1.08i Integration by parts

10

1. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } ( \sin 3 x - 3 x \cos 3 x ) = 9 x \sin 3 x\). The curve shown in the figure below is part of the graph of the function \(y = x \sin 3 x\). \includegraphics[max width=\textwidth, alt={}, center]{3e4281d1-dbad-46a2-bbb7-97706bda2dfa-3_508_1136_1939_466}
2. Show that \(\int _ { 0 } ^ { \frac { 2 \pi } { 3 } } | x \sin 3 x | \mathrm { d } x = \frac { 4 \pi } { 9 }\).

3 Show that $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } ( \pi - x ) \cos 2 x \mathrm {~d} x = \frac { 1 } { 4 } + \frac { 3 } { 8 } \pi$$

a) Find \(\int \left( \mathrm { e } ^ { 2 x } + 6 \sin 3 x \right) \mathrm { d } x\). b) Find \(\int 7 \left( x ^ { 2 } + \sin x \right) ^ { 6 } ( 2 x + \cos x ) \mathrm { d } x\).
c) Find \(\int \frac { 1 } { x ^ { 2 } } \ln x \mathrm {~d} x\).
d) Use the substitution \(u = 2 \cos x + 1\) to evaluate $$\int _ { 0 } ^ { \frac { \pi } { 3 } } \frac { \sin x } { ( 2 \cos x + 1 ) ^ { 2 } } d x$$

\includegraphics{figure_8} The diagram shows a sketch of the curve \(y = x^2\ln x\) and its minimum point \(M\). The curve cuts the \(x\)-axis at the point \((1, 0)\).

1. Find the exact value of the \(x\)-coordinate of \(M\). [4]
2. Use integration by parts to find the area of the shaded region enclosed by the curve, the \(x\)-axis and the line \(x = 4\). Give your answer correct to 2 decimal places. [5]

\includegraphics{figure_8} The diagram shows the curve \(y = x\cos\frac{1}{2}x\) for \(0 \leqslant x \leqslant \pi\).

1. Find \(\frac{dy}{dx}\) and show that \(4\frac{d^2y}{dx^2} + y + 4\sin\frac{1}{2}x = 0\). [5]
2. Find the exact value of the area of the region enclosed by this part of the curve and the \(x\)-axis. [5]

\includegraphics{figure_10} The diagram shows the curve \(y = x^2 \cos 2x\) for \(0 \leq x \leq \frac{1}{4}\pi\). The curve has a maximum point at \(M\) where \(x = p\).

1. Show that \(p\) satisfies the equation \(p = \frac{1}{2} \tan^{-1} \left(\frac{1}{p}\right)\). [3]
2. Use the iterative formula \(p_{n+1} = \frac{1}{2} \tan^{-1} \left(\frac{1}{p_n}\right)\) to determine the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
3. Find, showing all necessary working, the exact area of the region bounded by the curve and the \(x\)-axis. [5]

Find the exact value of \(\int_1^4 \frac{\ln x}{\sqrt{x}} dx\). [5]

\includegraphics{figure_9} The diagram shows the curve \(y = (1 + x^2)\text{e}^{-\frac{3x}{4}}\) for \(x \geqslant 0\). The shaded region \(R\) is enclosed by the curve, the \(x\)-axis and the lines \(x = 0\) and \(x = 2\).

1. Find the exact values of the \(x\)-coordinates of the stationary points of the curve. [4]
2. Show that the exact value of the area of \(R\) is \(18 - \frac{42}{\text{e}}\). [5]

The integral \(I_n\), where \(n\) is a positive integer, is defined by $$I_n = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^n \sin \pi x \, dx.$$

1. Show that $$n(n+1)I_{n+2} = 2^{n+1}n + \pi - \pi^2 I_n.$$ [5]
2. Find \(I_5\) in terms of \(\pi\) and \(I_1\). [2]

Let \(I_n = \int_0^1 (1+3x)^n e^{-3x} dx\), where \(n\) is an integer.

1. Show that \(3I_n = 1 - 4^n e^{-3} + 3nI_{n-1}\). [3]
2. Find the exact value of \(I_2\). [3]

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]

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]

\includegraphics{figure_1} Figure 1 shows the graph of the curve with equation $$y = xe^x, \quad x \geq 0.$$ The finite region \(R\) bounded by the lines \(x = 1\), the \(x\)-axis and the curve is shown shaded in Figure 1.

1. Use integration to find the exact value of the area for \(R\). [5]
2. Complete the table with the values of \(y\) corresponding to \(x = 0.4\) and \(0.8\).

   \(x\)	0	0.2	0.4	0.6	0.8
   \(y = xe^x\)	0	0.29836		1.99207

   [1]
3. Use the trapezium rule with all the values in the table to find an approximate value for this area, giving your answer to 4 significant figures. [4]

\includegraphics{figure_1} Figure 1 shows a sketch of part of the curve with equation \(y = xe^{-\frac{1}{2}x}\), \(x > 0\). The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis, and the line \(x = 4\). The table shows corresponding values of \(x\) and \(y\) for \(y = xe^{-\frac{1}{2}x}\).

1. Complete the table with the value of \(y\) corresponding to \(x = 2\) [1]
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. [4]
3. 1. Find \(\int xe^{-\frac{1}{2}x} \, dx\).
   2. Hence find the exact area of \(R\), giving your answer in the form \(a + be^{-2}\), where \(a\) and \(b\) are integers. [6]

\includegraphics{figure_1} Figure 1 shows a sketch of part of the curve with equation \(y = 4x - xe^{\frac{1}{x}}, x \geqslant 0\) The curve meets the \(x\)-axis at the origin \(O\) and cuts the \(x\)-axis at the point \(A\).

1. Find, in terms of \(\ln 2\), the \(x\) coordinate of the point \(A\). [2]
2. Find $$\int xe^{\frac{1}{x}} dx$$ [3]
3. Find, by integration, the exact value for the area of \(R\). Give your answer in terms of \(\ln 2\) [3]

The finite region \(R\), shown shaded in Figure 1, is bounded by the \(x\)-axis and the curve with equation $$y = 4x - xe^{\frac{1}{x}}, x \geqslant 0$$

\includegraphics{figure_2} Figure 2 shows the curve with equation $$y = x^2 \sin\left(\frac{1}{2}x\right), \quad 0 < x \leq 2\pi.$$ The finite region \(R\) bounded by the line \(x = \pi\), the \(x\)-axis, and the curve is shown shaded in Fig 2.

1. Find the exact value of the area of \(R\), by integration. Give your answer in terms of \(\pi\). [7]

The table shows corresponding values of \(x\) and \(y\).

1. Find the value of \(G\). [1]
2. Use the trapezium rule with values of \(x^2 \sin\left(\frac{1}{2}x\right)\)
   1. at \(x = \pi\), \(x = \frac{3\pi}{2}\) and \(x = 2\pi\) to find an approximate value for the area \(R\), giving your answer to 4 significant figures,
   2. at \(x = \pi\), \(x = \frac{5\pi}{4}\), \(x = \frac{3\pi}{2}\), \(x = \frac{7\pi}{4}\) and \(x = 2\pi\) to find an improved approximation for the area \(R\), giving your answer to 4 significant figures.
   [5]

1. Use integration by parts to show that $$\int_0^{\frac{\pi}{4}} x \sec^2 x \, dx = \frac{1}{4}\pi - \frac{1}{2}\ln 2.$$ [6]

\includegraphics{figure_1} The finite region \(R\), bounded by the equation \(y = x^{\frac{1}{2}} \sec x\), the line \(x = \frac{\pi}{4}\) and the \(x\)-axis is shown in Fig. 1. The region \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis.

1. Find the volume of the solid of revolution generated. [2]
2. Find the gradient of the curve with equation \(y = x^{\frac{1}{2}} \sec x\) at the point where \(x = \frac{\pi}{4}\). [3]

$$I_n = \int \sec^n x \, dx \quad n \geq 0$$

1. Prove that for \(n \geq 2\) $$(n-1)I_n = \tan x \sec^{n-2} x + (n-2)I_{n-2}$$ [6]
2. Hence, showing each step of your working, find the exact value of $$\int_0^{\pi/4} \sec^6 x \, dx$$ [4]

Given that \(y = \text{artanh}(\cos x)\)

1. show that $$\frac{dy}{dx} = -\text{cosec } x$$ [2]
2. Hence find the exact value of $$\int_{0}^{\frac{\pi}{4}} \cos x \, \text{artanh}(\cos x) \, dx$$ giving your answer in the form \(a \ln\left(b + c\sqrt{3}\right) + d\pi\), where \(a\), \(b\), \(c\) and \(d\) are rational numbers to be found. [5]

$$I_n = \int_1^e x^2 (\ln x)^n dx, \quad n \geq 0$$

1. Prove that, for \(n \geq 1\), $$I_n = \frac{e^3}{3} - \frac{n}{3} I_{n-1}$$ [4]
2. Find the exact value of \(I_3\). [4]

Using calculus, find the exact value of

1. \(\int_1^2 \frac{1}{\sqrt{x^2 - 2x + 3}} \, dx\) [4]
2. \(\int_0^1 e^{-x} \sinh x \, dx\) [4]