1.08h Integration by substitution

A curve has equation \(y = \frac{3 \cos x}{2 + \sin x}\), for \(-\frac{1}{2}\pi \leqslant x \leqslant \frac{1}{2}\pi\).

1. Find the exact coordinates of the stationary point of the curve. [6]
2. The constant \(a\) is such that \(\int_0^a \frac{3 \cos x}{2 + \sin x} \, dx = 1\). Find the value of \(a\), giving your answer correct to 3 significant figures. [4]

\includegraphics{figure_7} The diagram shows the curve \(y = 5\sin^2 x \cos^3 x\) for \(0 \leqslant x \leqslant \frac{1}{2}\pi\), and its maximum point \(M\). The shaded region \(R\) is bounded by the curve and the \(x\)-axis.

1. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places. [5]
2. Using the substitution \(u = \sin x\) and showing all necessary working, find the exact area of \(R\). [4]

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]

A particle \(P\) moves in a straight line and passes through a point \(O\) of the line with velocity \(2\,\text{m s}^{-1}\). At time \(t\) s after passing through \(O\), the velocity of \(P\) is \(v\,\text{m s}^{-1}\) and the acceleration of \(P\) is given by \(e^{-0.5t}\,\text{m s}^{-2}\). Calculate the velocity of \(P\) when \(t = 1.2\). [4]

A particle \(P\) is moving in a horizontal straight line. Initially \(P\) is at the point \(O\) on the line and is moving with velocity \(25 \text{ m s}^{-1}\). At time \(t\) s after passing through \(O\), the acceleration of \(P\) is \(-\frac{4000}{(5t + 4)^3} \text{ m s}^{-2}\) in the direction \(PO\). The displacement of \(P\) from \(O\) at time \(t\) is \(x\) m. Find an expression for \(x\) in terms of \(t\). [5]

A particle \(P\) moving in a straight line has displacement \(x\)m from a fixed point \(O\) on the line and velocity \(v\)m s\(^{-1}\) at time \(t\)s. The acceleration of \(P\), in m s\(^{-2}\), is given by \(6\sqrt{v + 9}\). When \(t = 0\), \(x = 2\) and \(v = 72\).

1. Find an expression for \(v\) in terms of \(x\). [4]
2. Find an expression for \(x\) in terms of \(t\). [5]

A particle of mass \(m\) kg falls vertically under gravity, from rest. At time \(t\) s, \(P\) has fallen \(x\) m and has velocity \(v\) m s\(^{-1}\). The only forces acting on \(P\) are its weight and a resistance of magnitude \(kmgv\) N, where \(k\) is a constant.

1. Find an expression for \(v\) in terms of \(t\), \(g\) and \(k\). [5]
2. Given that \(k = 0.05\), find, in metres, how far \(P\) has fallen when its speed is \(12\) m s\(^{-1}\). [5]

A particle \(P\) of mass \(2\) kg moving on a horizontal straight line has displacement \(x\) m from a fixed point \(O\) on the line and velocity \(v\) m s\(^{-1}\) at time \(t\) s. The only horizontal force acting on \(P\) has magnitude \(\frac{1}{10}(2v - 1)^2e^{-t}\) N and acts towards \(O\). When \(t = 0\), \(x = 1\) and \(v = 3\).

1. Find an expression for \(v\) in terms of \(t\). [5]
2. Find an expression for \(x\) in terms of \(t\). [4]

\includegraphics{figure_3} The curve \(C\), shown in Figure 3, has equation $$y = \frac{x^{-\frac{1}{4}}}{\sqrt{1+x}\left(\arctan\sqrt{x}\right)}$$ The region \(R\), shown shaded in Figure 3, is bounded by \(C\), the line with equation \(x = 3\), the \(x\)-axis and the line with equation \(x = \frac{1}{3}\) The region \(R\) is rotated through \(360°\) about the \(x\)-axis to form a solid. Using the substitution \(\tan u = \sqrt{x}\)

1. show that the volume \(V\) of the solid formed is given by $$k \int_a^b \frac{1}{u^2} du$$ where \(k\), \(a\) and \(b\) are constants to be found. [6]
2. Hence, using algebraic integration, find the value of \(V\) in simplest form. [3]

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

1. Use the substitution \(u = e^x - 3\) to show that $$\int_{\ln 5}^{\ln 7} \frac{4e^{3x}}{e^x - 3} \, dx = a + b \ln 2$$ where \(a\) and \(b\) are constants to be found. [7]
2. Show, by integration, that $$\int 3e^x \cos 2x \, dx = pe^x \sin 2x + qe^x \cos 2x + c$$ where \(p\) and \(q\) are constants to be found and \(c\) is an arbitrary constant. [5]

Use the substitution \(x = \sin \theta\) to find the exact value of $$\int_0^1 \frac{1}{(1-x^2)^{3/2}} dx.$$ [7]

\includegraphics{figure_2} Figure 2 shows a sketch of the curve with equation \(y = \sqrt{(3-x)(x+1)}\), \(0 \leqslant x \leqslant 3\) The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis, and the \(y\)-axis.

1. Use the substitution \(x = 1 + 2\sin\theta\) to show that $$\int_0^3 \sqrt{(3-x)(x+1)} dx = k \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2\theta d\theta$$ where \(k\) is a constant to be determined. [5]
2. Hence find, by integration, the exact area of \(R\). [3]

Use the substitution \(u = 4 + 3x^2\) to find the exact value of $$\int_0^2 \frac{2x}{(4 + 3x^2)^2} \, dx .$$ [6]

A curve has equation $$y = \sqrt{9 - x^2} \quad 0 \leq x \leq 3$$

1. Using calculus, show that the length of the curve is \(\frac{3\pi}{2}\) [4]

The curve is rotated through \(2\pi\) radians about the \(x\)-axis.

1. Using calculus, find the exact area of the surface generated. [3]

Using calculus, find the exact values of

1. \(\int_1^2 \frac{1}{x^2 - 4x + 5} \, dx\) [3]
2. \(\int_{\sqrt{3}}^3 \frac{\sqrt{x^2 - 3}}{x^2} \, dx\) [5]

Without using a calculator, find

1. \(\int_{-2}^{1} \frac{1}{x^2 + 4x + 13} \, dx\), giving your answer as a multiple of \(\pi\), [5]
2. \(\int_{-1}^{4} \frac{1}{\sqrt{4x^2 - 12x + 34}} \, dx\), giving your answer in the form \(p \ln\left(q + r\sqrt{2}\right)\), [7]

where \(p\), \(q\) and \(r\) are rational numbers to be found.

The curve \(C\) has parametric equations $$x = 3t^4, \quad y = 4t^3, \quad 0 \leq t \leq 1$$ The curve \(C\) is rotated through \(2\pi\) radians about the \(x\)-axis. The area of the curved surface generated is \(S\).

1. Show that $$S = k\pi \int_{0}^{1} t^2(t^2 + 1)^{\frac{1}{2}} dt$$ where \(k\) is a constant to be found. [4]
2. Use the substitution \(u^2 = t^2 + 1\) to find the value of \(S\), giving your answer in the form \(p\pi\left(11\sqrt{2} - 4\right)\) where \(p\) is a rational number to be found. [7]

Show that

1. \(\int_5^8 \frac{1}{x^2 - 10x + 34} dx = k\pi\), giving the value of the fraction \(k\), [5]
2. \(\int_5^8 \frac{1}{\sqrt{x^2 - 10x + 34}} dx = \ln(A + \sqrt{n})\), giving the values of the integers \(A\) and \(n\). [4]

$$4x^2 + 4x + 17 \equiv (ax + b)^2 + c, \quad a > 0.$$

1. Find the values of \(a\), \(b\) and \(c\). [3]
2. Find the exact value of $$\int_{-0.5}^{1.5} \frac{1}{4x^2 + 4x + 17} \, dx.$$ [4]

1. Find \(\int \frac{1+x}{\sqrt{1-4x^2}} \, dx\). [5]
2. Find, to 3 decimal places, the value of $$\int_0^{0.3} \frac{1+x}{\sqrt{1-4x^2}} \, dx.$$ [2]

(Total 7 marks)