1.08h Integration by substitution

1. A particle \(P\) of mass 0.8 kg moves along the \(x\)-axis in the positive \(x\) direction under the action of a resultant force. This force acts in the direction of \(x\) increasing. At time \(t\) seconds, \(t \geqslant 0 , P\) is \(x\) metres from the origin \(O , P\) is moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the force has magnitude \(\frac { 4 } { ( x + 1 ) ^ { 3 } } \mathrm {~N}\).

When \(t = 0 , P\) is at rest at \(O\).

1. Show that \(v ^ { 2 } = 5 \left( \frac { ( x + 1 ) ^ { 2 } - 1 } { ( x + 1 ) ^ { 2 } } \right)\) When \(t = 2 , P\) is at the point \(A\). When \(t = 4 , P\) is at the point \(B\).
2. Using algebraic integration, find the distance \(A B\).

7 Fig. 7 shows the curve \(y = \frac { x ^ { 2 } } { 1 + 2 x ^ { 3 } }\). It is undefined at \(x = a\); the line \(x = a\) is a vertical asymptote. \begin{figure}[h] \end{figure}

1. Calculate the value of \(a\), giving your answer correct to 3 significant figures.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x - 2 x ^ { 4 } } { \left( 1 + 2 x ^ { 3 } \right) ^ { 2 } }\). Hence determine the coordinates of the turning points of the curve.
3. Show that the area of the region between the curve and the \(x\)-axis from \(x = 0\) to \(x = 1\) is \(\frac { 1 } { 6 } \ln 3\).

4 Show that \(\int _ { 1 } ^ { 4 } \frac { x } { x ^ { 2 } + 2 } \mathrm {~d} x = \frac { 1 } { 2 } \ln 6\).

8 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 1 + \cos x }\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
P is the point on the curve with \(x\)-coordinate \(\frac { 1 } { 3 } \pi\). \begin{figure}[h] \end{figure}

1. Find the \(y\)-coordinate of P .
2. Find \(\mathrm { f } ^ { \prime } ( x )\). Hence find the gradient of the curve at the point P .
3. Show that the derivative of \(\frac { \sin x } { 1 + \cos x }\) is \(\frac { 1 } { 1 + \cos x }\). Hence find the exact area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 3 } \pi\).
4. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arccos \left( \frac { 1 } { x } - 1 \right)\). State the domain of this inverse function, and add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.

9 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }\). The curve crosses the \(y\)-axis at P. \begin{figure}[h] \end{figure}

1. Find the coordinates of P .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), simplifying your answer. Hence calculate the gradient of the curve at P .
3. Show that the area of the region enclosed by \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\) is \(\frac { 1 } { 2 } \ln \left( \frac { 1 + \mathrm { e } ^ { 2 } } { 2 } \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { 2 } \left( \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } \right)\).
4. Prove algebraically that \(\mathrm { g } ( x )\) is an odd function. Interpret this result graphically.
5. (A) Show that \(\mathrm { g } ( x ) + \frac { 1 } { 2 } = \mathrm { f } ( x )\).
   (B) Describe the transformation which maps the curve \(y = \mathrm { g } ( x )\) onto the curve \(y = \mathrm { f } ( x )\).
   (C) What can you conclude about the symmetry of the curve \(y = \mathrm { f } ( x )\) ?

8 You are given that \(\mathrm { f } ( x ) = \frac { x } { x ^ { 2 } + 1 }\) for all real values of \(x\).

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 1 - x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 2 } }\).
2. Hence show that there is a stationary value at \(\left( 1 , \frac { 1 } { 2 } \right)\) and find the coordinates of the other stationary point.
3. The graph of the curve is shown in Fig. 8. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-3_518_892_1612_705} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure} State whether the curve is odd or even and prove the result algebraically.
4. Show that \(\int _ { 1 } ^ { 4 } \frac { x } { x ^ { 2 } + 1 } \mathrm {~d} x = \int _ { a } ^ { b } k \frac { 1 } { u + 1 } \mathrm {~d} u\), where the values of \(a , b\) and \(k\) are to be determined.
5. Hence find the area of the shaded region in Fig. 8.

6

1. Find \(\int ( 2 x - 3 ) ^ { 7 } \mathrm {~d} x\).
2. Use the substitution \(u = x ^ { 2 } + 1\), or otherwise, to find \(\int _ { 1 } ^ { 2 } x \left( x ^ { 2 } + 1 \right) ^ { 3 } \mathrm {~d} x\).

6

1. Find \(\int x \cos 2 x d x\).
2. Using the substitution \(u = x ^ { 2 } + 1\), or otherwise, find the exact value of \(\int _ { 2 } ^ { 3 } \frac { x } { x ^ { 2 } + 1 } \mathrm {~d} x\).

6. \includegraphics[max width=\textwidth, alt={}, center]{687756c0-2038-4077-8c5c-fe0ca0f6ce65-2_444_825_1571_516} The diagram shows the curve with equation \(y = \sqrt { \frac { x } { x + 1 } }\).
The shaded region is bounded by the curve, the \(x\)-axis and the line \(x = 3\).

1. Use Simpson's rule with six strips to estimate the area of the shaded region. The shaded region is rotated through four right angles about the \(x\)-axis.
2. Show that the volume of the solid formed is \(\pi ( 3 - \ln 4 )\).

1 Find \(\int \sqrt [ 3 ] { 2 x - 1 } \mathrm {~d} x\).

4 Fig. 9 shows the curve \(y = x \mathrm { e } ^ { - 2 x }\) together with the straight line \(y = m x\), where \(m\) is a constant, with \(0 < m < 1\). The curve and the line meet at O and P . The dashed line is the tangent at P . \begin{figure}[h] \end{figure}

1. Show that the \(x\)-coordinate of P is \(- \frac { 1 } { 2 } \ln m\).
2. Find, in terms of \(m\), the gradient of the tangent to the curve at P . You are given that OP and this tangent are equally inclined to the \(x\)-axis.
3. Show that \(m = \mathrm { e } ^ { - 2 }\), and find the exact coordinates of P .
4. Find the exact area of the shaded region between the line OP and the curve.

5 Using a suitable substitution or otherwise, show that \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 2 x } { 3 + \cos 2 x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2\).

2 Fig. 9 shows the curve with equation \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\). It has an asymptote \(x = a\) and turning point P . \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).

3 Fig. 9 shows the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\). The function \(y = \mathrm { f } ( x )\) is given by $$f ( x ) = \ln \left( \frac { 2 x } { 1 + x } \right) , x > 0$$ The curve \(y = \mathrm { f } ( x )\) crosses the \(x\)-axis at P , and the line \(x = 2\) at Q . \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 1 . Find the exact \(y\)-coordinate of Q .
2. Find the gradient of the curve at P. [Hint: use \(\ln \frac { a } { b } = \ln a - \ln b\).] The function \(\mathrm { g } ( x )\) is given by $$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } } { 2 - \mathrm { e } ^ { x } } , \quad x < \ln 2 .$$ The curve \(y = \mathrm { g } ( x )\) crosses the \(y\)-axis at the point R .
3. Show that \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\). Write down the gradient of \(y = \mathrm { g } ( x )\) at R .
4. Show, using the substitution \(u = 2 - \mathrm { e } ^ { x }\) or otherwise, that \(\int _ { 0 } ^ { \ln \frac { 4 } { 3 } } \mathrm {~g} ( x ) \mathrm { d } x = \ln \frac { 3 } { 2 }\). 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 \(y = x\).]

1 Show that \(\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt { 3 x - 2 } } \mathrm {~d} x = \frac { 2 } { 3 }\).

1 Fig. 7 shows the curve \(y = _ { x - 1 }\). It has a minimum at the point P . The line \(l\) is an asymptote to the curve. \begin{figure}[h] \end{figure}

1. Write down the equation of the asymptote \(l\).
2. Find the coordinates of P .
3. Using the substitution \(u = x - 1\), show that the area of the region enclosed by the \(x\)-axis, the curve and the lines \(x = 2\) and \(x = 3\) is given by $$\int _ { 1 } ^ { 2 } \left( u + 2 + \frac { 4 } { u } \right) \mathrm { d } u$$ Evaluate this area exactly.
4. Another curve is defined by the equation \(\mathrm { e } ^ { y } = \frac { x ^ { 2 } + 3 } { x - 1 }\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\) by differentiating implicitly. Hence find the gradient of this curve at the point where \(x = 2\).

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 + x _ { 2 } } }\). \begin{figure}[h] \end{figure}

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

1 Fig. 9 shows the curve with equation \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\). It has an asymptote \(x = a\) and turning point P . \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).

2 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 \(( 11,11 )\). \begin{figure}[h] \end{figure}

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

4 Fig. 8 shows parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where \(\mathrm { f } ( x ) = \tan x\) and \(\mathrm { g } ( x ) = 1 + \mathrm { f } \left( x - \frac { 1 } { 4 } \pi \right)\). \begin{figure}[h] \end{figure}

1. Describe a sequence of two transformations which maps the curve \(y = \mathrm { f } ( x )\) to the curve \(y = \mathrm { g } ( x )\). [4] It can be shown that \(\mathrm { g } ( x ) = \frac { 2 \sin x } { \sin x + \cos x }\).
2. Show that \(\mathrm { g } ^ { \prime } ( x ) = \frac { 2 } { ( \sin x + \cos x ) ^ { 2 } }\). Hence verify that the gradient of \(y = \mathrm { g } ( x )\) at the point \(\left( \frac { 1 } { 4 } \pi , 1 \right)\) is the same as that of \(y = \mathrm { f } ( x )\) at the origin.
3. By writing \(\tan x = \frac { \sin x } { \cos x }\) and using the substitution \(u = \cos x\), show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \mathrm { f } ( x ) \mathrm { d } x = \int _ { \frac { 1 } { \sqrt { 2 } } } ^ { 1 } \frac { 1 } { u } \mathrm {~d} u\). Evaluate this integral exactly.
4. Hence find the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the lines \(x = \frac { 1 } { 4 } \pi\) and \(x = \frac { 1 } { 2 } \pi\).

1 \begin{figure}[h] \end{figure} Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } + 2 }\).

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

3

1. Use the substitution \(u = 1 + x\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$ where \(a\) and \(b\) are to be found.
   Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x\), giving your answer in exact form. Fig. 8 shows the curve \(y = x ^ { 2 } \ln ( 1 + x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{d1206ce8-7716-4205-b98e-664e7ead8a25-3_830_806_907_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
3. Using integration by parts, and the result of part (i), find the exact area enclosed by the curve \(y = x ^ { 2 } \ln ( 1 + x )\), the \(x\)-axis and the line \(x = 1\).

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

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence verify that the origin is a stationary point of the curve.
2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that the origin is a minimum point.
3. Using the substitution \(u = 1 + x\), show that \(\int \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x = \int \left( u - 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x\), giving your answer in an exact form.
4. Using integration by parts and your answer to part (iii), evaluate \(\int _ { 0 } ^ { 1 } 2 x \ln ( 1 + x ) \mathrm { d } x\).

2 Fig. 7 shows part of the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = x \sqrt { 1 + x }\). The curve meets the \(x\)-axis at the origin and at the point P . \begin{figure}[h] \end{figure}

1. Verify that the point P has coordinates \(( - 1,0 )\). Hence state the domain of the function \(\mathrm { f } ( x )\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + 3 x } { 2 \sqrt { 1 + x } }\).
3. Find the exact coordinates of the turning point of the curve. Hence write down the range of the function.
4. Use the substitution \(u = 1 + x\) to show that $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( \begin{array} { l l } u ^ { \frac { 3 } { 2 } } & u ^ { \frac { 1 } { 2 } } \end{array} \right) \mathrm { d } u .$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.