1.08h Integration by substitution

1

1. A curve has polar equation \(r = a \cos 3 \theta\) for \(- \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\), where \(a\) is a positive constant.
   1. Sketch the curve, using a continuous line for sections where \(r > 0\) and a broken line for sections where \(r < 0\).
   2. Find the area enclosed by one of the loops.
2. Find the exact value of \(\int _ { 0 } ^ { \frac { 3 } { 4 } } \frac { 1 } { \sqrt { 3 - 4 x ^ { 2 } } } \mathrm {~d} x\).
3. Use a trigonometric substitution to find \(\int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + 3 x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x\).

4

1. Show that \(\operatorname { arcosh } x = \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)\).
2. Find \(\int _ { 2.5 } ^ { 3.9 } \frac { 1 } { \sqrt { 4 x ^ { 2 } - 9 } } \mathrm {~d} x\), giving your answer in the form \(a \ln b\), where \(a\) and \(b\) are rational numbers.
3. There are two points on the curve \(y = \frac { \cosh x } { 2 + \sinh x }\) at which the gradient is \(\frac { 1 } { 9 }\). Show that one of these points is \(\left( \ln ( 1 + \sqrt { 2 } ) , \frac { 1 } { 3 } \sqrt { 2 } \right)\), and find the coordinates of the other point, in a similar form.

4

1. Given that \(k \geqslant 1\) and \(\cosh x = k\), show that \(x = \pm \ln \left( k + \sqrt { k ^ { 2 } - 1 } \right)\).
2. Find \(\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt { 4 x ^ { 2 } - 1 } } \mathrm {~d} x\), giving the answer in an exact logarithmic form.
3. Solve the equation \(6 \sinh x - \sinh 2 x = 0\), giving the answers in an exact form, using logarithms where appropriate.
4. Show that there is no point on the curve \(y = 6 \sinh x - \sinh 2 x\) at which the gradient is 5 .

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 \(\int _ { 0 } ^ { \frac { \pi } { 4 } } \sin x \sqrt { } ( \cos 2 x ) \mathrm { d } x\).

5. $$I = \int \frac { 1 } { ( x - 1 ) \sqrt { } \left( x ^ { 2 } - 1 \right) } \mathrm { d } x , \quad x > 1$$

1. Use the substitution \(x = 1 + u ^ { - 1 }\) to show that $$I = - \left( \frac { x + 1 } { x - 1 } \right) ^ { \frac { 1 } { 2 } } + c$$
2. Hence show that $$\int _ { \sec \alpha } ^ { \sec \beta } \frac { 1 } { ( x - 1 ) \sqrt { } \left( x ^ { 2 } - 1 \right) } \mathrm { d } x = \cot \left( \frac { \alpha } { 2 } \right) - \cot \left( \frac { \beta } { 2 } \right) , \quad 0 < \alpha < \beta < \frac { \pi } { 2 }$$

2.(a)Show that $$\sin 3 x = 3 \sin x - 4 \sin ^ { 3 } x$$ Hence find
(b) \(\int \cos x ( 6 \sin x - 2 \sin 3 x ) ^ { \frac { 2 } { 3 } } \mathrm {~d} x\) (c) \(\int ( 3 \sin 2 x - 2 \sin 3 x \cos x ) ^ { \frac { 1 } { 3 } } \mathrm {~d} x\)

6.(a)Starting from \([ \mathrm { f } ( x ) - \lambda \mathrm { g } ( x ) ] ^ { 2 } \geqslant 0\) show that \(\lambda\) satisfies the quadratic inequality $$\left( \int _ { a } ^ { b } [ \operatorname { g } ( x ) ] ^ { 2 } \mathrm {~d} x \right) \lambda ^ { 2 } - 2 \left( \int _ { a } ^ { b } \mathrm { f } ( x ) \mathrm { g } ( x ) \mathrm { d } x \right) \lambda + \int _ { a } ^ { b } [ \mathrm { f } ( x ) ] ^ { 2 } \mathrm {~d} x \geqslant 0$$ where \(a\) and \(b\) are constants and \(\lambda\) can take any real value.
(2)
(b)Hence prove that $$\left[ \int _ { a } ^ { b } \mathrm { f } ( x ) \mathrm { g } ( x ) \mathrm { d } x \right] ^ { 2 } \leqslant \left[ \int _ { a } ^ { b } [ \mathrm { f } ( x ) ] ^ { 2 } \mathrm {~d} x \right] \times \left[ \int _ { a } ^ { b } [ \mathrm {~g} ( x ) ] ^ { 2 } \mathrm {~d} x \right]$$ (c)By letting \(\mathrm { f } ( x ) = 1\) and \(\mathrm { g } ( x ) = \left( 1 + x ^ { 3 } \right) ^ { \frac { 1 } { 2 } }\) show that $$\int _ { - 1 } ^ { 2 } \left( 1 + x ^ { 3 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x \leqslant \frac { 9 } { 2 }$$ (d)Show that \(\int _ { - 1 } ^ { 2 } x ^ { 2 } \left( 1 + x ^ { 3 } \right) ^ { \frac { 1 } { 4 } } \mathrm {~d} x = \frac { 12 \sqrt { } 3 } { 5 }\) (e)Hence show that $$\frac { 144 } { 55 } \leqslant \int _ { - 1 } ^ { 2 } \left( 1 + x ^ { 3 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x$$

6.(a)Show that $$\frac { \mathrm { d } } { \mathrm {~d} u } \ln \left( u + \sqrt { u ^ { 2 } - 1 } \right) = \frac { 1 } { \sqrt { u ^ { 2 } - 1 } }$$ (b)Use the result from part(a)and the substitution \(x + 3 = \frac { 1 } { t }\) to find $$\int \frac { 1 } { ( x + 3 ) \sqrt { 2 x + 7 } } \mathrm {~d} x$$ (6)
(c)Express \(\frac { 1 } { 2 x ^ { 2 } + 13 x + 21 }\) in partial fractions.
(d)Find $$\int _ { 1 } ^ { 9 } \frac { 1 } { \left( 2 x ^ { 2 } + 13 x + 21 \right) \sqrt { 2 x + 7 } } \mathrm {~d} x$$ giving your answer in the form \(\ln r - s\) where \(r\) and \(s\) are rational numbers.

12. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve with parametric equations $$x = 3 \sin t , \quad y = 2 \sin 2 t , \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$ The finite region \(S\), shown shaded in Figure 3, is bounded by the curve, the \(x\)-axis and the line with equation \(x = \frac { 3 } { 2 }\) The shaded region \(S\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.

1. Show that the volume of the solid of revolution is given by $$k \int _ { 0 } ^ { a } \sin ^ { 2 } t \cos ^ { 3 } t \mathrm {~d} t$$ where \(k\) and \(a\) are constants to be given in terms of \(\pi\).
2. Use the substitution \(u = \sin t\), or otherwise, to find the exact value of this volume, giving your answer in the form \(\frac { p \pi } { q }\) where \(p\) and \(q\) are integers. (Solutions based entirely on graphical or numerical methods are not acceptable.)

1

1. Given that \(y = \arctan \sqrt { x }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer in terms of \(x\). Hence show that $$\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { x } ( x + 1 ) } \mathrm { d } x = \frac { \pi } { 2 }$$
2. A curve has cartesian equation $$x ^ { 2 } + y ^ { 2 } = x y + 1$$
   1. Show that the polar equation of the curve is $$r ^ { 2 } = \frac { 2 } { 2 - \sin 2 \theta }$$
   2. Determine the greatest and least positive values of \(r\) and the values of \(\theta\) between 0 and \(2 \pi\) for which they occur.
   3. Sketch the curve.

4

1. Given that \(\sinh y = x\), show that $$y = \ln \left( x + \sqrt { 1 + x ^ { 2 } } \right)$$ Differentiate (*) to show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 1 + x ^ { 2 } } }$$
2. Find \(\int \frac { 1 } { \sqrt { 25 + 4 x ^ { 2 } } } \mathrm {~d} x\), expressing your answer in logarithmic form.
3. Use integration by substitution with \(2 x = 5 \sinh u\) to show that $$\int \sqrt { 25 + 4 x ^ { 2 } } \mathrm {~d} x = \frac { 25 } { 4 } \left( \ln \left( \frac { 2 x } { 5 } + \sqrt { 1 + \frac { 4 x ^ { 2 } } { 25 } } \right) + \frac { 2 x } { 5 } \sqrt { 1 + \frac { 4 x ^ { 2 } } { 25 } } \right) + c$$ where \(c\) is an arbitrary constant. \section*{OCR}

9 \includegraphics[max width=\textwidth, alt={}, center]{c1eee696-3d7f-410a-91a8-fa902309c117-16_307_593_269_735} The diagram shows the curve \(y = \sin ^ { 2 } 2 x \cos x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Using the substitution \(u = \sin x\), find the area of the shaded region bounded by the curve and the \(x\)-axis.

1 Find

1. \(\int 8 \mathrm { e } ^ { - 2 x } \mathrm {~d} x\),
2. \(\int ( 4 x + 5 ) ^ { 6 } \mathrm {~d} x\).

2 \includegraphics[max width=\textwidth, alt={}, center]{89e54367-bb83-483a-add5-0527b71a5cac-2_490_713_447_660} The diagram shows part of the curve \(y = \frac { 6 } { ( 2 x + 1 ) ^ { 2 } }\). The shaded region is bounded by the curve and the lines \(x = 0 , x = 1\) and \(y = 0\). Find the exact volume of the solid produced when this shaded region is rotated completely about the \(x\)-axis.

4

1. Show that \(\int _ { 0 } ^ { 4 } \frac { 18 } { \sqrt { 6 x + 1 } } \mathrm {~d} x = 24\).
2. Find \(\int _ { 0 } ^ { 1 } \left( \mathrm { e } ^ { x } + 2 \right) ^ { 2 } \mathrm {~d} x\), giving your answer in terms of e .

1 Find

1. \(\quad \int ( 4 - 3 x ) ^ { 7 } \mathrm {~d} x\),
2. \(\quad \int ( 4 - 3 x ) ^ { - 1 } \mathrm {~d} x\).

7

1. Find the exact value of \(\int _ { 1 } ^ { 9 } ( 7 x + 1 ) ^ { \frac { 1 } { 3 } } \mathrm {~d} x\).
2. Use Simpson's rule with two strips to show that an approximate value of \(\int _ { 1 } ^ { 9 } ( 7 x + 1 ) ^ { \frac { 1 } { 3 } } \mathrm {~d} x\) can be expressed in the form \(m + n \sqrt [ 3 ] { 36 }\), where the values of the constants \(m\) and \(n\) are to be stated.
3. Use the results from parts (i) and (ii) to find an approximate value of \(\sqrt [ 3 ] { 36 }\), giving your answer in the form \(\frac { p } { q }\) where \(p\) and \(q\) are integers. \section*{Question 8 begins on page 4.}

2 Find

1. \(\int \left( 2 - \frac { 1 } { x } \right) ^ { 2 } \mathrm {~d} x\),
2. \(\int ( 4 x + 1 ) ^ { \frac { 1 } { 3 } } \mathrm {~d} x\).

4 Find the exact value of \(\int _ { 0 } ^ { 2 } \sqrt { 1 + 4 x } \mathrm {~d} x\), showing your working.

4 Evaluate the following integrals, giving your answers in exact form.

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

9 Fig. 9 shows the curve \(y = \frac { x ^ { 2 } } { 3 x - 1 }\).
P is a turning point, and the curve has a vertical asymptote \(x = a\). \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ( 3 x - 2 ) } { ( 3 x - 1 ) ^ { 2 } }\).
3. Find the exact coordinates of the turning point P . Calculate the gradient of the curve when \(x = 0.6\) and \(x = 0.8\), and hence verify that P is a minimum point.
4. Using the substitution \(u = 3 x - 1\), show that \(\int \frac { x ^ { 2 } } { 3 x - 1 } \mathrm {~d} x = \frac { 1 } { 27 } \int \left( u + 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines \(x = \frac { 2 } { 3 }\) and \(x = 1\).

6 Using a suitable substitution or otherwise, show that \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 2 x } { 3 + \cos 2 x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2\).

9 Fig. 9 shows the curve with equation \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\). It has an asymptote \(x = a\) and turning point P . \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).

2 Find \(\int \sqrt [ 3 ] { 2 x - 1 } \mathrm {~d} x\).