1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)

5.

1. Find $$\int \left( ( 3 x + 5 ) ^ { 9 } + \mathrm { e } ^ { 5 x } \right) \mathrm { d } x$$
2. Given that \(b\) is a constant greater than 2 , and $$\int _ { 2 } ^ { b } \frac { x } { x ^ { 2 } + 5 } \mathrm {~d} x = \ln ( \sqrt { 6 } )$$ use integration to find the value of \(b\).

6.

1. Find \(\int \tan ^ { 2 } x \mathrm {~d} x\).
2. Use integration by parts to find \(\int \frac { 1 } { x ^ { 3 } } \ln x \mathrm {~d} x\).
3. Use the substitution \(u = 1 + e ^ { x }\) to show that $$\int \frac { \mathrm { e } ^ { 3 x } } { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - \mathrm { e } ^ { x } + \ln \left( 1 + \mathrm { e } ^ { x } \right) + k$$ where \(k\) is a constant.
   

8.

1. Use de Moivre's theorem to show that $$\cos ^ { 5 } \theta \equiv p \cos 5 \theta + q \cos 3 \theta + r \cos \theta$$ where \(p , q\) and \(r\) are rational numbers to be found.
2. Hence, showing all your working, find the exact value of $$\int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 3 } } \cos ^ { 5 } \theta \mathrm {~d} \theta$$

Find $$\int \left( 3 x ^ { 2 } - \frac { 4 } { x ^ { 2 } } \right) \mathrm { d } x$$ giving each term in its simplest form.

Use calculus to find the exact value of \(\int _ { 1 } ^ { 2 } \left( 3 x ^ { 2 } + 5 + \frac { 4 } { x ^ { 2 } } \right) \mathrm { d } x\).

6. Given that $$\mathrm { f } ^ { \prime } ( x ) = 5 + \frac { 4 } { x ^ { 2 } } , \quad x \neq 0$$

1. find an expression for \(\mathrm { f } ( x )\). Given also that $$\mathrm { f } ( 2 ) = 2 \mathrm { f } ( 1 ) ,$$
2. find \(\mathrm { f } ( 4 )\).

5.

1. Find $$\int \left( 8 x - \frac { 2 } { x ^ { 3 } } \right) \mathrm { d } x$$ The gradient of a curve is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 8 x - \frac { 2 } { x ^ { 3 } } , \quad x \neq 0$$ and the curve passes through the point \(( 1,1 )\).
2. Show that the equation of the curve can be written in the form $$y = \left( a x + \frac { b } { x } \right) ^ { 2 }$$ where \(a\) and \(b\) are integers to be found.

6. Find the value of each of the following integrals in exact, simplified form.

1. \(\quad \int _ { - 1 } ^ { 0 } \mathrm { e } ^ { 1 - 2 x } \mathrm {~d} x\)
2. \(\int _ { 2 } ^ { 4 } \frac { 3 x ^ { 2 } - 2 } { x } \mathrm {~d} x\)

1 Show that \(\int _ { 2 } ^ { 8 } \frac { 3 } { x } \mathrm {~d} x = \ln 64\).

8 The definite integral \(I\) is defined by $$I = \int _ { 0 } ^ { 6 } 2 ^ { x } \mathrm {~d} x$$

1. Use Simpson's rule with 6 strips to find an approximate value of \(I\).
2. By first writing \(2 ^ { x }\) in the form \(\mathrm { e } ^ { k x }\), where the constant \(k\) is to be determined, find the exact value of \(I\).
3. Use the answers to parts (i) and (ii) to deduce that \(\ln 2 \approx \frac { 9 } { 13 }\).

1 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \cos 3 x \mathrm {~d} x\).

A particle \(P\) of mass 0.5 kg moves along the positive \(x\)-axis. It moves away from the origin \(O\) under the action of a single force directed away from \(O\). When \(O P = x\) metres, the magnitude of the force is \(\frac { 3 } { ( x + 1 ) ^ { 3 } } \mathrm {~N}\) and the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Initially \(P\) is at rest at \(O\).

1. Show that \(v ^ { 2 } = 6 \left( 1 - \frac { 1 } { ( x + 1 ) ^ { 2 } } \right)\).
2. Show that the speed of \(P\) never reaches \(\sqrt { } 6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
3. Find \(x\) when \(P\) has been moving for 2 seconds.
   \section*{LL \(\_\_\_\_\)}

At time \(t = 0\), a particle \(P\) is at the origin \(O\) moving with speed \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) along the \(x\)-axis in the positive \(x\)-direction. At time \(t\) seconds \(( t > 0 )\), the acceleration of \(P\) has magnitude \(\frac { 3 } { ( t + 1 ) ^ { 2 } } \mathrm {~m} \mathrm {~s} ^ { - 2 }\) and is directed towards \(O\).

1. Show that at time \(t\) seconds the velocity of \(P\) is \(\left( \frac { 3 } { t + 1 } - 1 \right) \mathrm { m } \mathrm { s } ^ { - 1 }\).
2. Find, to 3 significant figures, the distance of \(P\) from \(O\) when \(P\) is instantaneously at rest.
   

3 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( 1 - \sin 3 x ) \mathrm { d } x\), giving your answer in exact form.

3 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \sin 3 x \mathrm {~d} x\).
[0pt] [3]

4 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.

3 Fig. 8 shows the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\). P is the point on this curve with \(x\)-coordinate 1 , and R is the point \(\left( 0 , - \frac { 7 } { 8 } \right)\). \begin{figure}[h] \end{figure}

1. Find the gradient of PR.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that PR is a tangent to the curve.
3. Find the exact coordinates of the turning point Q .
4. Differentiate \(x \ln x - x\). Hence, or otherwise, show that the area of the region enclosed by the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) is \(\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2\).

1 Fig. 8 shows part of the curve \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \text { for } x \geqslant 0 .$$ \begin{figure}[h] \end{figure}

1. Find \(\mathrm { f } ^ { \prime } ( x )\), and hence calculate the gradient of the curve \(y = \mathrm { f } ( x )\) at the origin and at the point \(( \ln 2,1 )\). The function \(\mathrm { g } ( x )\) is defined by $$\sqrt { } \text { for } x \geqslant 0 \text {. }$$
2. Show that \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are inverse functions. Hence sketch the graph of \(y = \mathrm { g } ( x )\). Write down the gradient of the curve \(y = \mathrm { g } ( x )\) at the point \(( 1 , \ln 2 )\).
3. Show that \(\int \left( \mathrm { e } ^ { x } 1 \right) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } \quad 2 \mathrm { e } ^ { x } + x + c\). Hence evaluate \(\int _ { 0 } ^ { \ln 2 } \left( \mathrm { e } ^ { x } \quad 1 \right) ^ { 2 } \mathrm {~d} x\), giving your answer in an exact form.
4. Using your answer to part (iii), calculate the area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the line \(x = 1\).

2 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \cos 3 x \mathrm {~d} x\).

5 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \cos ^ { 2 } x } , - \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\), together with its asymptotes \(x = \frac { 1 } { 2 } \pi\) and \(x = - \frac { 1 } { 2 } \pi\). \begin{figure}[h] \end{figure}

1. Use the quotient rule to show that the derivative of \(\frac { \sin x } { \cos x }\) is \(\frac { 1 } { \cos ^ { 2 } x }\).
2. Find the area bounded by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 4 } \pi\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { 2 } \mathrm { f } \left( x + \frac { 1 } { 4 } \pi \right)\).
3. Verify that the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) cross at \(( 0,1 )\).
4. State a sequence of two transformations such that the curve \(y = \mathrm { f } ( x )\) is mapped to the curve \(y = \mathrm { g } ( x )\). On the copy of Fig. 9, sketch the curve \(y = \mathrm { g } ( x )\), indicating clearly the coordinates of the minimum point and the equations of the asymptotes to the curve.
5. Use your result from part (ii) to write down the area bounded by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = - \frac { 1 } { 4 } \pi\).

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + \sin 2 x\) for \(- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi\). \begin{figure}[h] \end{figure}

1. State a sequence of two transformations that would map part of the curve \(y = \sin x\) onto the curve \(y = \mathrm { f } ( x )\).
2. Find the area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the line \(x = \frac { 1 } { 4 } \pi\).
3. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point ( 0,1 ). Hence write down the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).
4. State the domain of \(\mathrm { f } ^ { - 1 } ( x )\). Add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.
5. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).

4 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.

4 Evaluate the following integrals, giving your answers in exact form. \begin{displayquote}

1. \(\int _ { 0 } ^ { 1 } \frac { 2 x } { x ^ { 2 } + 1 } \mathrm {~d} x\)
2. \(\int _ { 0 } ^ { 1 } \frac { 2 x } { x + 1 } \mathrm {~d} x\) \end{displayquote}

3 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 \(y = \mathrm { f } ( x )\) ?