1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2

4 Find

1. \(\quad \int ( 2 x + 3 ) ^ { 4 } \mathrm {~d} x\)
2. \(\quad \int \left( 1 + \tan ^ { 2 } 2 x \right) \mathrm { d } x\)

9

1. Prove that \(\operatorname { cosec } 2 x - \cot 2 x \equiv \tan x\) and hence find an exact value for \(\tan \left( \frac { 3 } { 8 } \pi \right)\).
2. Find the exact value of \(\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 3 } { 8 } \pi } ( \operatorname { cosec } 2 x - \cot 2 x ) ^ { 2 } \mathrm {~d} x\).

10 A curve has equation $$y = \mathrm { e } ^ { a x } \cos b x$$ where \(a\) and \(b\) are constants.

1. Show that, at any stationary points on the curve, \(\tan b x = \frac { a } { b }\).
2. \includegraphics[max width=\textwidth, alt={}, center]{48b63de9-f022-4881-a187-f08e3c7d9f1a-4_620_894_1064_342} Values of related quantities \(x\) and \(y\) were measured in an experiment and plotted on a graph of \(y\) against \(x\), as shown in the diagram. Two of the points, labelled \(A\) and \(B\), have coordinates \(( 0,1 )\) and \(( 0.2 , - 0.8 )\) respectively. A third point labelled \(C\) has coordinates ( \(0.3,0.04\) ). Attempts were then made to find the equation of a curve which fitted closely to these three points, and two models were proposed. In the first model the equation is \(y = \mathrm { e } ^ { - x } \cos 15 x\). In the second model the equation is \(y = f \cos ( \lambda x ) + g\), where the constants \(f , \lambda\), and \(g\) are chosen to give a maximum precisely at the point \(A ( 0,1 )\) and a minimum precisely at the point \(B ( 0.2 , - 0.8 )\). By calculating suitable values evaluate the suitability of the two models.

It is given that \(\theta\) is an acute angle measured in degrees such that $$2\sec^2\theta + 3\tan\theta = 22.$$

1. Find the value of \(\tan\theta\). [3]
2. Use an appropriate formula to find the exact value of \(\tan(\theta + 135°)\). [3]

1. By differentiating \(\frac{\cos x}{\sin x}\), show that if \(y = \cot x\) then \(\frac{dy}{dx} = -\cosec^2 x\). [3]
2. Hence show that \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cosec^2 x \, dx = \sqrt{3}\). [2]

By using appropriate trigonometrical identities, find the exact value of

1. \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cot^2 x \, dx\), [3]
2. \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{1 - \cos 2x} \, dx\). [3]

Solve the equation \(\sec^2 \theta = 3 \cosec \theta\) for \(0° < \theta < 180°\). [5]

\includegraphics{figure_6} The diagram shows the curve with equation \(y = \sqrt{1 + 3\cos^2(\frac{1}{2}x)}\) for \(0 \leqslant x \leqslant \pi\). The region \(R\) is bounded by the curve, the axes and the line \(x = \pi\).

1. Use the trapezium rule with two intervals to find an approximation to the area of \(R\), giving your answer correct to 3 significant figures. [3]
2. The region \(R\) is rotated completely about the \(x\)-axis. Without using a calculator, find the exact volume of the solid produced. [5]

Answer only one of the following two alternatives. EITHER The curve \(C\) is defined parametrically by $$x = 18t - t^2 \quad \text{and} \quad y = 8t^{\frac{1}{2}},$$ where \(0 < t \leqslant 4\).

1. Show that at all points of \(C\), $$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{-3(9 + t)}{2t^2(9 - t)^3}.$$ [4]
2. Show that the mean value of \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\) with respect to \(x\) over the interval \(0 < x \leqslant 56\) is \(\frac{3}{70}\). [4]
3. Find the area of the surface generated when \(C\) is rotated through \(2\pi\) radians about the \(x\)-axis, showing full working. [6]

OR Let \(I_n = \int_1^{\sqrt{2}} (x^2 - 1)^n \mathrm{d}x\).

1. Show that, for \(n \geqslant 1\), $$(2n + 1)I_n = \sqrt{2} - 2nI_{n-1}.$$ [5]
2. Using the substitution \(x = \sec \theta\), show that $$I_n = \int_0^{\frac{1}{4}\pi} \tan^{2n+1} \theta \sec \theta \, \mathrm{d}\theta.$$ [4]
3. Deduce the exact value of $$\int_0^{\frac{1}{4}\pi} \frac{\sin^7 \theta}{\cos^8 \theta} \, \mathrm{d}\theta.$$ [5]

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Find $$\int_0^{\pi/6} x \cos 3x \, dx$$ giving your answer in simplest form. [5]

\includegraphics{figure_3} Figure 3 shows a sketch of the curve \(C\) with parametric equations $$x = 1 + 3\tan t, \quad y = 2\cos 2t, \quad -\frac{\pi}{6} \leq t \leq \frac{\pi}{3}$$ The curve crosses the \(x\)-axis at point \(P\), as shown in Figure 3.

1. Find the equation of the tangent to \(C\) at \(P\), writing your answer in the form \(y = mx + c\), where \(m\) and \(c\) are constants to be found. [5]

The curve \(C\) has equation \(y = f(x)\), where \(f\) is a function with domain \(\left[k, 1 + 3\sqrt{3}\right]\)

1. Find the exact value of the constant \(k\). [1]
2. Find the range of \(f\). [2]

\includegraphics{figure_2} Figure 2 shows a sketch of part of the curve \(C\) with parametric equations $$x = \tan \theta, \quad y = 1 + 2\cos 2\theta, \quad 0 \leq \theta < \frac{\pi}{2}$$ The curve \(C\) crosses the \(x\)-axis at \((\sqrt{3}, 0)\). The finite shaded region \(S\) shown in Figure 2 is bounded by \(C\), the line \(x = 1\) and the \(x\)-axis. This shaded region 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 formed is given by the integral $$k \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (16 \cos^2 \theta - 8 + \sec^2 \theta) \, d\theta$$ where \(k\) is a constant. [5]
2. Hence, use integration to find the exact value for this volume. [5]

$$f(x) = 5\sin 3x°, \quad 0 \leq x \leq 180.$$

1. Sketch the graph of \(f(x)\), indicating the value of \(x\) at each point where the graph intersects the \(x\)-axis [3]
2. Write down the coordinates of all the maximum and minimum points of \(f(x)\). [3]
3. Calculate the values of \(x\) for which \(f(x) = 2.5\) [4]

Solve, for \(0° < \theta < 360°\), the equation \(\sec^2 \theta = 4 \tan \theta - 2\). [5]

The angles \(\alpha\) and \(\beta\) are such that $$\tan \alpha = m + 2 \quad \text{and} \quad \tan \beta = m,$$ where \(m\) is a constant.

1. Given that \(\sec^2 \alpha - \sec^2 \beta = 16\), find the value of \(m\). [3]
2. Hence find the exact value of \(\tan(\alpha + \beta)\). [3]

Fig. 8 shows part of the curve \(y = x \sin 3x\). It crosses the \(x\)-axis at P. The point on the curve with \(x\)-coordinate \(\frac{1}{6}\pi\) is Q. \includegraphics{figure_8}

1. Find the \(x\)-coordinate of P. [3]
2. Show that Q lies on the line \(y = x\). [1]
3. Differentiate \(x \sin 3x\). Hence prove that the line \(y = x\) touches the curve at Q. [6]
4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac{1}{72}(\pi^2 - 8)\). [7]

Fig. 8 shows parts of the curves \(y = f(x)\) and \(y = g(x)\), where \(f(x) = \tan x\) and \(g(x) = 1 + f(x - \frac{1}{4}\pi)\). \includegraphics{figure_8}

1. Describe a sequence of two transformations which maps the curve \(y = f(x)\) to the curve \(y = g(x)\). [4]

It can be shown that \(g(x) = \frac{2\sin x}{\sin x + \cos x}\).

1. Show that \(g'(x) = \frac{2}{(\sin x + \cos x)^2}\). Hence verify that the gradient of \(y = g(x)\) at the point \((\frac{1}{4}\pi, 1)\) is the same as that of \(y = f(x)\) at the origin. [7]
2. By writing \(\tan x = \frac{\sin x}{\cos x}\) and using the substitution \(u = \cos x\), show that \(\int_0^{\frac{1}{4}\pi} f(x)dx = \int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{u}du\). Evaluate this integral exactly. [4]
3. Hence find the exact area of the region enclosed by the curve \(y = g(x)\), the \(x\)-axis and the lines \(x = \frac{1}{4}\pi\) and \(x = \frac{1}{2}\pi\). [2]

A curve is defined by the equation \(\sin 2x + \cos y = \sqrt{3}\).

1. Verify that the point P \((\frac{\pi}{6}, \frac{\pi}{6})\) lies on the curve. [1]
2. Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at the point P. [5]

Find the exact gradient of the curve \(y = \ln(1 - \cos 2x)\) at the point with \(x\)-coordinate \(\frac{1}{4}\pi\). [5]

The function \(\text{f}(x) = \frac{\sin x}{2 - \cos x}\) has domain \(-\pi \leqslant x \leqslant \pi\). Fig. 8 shows the graph of \(y = \text{f}(x)\) for \(0 \leqslant x \leqslant \pi\). \includegraphics{figure_6}

1. Find \(\text{f}(-x)\) in terms of \(\text{f}(x)\). Hence sketch the graph of \(y = \text{f}(x)\) for the complete domain \(-\pi \leqslant x \leqslant \pi\). [3]
2. Show that \(\text{f}'(x) = \frac{2\cos x - 1}{(2 - \cos x)^2}\). Hence find the exact coordinates of the turning point P. State the range of the function \(\text{f}(x)\), giving your answer exactly. [8]
3. Using the substitution \(u = 2 - \cos x\) or otherwise, find the exact value of \(\int_0^\pi \frac{\sin x}{2 - \cos x} dx\). [4]
4. Sketch the graph of \(y = \text{f}(2x)\). [1]
5. Using your answers to parts (iii) and (iv), write down the exact value of \(\int_0^{\frac{\pi}{2}} \frac{\sin 2x}{2 - \cos 2x} dx\). [2]