1.08h Integration by substitution

Evaluate the integral $$\int_0^1 \frac{dx}{\sqrt{2x^2 + 4x + 6}}.$$ [6]

Show that $$\int_0^{\pi/2} \frac{\sin 2\theta}{1 + \cos \theta} \, d\theta = 2 - 2\ln 2$$ [7]

In this question you must show detailed reasoning. \includegraphics{figure_9} The diagram shows the curve \(y = \frac{4\cos 2x}{3 - \sin 2x}\) for \(x > 0\), and the normal to the curve at the point \((\frac{1}{4}\pi, 0)\). Show that the exact area of the shaded region enclosed by the curve, the normal to the curve and the \(y\)-axis is \(\ln \frac{2}{3} + \frac{1}{128}\pi^2\). [10]

1. Show that the two non-stationary points of inflection on the curve \(y = \ln(1 + 4x^2)\) are at \(x = \pm\frac{1}{2}\). [6]

\includegraphics{figure_9} The diagram shows the curve \(y = \ln(1 + 4x^2)\). The shaded region is bounded by the curve and a line parallel to the \(x\)-axis which meets the curve where \(x = \frac{1}{2}\) and \(x = -\frac{1}{2}\).

1. Show that the area of the shaded region is given by $$\int_0^{\ln 2} \sqrt{e^y - 1} \, dy.$$ [3]
2. Show that the substitution \(e^y = \sec^2\theta\) transforms the integral in part (ii) to \(\int_0^{\frac{\pi}{4}} 2\tan^2\theta \, d\theta\). [2]
3. Hence find the exact area of the shaded region. [3]

Using an appropriate substitution, or otherwise, show that $$\int_0^{\frac{\pi}{2}} \frac{\sin 2\theta}{1 + \cos \theta} d\theta = 2 - 2\ln 2$$ [7]

A curve \(C\) has equation $$x^3 + y^3 = 3xy + 48$$ Prove that \(C\) has two stationary points and find their coordinates. [7]

\includegraphics{figure_1} Figure 1 shows part of the curve with equation \(y = e^{\frac{1}{5}x^2}\) for \(x \geq 0\) 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 \(x = 2\) The table below shows corresponding values of \(x\) and \(y\) for \(y = e^{\frac{1}{5}x^2}\)

\(x\)	0	0.5	1	1.5	2
\(y\)	1	\(e^{0.05}\)	\(e^{0.2}\)	\(e^{0.45}\)	\(e^{0.8}\)

1. Use the trapezium rule, with all the values of \(y\) in the table, to find an estimate for the area of \(R\), giving your answer to 2 decimal places. [3]
2. Use your answer to part (a) to deduce an estimate for
   1. \(\int_0^2 \left( 4 + e^{\frac{1}{5}x^2} \right) dx\)
   2. \(\int_1^3 e^{\frac{1}{5}(x-1)^2} dx\) giving your answers to 2 decimal places. [2]

1. Determine the values of \(p\) and \(q\) for which $$x^2 - 6x + 10 \equiv (x - p)^2 + q.$$ [2]

1. Use the substitution \(x - p = \tan u\), where \(p\) takes the value found in part (i), to evaluate $$\int_3^4 \frac{1}{x^2 - 6x + 10} \, dx.$$ [3]

1. Determine the value of $$\int_3^4 \frac{x}{x^2 - 6x + 10} \, dx,$$ giving your answer in the form \(a + \ln b\), where \(a\) and \(b\) are constants to be determined. [5]

Use a suitable trigonometric substitution to find \(\int \frac{x^2}{\sqrt{1-x^2}} \text{d}x\). [7]

\includegraphics{figure_6} The diagram shows the curve with parametric equations \(x = \ln(t^2 - 4)\), \(y = \frac{4}{t}\), where \(t > 2\). The shaded region \(R\) is enclosed by the curve, the \(x\)-axis and the lines \(x = \ln 5\) and \(x = \ln 12\).

1. Show that the area of \(R\) is given by $$\int_a^b \frac{8}{t(t^2 - 4)} dt,$$ where \(a\) and \(b\) are constants to be determined. [4]
2. In this question you must show detailed reasoning. Hence find the exact area of \(R\), giving your answer in the form \(\ln k\), where \(k\) is a constant to be determined. [8]
3. Find a cartesian equation of the curve in the form \(y = \text{f}(x)\). [3]

Using the substitution \(u = 1 + \sqrt{x}\), or otherwise, find \(\int \frac{1}{1 + \sqrt{x}} dx\) giving your answer in terms of \(x\). [5]

Given that $$\int_0^{\frac{\pi}{2}} (1 + \tan\left[\frac{1}{2}x\right])^2 \, dx = a + \ln b$$ find the value of \(a\) and the value of \(b\). [Total 7 marks]

7 A small ball \(B\) of mass 0.2 kg moves in a narrow fixed smooth cylindrical tube \(O A\) of length 1 m , closed at the end \(A\). When the ball has displacement \(x \mathrm {~m}\) from \(O\), it has velocity \(v \mathrm {~ms} ^ { - 1 }\) in the direction \(O A\) and experiences a resisting force of magnitude \(\frac { k } { 1 - x } \mathrm {~N}\).

1. \includegraphics[max width=\textwidth, alt={}, center]{10abedc3-c814-47c0-8ed4-849ef325feca-4_186_805_488_715} The tube is fixed in a horizontal position and \(B\) is projected from \(O\) towards \(A\) with velocity \(1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) (see diagram). Given that \(B\) comes to instantaneous rest after travelling 0.55 m , show that \(k = 0.1803\), correct to 4 significant figures.
2. The tube is now fixed in a vertical position with \(O\) above \(A\). The ball \(B\) is released from rest at \(O\). Calculate the speed of \(B\) after it has descended 0.1 m . \end{document}