1.08h Integration by substitution

1 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 \(\ln \frac { 32 } { 27 }\).
   [0pt] [Hint: consider its reflection in \(y = x\).]

3 Evaluate \(\int _ { 0 } ^ { 3 } x ( x + 1 ) ^ { - \frac { 1 } { 2 } } \mathrm {~d} x\), giving your answer as an exact fraction.

5 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 \(x = 11\).

2

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]{35646966-3747-4f1d-bf94-60e9e3130afe-2_829_806_944_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 \(x = 1\).

2 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 \(\frac { 1 } { 6 } \ln 3\).

6

1. Show that the substitution \(x = \sin ^ { 2 } \theta\) transforms \(\int \sqrt { \frac { x } { 1 - x } } \mathrm {~d} x\) to \(\int 2 \sin ^ { 2 } \theta \mathrm {~d} \theta\).
2. Hence find \(\int _ { 0 } ^ { 1 } \sqrt { \frac { x } { 1 - x } } \mathrm {~d} x\).

10

1. Use the substitution \(x = \sin \theta\) to find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x$$
2. Find the exact value of $$\int _ { 1 } ^ { 3 } \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x$$ 4

8

1. Given that \(\frac { 2 t } { ( t + 1 ) ^ { 2 } }\) can be expressed in the form \(\frac { A } { t + 1 } + \frac { B } { ( t + 1 ) ^ { 2 } }\), find the values of the constants \(A\) and \(B\).
2. Show that the substitution \(t = \sqrt { 2 x - 1 }\) transforms \(\int \frac { 1 } { x + \sqrt { 2 x - 1 } } \mathrm {~d} x\) to \(\int \frac { 2 t } { ( t + 1 ) ^ { 2 } } \mathrm {~d} t\).
3. Hence find the exact value of \(\int _ { 1 } ^ { 5 } \frac { 1 } { x + \sqrt { 2 x - 1 } } \mathrm {~d} x\).

8 Let \(I = \int \frac { 1 } { x ( 1 + \sqrt { } x ) ^ { 2 } } \mathrm {~d} x\).

1. Show that the substitution \(u = \sqrt { } x\) transforms \(I\) to \(\int \frac { 2 } { u ( 1 + u ) ^ { 2 } } \mathrm {~d} u\).
2. Express \(\frac { 2 } { u ( 1 + u ) ^ { 2 } }\) in the form \(\frac { A } { u } + \frac { B } { 1 + u } + \frac { C } { ( 1 + u ) ^ { 2 } }\).
3. Hence find \(I\).

6. (i) Find $$\int \cot ^ { 2 } 2 x \mathrm {~d} x$$ (ii) Use the substitution \(u ^ { 2 } = x + 1\) to evaluate $$\int _ { 0 } ^ { 3 } \frac { x ^ { 2 } } { \sqrt { x + 1 } } \mathrm {~d} x$$

6. (i) Find \(\int \tan ^ { 2 } 3 x \mathrm {~d} x\).
(ii) Using the substitution \(u = x ^ { 2 } + 4\), evaluate $$\int _ { 0 } ^ { 2 } \frac { 5 x } { \left( x ^ { 2 } + 4 \right) ^ { 2 } } d x$$

3. Find

1. \(\int \frac { x } { 2 - x ^ { 2 } } \mathrm {~d} x\),
2. \(\int x ^ { 2 } \mathrm { e } ^ { - x } \mathrm {~d} x\).

4. \includegraphics[max width=\textwidth, alt={}, center]{c7b867af-0730-459e-9c76-15eb07b9e476-1_465_976_1539_388} The diagram shows the curve with parametric equations $$x = \tan \theta , \quad y = \cos ^ { 2 } \theta , \quad - \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 }$$

1. Find a cartesian equation for the curve. The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = - 1\) and \(x = 1\).
2. Using integration, with the substitution \(x = \tan u\), find the area of the shaded region.

1. Use the substitution \(u = 1 - x ^ { \frac { 1 } { 2 } }\) to find

$$\int \frac { 1 } { 1 - x ^ { \frac { 1 } { 2 } } } \mathrm {~d} x$$

3. Using the substitution \(u = \mathrm { e } ^ { x } - 1\), show that $$\int _ { \ln 2 } ^ { \ln 5 } \frac { \mathrm { e } ^ { 2 x } } { \sqrt { \mathrm { e } ^ { x } - 1 } } \mathrm {~d} x = \frac { 20 } { 3 }$$

1. (i) Find

$$\int x ^ { 2 } \sin x \mathrm {~d} x$$ (ii) Use the substitution \(u = 1 + \sin x\) to find the value of $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \cos x ( 1 + \sin x ) ^ { 3 } d x$$

1. Show that

$$\int _ { 2 } ^ { 4 } x \left( x ^ { 2 } - 4 \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x = 8 \sqrt { 3 }$$

8. (i) Find $$\int x ^ { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } x } \mathrm {~d} x$$ (ii) Using the substitution \(u = \sin t\), evaluate $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { 2 } 2 t \cos t \mathrm {~d} t$$

5. \includegraphics[max width=\textwidth, alt={}, center]{5840974b-b08a-4818-9a59-97b2d3ce9890-1_469_809_1777_484} The diagram shows the curve with equation \(y = x \sqrt { 1 - x } , 0 \leq x \leq 1\).
Use the substitution \(u ^ { 2 } = 1 - x\) to show that the area of the region bounded by the curve and the \(x\)-axis is \(\frac { 4 } { 15 }\).

4

1. On separate diagrams, sketch the graphs of \(y = \sinh x\) and \(y = \operatorname { cosech } x\).
2. Show that \(\operatorname { cosech } x = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } - 1 }\), and hence, using the substitution \(u = \mathrm { e } ^ { x }\), find \(\int \operatorname { cosech } x \mathrm {~d} x\).

7

1. Express \(\frac { 1 - t ^ { 2 } } { t ^ { 2 } \left( 1 + t ^ { 2 } \right) }\) in partial fractions.
2. Use the substitution \(t = \tan \frac { 1 } { 2 } x\) to show that $$\int _ { \frac { 1 } { 3 } \pi } ^ { \frac { 1 } { 2 } \pi } \frac { \cos x } { 1 - \cos x } \mathrm {~d} x = \sqrt { 3 } - 1 - \frac { 1 } { 6 } \pi$$

9

1. Prove that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \cosh ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 2 } - 1 } }\).
2. Hence, or otherwise, find \(\int \frac { 1 } { \sqrt { 4 x ^ { 2 } - 1 } } \mathrm {~d} x\).
3. By means of a suitable substitution, find \(\int \sqrt { 4 x ^ { 2 } - 1 } \mathrm {~d} x\).

5

1. Express \(t ^ { 2 } + t + 1\) in the form \(( t + a ) ^ { 2 } + b\).
2. By using the substitution \(\tan \frac { 1 } { 2 } x = t\), show that $$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { 2 + \sin x } \mathrm {~d} x = \frac { \sqrt { 3 } } { 9 } \pi$$

3 By using the substitution \(t = \tan \frac { 1 } { 2 } x\), find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { 2 - \cos x } \mathrm {~d} x$$ giving the answer in terms of \(\pi\).

8

1. Use the substitution \(t = \tan \frac { 1 } { 2 } x\) to show that $$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sqrt { \frac { 1 - \cos x } { 1 + \sin x } } \mathrm {~d} x = 2 \sqrt { } 2 \int _ { 0 } ^ { 1 } \frac { t } { ( 1 + t ) \left( 1 + t ^ { 2 } \right) } \mathrm { d } t$$
2. Express \(\frac { t } { ( 1 + t ) \left( 1 + t ^ { 2 } \right) }\) in partial fractions.
3. Hence find \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sqrt { \frac { 1 - \cos x } { 1 + \sin x } } \mathrm {~d} x\), expressing your answer in an exact form.