitution \(x = u ^ { 2 } , u > 0\), to show that

$$\int \frac { 1 } { x ( 2 \sqrt { x } - 1 ) } \mathrm { d } x = \int \frac { 2 } { u ( 2 u - 1 ) } \mathrm { d } u$$ (b) Hence show that $$\int _ { 1 } ^ { 9 } \frac { 1 } { x ( 2 \sqrt { x } - 1 ) } \mathrm { d } x = 2 \ln \left( \frac { a } { b } \right)$$ where \(a\) and \(b\) are integers to be determined.

6. (i) Given that \(y > 0\), find $$\int \frac { 3 y - 4 } { y ( 3 y + 2 ) } d y$$ (ii) (a) Use the substitution \(x = 4 \sin ^ { 2 } \theta\) to show that $$\int _ { 0 } ^ { 3 } \sqrt { \left( \frac { x } { 4 - x } \right) } \mathrm { d } x = \lambda \int _ { 0 } ^ { \frac { \pi } { 3 } } \sin ^ { 2 } \theta \mathrm {~d} \theta$$ where \(\lambda\) is a constant to be determined.
(b) Hence use integration to find $$\int _ { 0 } ^ { 3 } \sqrt { \left( \frac { x } { 4 - x } \right) } d x$$ giving your answer in the form \(a \pi + b\), where \(a\) and \(b\) are exact constants.

3. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \frac { 6 } { \left( \mathrm { e } ^ { x } + 2 \right) } , x \in \mathbb { R }\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(y\)-axis, the \(x\)-axis and the line with equation \(x = 1\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 6 } { \left( \mathrm { e } ^ { x } + 2 \right) }\)

1. Complete the table above by giving the missing value of \(y\) to 5 decimal places.
2. Use the trapezium rule, with all the values of \(y\) in the completed table, to find an estimate for the area of \(R\), giving your answer to 4 decimal places.
3. Use the substitution \(u = \mathrm { e } ^ { x }\) to show that the area of \(R\) can be given by $$\int _ { a } ^ { b } \frac { 6 } { u ( u + 2 ) } \mathrm { d } u$$ where \(a\) and \(b\) are constants to be determined.
4. Hence use calculus to find the exact area of \(R\). [Solutions based entirely on graphical or numerical methods are not acceptable.]

3. (i) Given that $$\frac { 13 - 4 x } { ( 2 x + 1 ) ^ { 2 } ( x + 3 ) } \equiv \frac { A } { ( 2 x + 1 ) } + \frac { B } { ( 2 x + 1 ) ^ { 2 } } + \frac { C } { ( x + 3 ) }$$

1. find the values of the constants \(A , B\) and \(C\).
2. Hence find $$\int \frac { 13 - 4 x } { ( 2 x + 1 ) ^ { 2 } ( x + 3 ) } \mathrm { d } x , \quad x > - \frac { 1 } { 2 }$$ (ii) Find $$\int \left( \mathrm { e } ^ { x } + 1 \right) ^ { 3 } \mathrm {~d} x$$ (iii) Using the substitution \(u ^ { 3 } = x\), or otherwise, find $$\int \frac { 1 } { 4 x + 5 x ^ { \frac { 1 } { 3 } } } \mathrm {~d} x , \quad x > 0$$

3. Use the substitution \(x = \tan \theta\) to show that $$\int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x = \frac { \pi } { 8 } + \frac { 1 } { 4 }$$ (8)

8. \begin{figure}[h] \end{figure} The curve \(C\) shown in Figure 1 has polar equation $$r = 1 - \sin \theta \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$ The point \(P\) lies on \(C\), such that the tangent to \(C\) at \(P\) is parallel to the initial line.

1. Use calculus to determine the polar coordinates of \(P\) The finite region \(R\), shown shaded in Figure 1, is bounded by
   $$\frac { 1 } { 32 } ( a \pi + b \sqrt { 3 } + c )$$ where \(a\), \(b\) and \(c\) are integers to be determined.

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve with polar equation $$r = 10 \cos \theta + \tan \theta \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$ The point \(P\) lies on the curve where \(\theta = \frac { \pi } { 3 }\) The region \(R\), shown shaded in Figure 1, is bounded by the initial line, the curve and the line \(O P\), where \(O\) is the pole. Use algebraic integration to show that the exact area of \(R\) is $$\frac { 1 } { 12 } ( a \pi + b \sqrt { 3 } + c )$$ where \(a\), \(b\) and \(c\) are integers to be determined.

1. (a) For all the values of \(x\) where the identity is defined, prove that

$$\cot 2 x + \tan x \equiv \operatorname { cosec } 2 x$$ (b) Show that the substitution \(y ^ { 2 } = w \sin 2 x\), where \(w\) is a function of \(x\), transforms the differential equation $$y \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } \tan x = \sin x \quad 0 < x < \frac { \pi } { 2 }$$ into the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} x } + 2 w \operatorname { cosec } 2 x = \sec x \quad 0 < x < \frac { \pi } { 2 }$$ (c) By solving differential equation (II), determine a general solution of differential equation (I) in the form \(y ^ { 2 } = \mathrm { f } ( x )\), where \(\mathrm { f } ( x )\) is a function in terms of \(\cos x\) $$\text { [You may use without proof } \left. \int \operatorname { cosec } 2 x \mathrm {~d} x = \frac { 1 } { 2 } \ln | \tan x | \text { (+ constant) } \right]$$

8. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve \(C\) with equation $$r = 6 ( 1 + \cos \theta ) \quad 0 \leqslant \theta \leqslant \pi$$ Given that \(C\) meets the initial line at the point \(A\), as shown in Figure 1,

1. write down the polar coordinates of \(A\). The line \(l _ { 1 }\), also shown in Figure 1, is the tangent to \(C\) at the point \(B\) and is parallel to the initial line.
2. Use calculus to determine the polar coordinates of \(B\). The line \(l _ { 2 }\), also shown in Figure 1, is the tangent to \(C\) at \(A\) and is perpendicular to the initial line. The region \(R\), shown shaded in Figure 1, is bounded by \(C , l _ { 1 }\) and \(l _ { 2 }\)
3. Use algebraic integration to find the exact area of \(R\), giving your answer in the form \(p \sqrt { 3 } + q \pi\) where \(p\) and \(q\) are constants to be determined.

4. Use algebraic integration and the substitution \(u = \sqrt { x }\) to find the exact value of $$\int _ { 1 } ^ { 4 } \frac { 10 } { 5 x + 2 x \sqrt { x } } \mathrm {~d} x$$ Write your answer in the form \(4 \ln \left( \frac { a } { b } \right)\), where \(a\) and \(b\) are integers to be found.
(Solutions relying entirely on calculator technology are not acceptable.)

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 \(x = 2 \sin u\) to show that $$\int _ { 0 } ^ { 1 } \frac { 3 x + 2 } { \left( 4 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } d x = \int _ { 0 } ^ { p } \left( \frac { 3 } { 2 } \operatorname { secutanu } + \frac { 1 } { 2 } \sec ^ { 2 } u \right) d u$$ where \(p\) is a constant to be found.
2. Hence find the exact value of $$\int _ { 0 } ^ { 1 } \frac { 3 x + 2 } { \left( 4 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } d x$$

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.

7. (i) Using a suitable substitution, find, using calculus, the value of $$\int _ { 1 } ^ { 5 } \frac { 3 x } { \sqrt { 2 x - 1 } } \mathrm {~d} x$$ (Solutions relying entirely on calculator technology are not acceptable.)
(ii) Find $$\int \frac { 6 x ^ { 2 } - 16 } { ( x + 1 ) ( 2 x - 3 ) } d x$$

6. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation $$y = \frac { 16 \sin 2 x } { ( 3 + 4 \sin x ) ^ { 2 } } \quad 0 \leqslant x \leqslant \frac { \pi } { 2 }$$ The region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the line with equation \(x = \frac { \pi } { 6 }\) Using the substitution \(u = 3 + 4 \sin x\), show that the area of \(R\) can be written in the form \(a + \ln b\), where \(a\) and \(b\) are rational constants to be found.

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.

6. (a) Using the substitution \(t = x ^ { 2 }\), or otherwise, find $$\int x ^ { 3 } \mathrm { e } ^ { - x ^ { 2 } } \mathrm {~d} x$$ (b) Find the general solution of the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = x \mathrm { e } ^ { - x ^ { 2 } } , \quad x > 0$$