1.08h Integration by substitution

Evaluate \(\int_1^4 \frac{1}{\sqrt{x^2 - 2x + 17}} \, dx\), giving your answer as an exact logarithm. [5]

Find \(\int \sqrt{x^2 + 4} \, dx\). (Total 7 marks)

Use the substitution \(u = 1 + 2\tan x\) to find $$\int \frac{1}{(1 + 2\tan x)^2 \cos^2 x} \, dx$$ [5]

The integral \(I\) is defined by $$I = \int_0^{13} (2x + 1)^{\frac{3}{2}} \, dx.$$

1. Use integration to find the exact value of \(I\). [4]
2. Use Simpson's rule with two strips to find an approximate value for \(I\). Give your answer correct to 3 significant figures. [3]

\includegraphics{figure_5} The diagram shows the curve \(y = \frac{6}{\sqrt{3x + 1}}\). The shaded region is bounded by the curve and the lines \(x = 2\), \(x = 9\) and \(y = 0\).

1. Show that the area of the shaded region is \(4\sqrt{7}\) square units. [4]
2. The shaded region is rotated completely about the \(x\)-axis. Show that the volume of the solid produced can be written in the form \(k\ln 2\), where the exact value of the constant \(k\) is to be determined. [5]

Using the substitution \(u = 2x + 1\), show that \(\int_0^1 \frac{x}{2x + 1} dx = \frac{1}{4}(2 - \ln 3)\).

Fig. 8 shows the curve \(y = \frac{x}{\sqrt{x-2}}\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \((11, 11)\). \includegraphics{figure_8}

1. Verify that the \(x\)-coordinate of P is 3. [2]
2. Show that, for the curve, \(\frac{dy}{dx} = \frac{x-4}{2(x-2)^{\frac{3}{2}}}\). Hence find the gradient of the curve at P. Use the result to show that the curve is not symmetrical about \(y = x\). [7]
3. Using the substitution \(u = x - 2\), show that \(\int_3^{11} \frac{x}{\sqrt{x-2}} \, dx = 25\frac{1}{3}\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\). [9]

Fig. 9 shows the curves \(y = \text{f}(x)\) and \(y = \text{g}(x)\). The function \(y = \text{f}(x)\) is given by $$\text{f}(x) = \ln \left( \frac{2x}{1+x} \right), \quad x > 0.$$ The curve \(y = \text{f}(x)\) crosses the \(x\)-axis at P, and the line \(x = 2\) at Q. \includegraphics{figure_9}

1. Verify that the \(x\)-coordinate of P is 1. Find the exact \(y\)-coordinate of Q. [2]
2. Find the gradient of the curve at P. [Hint: use \(\frac{a}{b} = \ln a - \ln b\).] [4]

The function \(\text{g}(x)\) is given by $$\text{g}(x) = \frac{e^x}{2-e^x}, \quad x < \ln 2.$$ The curve \(y = \text{g}(x)\) crosses the \(y\)-axis at the point R.

1. Show that \(\text{g}(x)\) is the inverse function of \(\text{f}(x)\). Write down the gradient of \(y = \text{g}(x)\) at R. [5]
2. Show, using the substitution \(u = 2 - e^x\) or otherwise, that \(\int_0^{\ln \frac{4}{3}} \text{g}(x) dx = \ln \frac{3}{2}\). Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac{32}{27}\). [Hint: consider its reflection in \(y = x\).] [7]

Evaluate \(\int_0^3 x(x + 1)^{-\frac{1}{2}} dx\), giving your answer as an exact fraction. [5]

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]

\includegraphics{figure_8} Fig. 8 shows the curve \(y = f(x)\), where \(f(x) = \frac{1}{e^x + e^{-x} + 2}\).

1. Show algebraically that \(f(x)\) is an even function, and state how this property relates to the curve \(y = f(x)\). [3]
2. Find \(f'(x)\). [3]
3. Show that \(f(x) = \frac{e^x}{(e^x + 1)^2}\). [2]
4. Hence, using the substitution \(u = e^x + 1\), or otherwise, find the exact area enclosed by the curve \(y = f(x)\), the \(x\)-axis, and the lines \(x = 0\) and \(x = 1\). [5]
5. Show that there is only one point of intersection of the curves \(y = f(x)\) and \(y = \frac{1}{4}e^x\), and find its coordinates. [5]

Fig. 8 shows the curve \(y = f(x)\), where \(f(x) = \frac{x}{\sqrt{2 + x^2}}\). \includegraphics{figure_8}

1. Show algebraically that \(f(x)\) is an odd function. Interpret this result geometrically. [3]
2. Show that \(f'(x) = \frac{2}{(2 + x^2)^{\frac{3}{2}}}\). Hence find the exact gradient of the curve at the origin. [5]
3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\). [4]
4. 1. Show that if \(y = \frac{x}{\sqrt{2 + x^2}}\), then \(\frac{1}{y^2} = \frac{2}{x^2} + 1\). [2]
   2. Differentiate \(\frac{1}{y^2} = \frac{2}{x^2} + 1\) implicitly to show that \(\frac{dy}{dx} = \frac{2y^3}{x^3}\). Explain why this expression cannot be used to find the gradient of the curve at the origin. [4]

Fig. 8 shows the curve \(y = \frac{x}{\sqrt{x+4}}\) and the line \(x = 5\). The curve has an asymptote \(l\). The tangent to the curve at the origin O crosses the line \(l\) at P and the line \(x = 5\) at Q. \includegraphics{figure_8}

1. Show that for this curve \(\frac{dy}{dx} = \frac{x + 8}{2(x + 4)^{\frac{3}{2}}}\). [5]
2. Find the coordinates of the point P. [4]
3. Using integration by substitution, find the exact area of the region enclosed by the curve, the tangent OQ and the line \(x = 5\). [9]

Fig. 8 shows the graph of \(y = x\sqrt{1 + x}\). The point P on the curve is on the \(x\)-axis. \includegraphics{figure_8}

1. Write down the coordinates of P. [1]
2. Show that \(\frac{dy}{dx} = \frac{3x + 2}{2\sqrt{1 + x}}\). [4]
3. Hence find the coordinates of the turning point on the curve. What can you say about the gradient of the curve at P? [4]
4. By using a suitable substitution, show that \(\int_0^0 x\sqrt{1 + x} dx = \int_0^1 \left(u^{\frac{3}{2}} - u^{\frac{1}{2}}\right) du\). Evaluate this integral, giving your answer in an exact form. What does this value represent? [7]
5. Use your answer to part (ii) to differentiate \(y = x\sqrt{1 + x} \sin 2x\) with respect to \(x\). (You need not simplify your result.) [2]

Evaluate $$\int_2^6 \sqrt{3x-2} \, dx.$$ [4]

Fig. 9 shows the curve \(y = \frac{x^2}{3x - 1}\). P is a turning point, and the curve has a vertical asymptote \(x = a\). \includegraphics{figure_1}

1. Write down the value of \(a\). [1]
2. Show that \(\frac{dy}{dx} = \frac{x(3x - 2)}{(3x - 1)^2}\) [3]
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. [7]
4. Using the substitution \(u = 3x - 1\), show that \(\int \frac{x^2}{3x - 1} dx = \frac{1}{27} \int \left( u + 2 + \frac{1}{u} \right) du\). Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines \(x = \frac{2}{3}\) and \(x = 1\). [7]

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]

Find the exact value of \(\int_0^2 \sqrt{1+4x} \, dx\), showing your working. [5]

A curve is defined by the equation \(y = 2x \ln(1 + x)\).

1. Find \(\frac{dy}{dx}\) and hence verify that the origin is a stationary point of the curve. [4]
2. Find \(\frac{d^2y}{dx^2}\) and use this to verify that the origin is a minimum point. [5]
3. Using the substitution \(u = 1 + x\), show that \(\int \frac{x^2}{1+x} \, dx = \int \left(u - 2 + \frac{1}{u}\right) du\). Hence evaluate \(\int_0^1 \frac{x^2}{1+x} \, dx\), giving your answer in an exact form. [6]
4. Using integration by parts and your answer to part (iii), evaluate \(\int_0^1 2x \ln(1 + x) \, dx\). [4]

1. Use the identity for \(\cos(A + B)\) to prove that \(\cos 2A = 2\cos^2 A - 1\). [2]
2. Use the substitution \(x = 2\sqrt{2} \sin \theta\) to prove that $$\int_2^{\sqrt{6}} \sqrt{(8 - x^2)} \, dx = \frac{1}{3}(\pi + 3\sqrt{3} - 6).$$ [7]

A curve is given by the parametric equations $$x = \sec \theta, \quad y = \ln(1 + \cos 2\theta), \quad 0 \leq \theta < \frac{\pi}{2}.$$

1. Find an equation of the tangent to the curve at the point where \(\theta = \frac{\pi}{3}\). [5]

Use the substitution \(u = 2x - 5\) to show that \(\int_2^3 (4x - 8)(2x - 5)^7 \, dx = \frac{17}{72}\). [5]