1.08h Integration by substitution

8 \includegraphics[max width=\textwidth, alt={}, center]{05e75fa2-81ae-44b1-b073-4100f5d911e0-12_771_839_262_651} The diagram shows the circle with equation \(( x - 2 ) ^ { 2 } + y ^ { 2 } = 8\). The chord \(A B\) of the circle intersects the positive \(y\)-axis at \(A\) and is parallel to the \(x\)-axis.

1. Find, by calculation, the coordinates of \(A\) and \(B\).
2. Find the volume of revolution when the shaded segment, bounded by the circle and the chord \(A B\), is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

2 Evaluate \(\int _ { 0 } ^ { 1 } \sqrt { } ( 3 x + 1 ) \mathrm { d } x\).

9 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 } { \sqrt { } ( 6 - 2 x ) }\), and \(P ( 1,8 )\) is a point on the curve.

1. The normal to the curve at the point \(P\) meets the coordinate axes at \(Q\) and at \(R\). Find the coordinates of the mid-point of \(Q R\).
2. Find the equation of the curve.

7 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 } { \sqrt { } ( 4 x - 3 ) }\) and \(P ( 3,3 )\) is a point on the curve.

1. Find the equation of the normal to the curve at \(P\), giving your answer in the form \(a x + b y = c\).
2. Find the equation of the curve.

7 \includegraphics[max width=\textwidth, alt={}, center]{b566719c-216e-41e5-8431-da77e1dad73e-3_301_485_264_829} A piece of wire of length 50 cm is bent to form the perimeter of a sector \(P O Q\) of a circle. The radius of the circle is \(r \mathrm {~cm}\) and the angle \(P O Q\) is \(\theta\) radians (see diagram).

1. Express \(\theta\) in terms of \(r\) and show that the area, \(A \mathrm {~cm} ^ { 2 }\), of the sector is given by $$A = 25 r - r ^ { 2 } .$$
2. Given that \(r\) can vary, find the stationary value of \(A\) and determine its nature.

10 \includegraphics[max width=\textwidth, alt={}, center]{56d376c5-b91f-488d-89e2-18edcb14052d-4_799_1390_255_376} The diagram shows the curve \(y = \sqrt { } ( 1 + 2 x )\) meeting the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\). The \(y\)-coordinate of the point \(C\) on the curve is 3 .

1. Find the coordinates of \(B\) and \(C\).
2. Find the equation of the normal to the curve at \(C\).
3. Find the volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(\boldsymbol { y }\)-axis.

9 \includegraphics[max width=\textwidth, alt={}, center]{11bfe5bd-604c-43e5-81e7-4c1f5676bcbb-4_502_663_255_740} The diagram shows part of the curve \(y = \frac { 9 } { 2 x + 3 }\), crossing the \(y\)-axis at the point \(B ( 0,3 )\). The point \(A\) on the curve has coordinates \(( 3,1 )\) and the tangent to the curve at \(A\) crosses the \(y\)-axis at \(C\).

1. Find the equation of the tangent to the curve at \(A\).
2. Determine, showing all necessary working, whether \(C\) is nearer to \(B\) or to \(O\).
3. Find, showing all necessary working, the exact volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

9 \includegraphics[max width=\textwidth, alt={}, center]{d5f66324-e1fc-40e1-98e7-625187e24d3d-4_584_670_881_740} The diagram shows part of the curve \(y = \frac { 8 } { x } + 2 x\) and three points \(A , B\) and \(C\) on the curve with \(x\)-coordinates 1, 2 and 5 respectively.

1. A point \(P\) moves along the curve in such a way that its \(x\)-coordinate increases at a constant rate of 0.04 units per second. Find the rate at which the \(y\)-coordinate of \(P\) is changing as \(P\) passes through \(A\).
2. Find the volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

6

1. Use the trapezium rule with three intervals to find an approximation to \(\int _ { 1 } ^ { 4 } \frac { 6 } { 1 + \sqrt { x } } \mathrm {~d} x\). Give your answer correct to 5 significant figures.
2. Find the exact value of \(\int _ { 1 } ^ { 4 } 2 \mathrm { e } ^ { \frac { 1 } { 2 } x - 2 } \mathrm {~d} x\).
3. \includegraphics[max width=\textwidth, alt={}, center]{2d6fc4c5-70ec-4cd8-9b48-59d5ce0e39b7-11_556_805_262_705} The diagram shows the curves \(y = \frac { 6 } { 1 + \sqrt { x } }\) and \(y = 2 \mathrm { e } ^ { \frac { 1 } { 2 } x - 2 }\) which meet at a point with \(x\)-coordinate 4. The shaded region is bounded by the two curves and the line \(x = 1\). Use your answers to parts (a) and (b) to find an approximation to the area of the shaded region. Give your answer correct to 3 significant figures.
4. State, with a reason, whether your answer to part (c) is an over-estimate or under-estimate of the exact area of the shaded region.

10 \includegraphics[max width=\textwidth, alt={}, center]{0f081749-4fe0-46e3-96c2-466e69cf49d3-4_620_894_338_687} The function f is defined by \(\mathrm { f } ( x ) = ( \ln x ) ^ { 2 }\) for \(x > 0\). The diagram shows a sketch of the graph of \(y = \mathrm { f } ( x )\). The minimum point of the graph is \(A\). The point \(B\) has \(x\)-coordinate e .

1. State the \(x\)-coordinate of \(A\).
2. Show that \(\mathrm { f } ^ { \prime \prime } ( x ) = 0\) at \(B\).
3. Use the substitution \(x = \mathrm { e } ^ { u }\) to show that the area of the region bounded by the \(x\)-axis, the line \(x = \mathrm { e }\), and the part of the curve between \(A\) and \(B\) is given by $$\int _ { 0 } ^ { 1 } u ^ { 2 } \mathrm { e } ^ { u } \mathrm {~d} u .$$
4. Hence, or otherwise, find the exact value of this area.

4

1. Use the substitution \(x = \tan \theta\) to show that $$\int \frac { 1 - x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x = \int \cos 2 \theta \mathrm {~d} \theta$$
2. Hence find the value of $$\int _ { 0 } ^ { 1 } \frac { 1 - x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x$$

8

1. Using partial fractions, find $$\int \frac { 1 } { y ( 4 - y ) } \mathrm { d } y$$
2. Given that \(y = 1\) when \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 4 - y ) ,$$ obtaining an expression for \(y\) in terms of \(x\).
3. State what happens to the value of \(y\) if \(x\) becomes very large and positive.

7 Let \(I = \int _ { 1 } ^ { 4 } \frac { 1 } { x ( 4 - \sqrt { } x ) } \mathrm { d } x\).

1. Use the substitution \(u = \sqrt { } x\) to show that \(I = \int _ { 1 } ^ { 2 } \frac { 2 } { u ( 4 - u ) } \mathrm { d } u\).
2. Hence show that \(I = \frac { 1 } { 2 } \ln 3\).

10 \includegraphics[max width=\textwidth, alt={}, center]{0f73e750-18a0-49ad-b4cb-fd6d14f0789e-4_424_713_262_715} The diagram shows the curve \(y = x ^ { 2 } \sqrt { } \left( 1 - x ^ { 2 } \right)\) for \(x \geqslant 0\) and its maximum point \(M\).

1. Find the exact value of the \(x\)-coordinate of \(M\).
2. Show, by means of the substitution \(x = \sin \theta\), that the area \(A\) of the shaded region between the curve and the \(x\)-axis is given by $$A = \frac { 1 } { 4 } \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { 2 } 2 \theta \mathrm {~d} \theta$$
3. Hence obtain the exact value of \(A\).

7 The integral \(I\) is defined by \(I = \int _ { 0 } ^ { 2 } 4 t ^ { 3 } \ln \left( t ^ { 2 } + 1 \right) \mathrm { d } t\).

1. Use the substitution \(x = t ^ { 2 } + 1\) to show that \(I = \int _ { 1 } ^ { 5 } ( 2 x - 2 ) \ln x \mathrm {~d} x\).
2. Hence find the exact value of \(I\).

8 \includegraphics[max width=\textwidth, alt={}, center]{5b219e1c-e5a0-4f75-910d-fca9761e5088-3_435_895_799_625} The diagram shows the curve \(y = 5 \sin ^ { 3 } x \cos ^ { 2 } x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Using the substitution \(u = \cos x\), find by integration the area of the shaded region bounded by the curve and the \(x\)-axis.

8 Let \(I = \int _ { 2 } ^ { 5 } \frac { 5 } { x + \sqrt { } ( 6 - x ) } \mathrm { d } x\).

1. Using the substitution \(u = \sqrt { } ( 6 - x )\), show that $$I = \int _ { 1 } ^ { 2 } \frac { 10 u } { ( 3 - u ) ( 2 + u ) } \mathrm { d } u$$
2. Hence show that \(I = 2 \ln \left( \frac { 9 } { 2 } \right)\).

7 \includegraphics[max width=\textwidth, alt={}, center]{e2cc23d2-f3ac-488b-97e1-79e2a98a87ba-3_421_885_251_628} The diagram shows part of the curve \(y = \cos ( \sqrt { } x )\) for \(x \geqslant 0\), where \(x\) is in radians. The shaded region between the curve, the axes and the line \(x = p ^ { 2 }\), where \(p > 0\), is denoted by \(R\). The area of \(R\) is equal to 1 .

1. Use the substitution \(x = u ^ { 2 }\) to find \(\int _ { 0 } ^ { p ^ { 2 } } \cos ( \sqrt { } x ) \mathrm { d } x\). Hence show that \(\sin p = \frac { 3 - 2 \cos p } { 2 p }\).
2. Use the iterative formula \(p _ { n + 1 } = \sin ^ { - 1 } \left( \frac { 3 - 2 \cos p _ { n } } { 2 p _ { n } } \right)\), with initial value \(p _ { 1 } = 1\), to find the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

8

1. Show that \(\int _ { 2 } ^ { 4 } 4 x \ln x \mathrm {~d} x = 56 \ln 2 - 12\).
2. Use the substitution \(u = \sin 4 x\) to find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 24 } \pi } \cos ^ { 3 } 4 x \mathrm {~d} x\).

4

1. Express \(( \sqrt { } 3 ) \cos x + \sin x\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), giving the exact values of \(R\) and \(\alpha\).
2. Hence show that $$\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { ( ( \sqrt { } 3 ) \cos x + \sin x ) ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 4 } \sqrt { } 3$$

9 \includegraphics[max width=\textwidth, alt={}, center]{436d891d-92ee-4076-8369-db756d413979-3_307_601_1553_772} The diagram shows the curve \(y = \sin ^ { 2 } 2 x \cos x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Using the substitution \(u = \sin x\), find by integration the area of the shaded region bounded by the curve and the \(x\)-axis.

2 Use the substitution \(u = 1 + 3 \tan x\) to find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { \sqrt { } ( 1 + 3 \tan x ) } { \cos ^ { 2 } x } d x$$

9 \includegraphics[max width=\textwidth, alt={}, center]{b2136f5d-0d66-4524-bb76-fcc4cb59150c-3_639_387_1749_879} The diagram shows the curve \(y = \mathrm { e } ^ { 2 \sin x } \cos x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Using the substitution \(u = \sin x\), find the exact value of the area of the shaded region bounded by the curve and the axes.
2. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.

5

1. Find \(\int \left( 4 + \tan ^ { 2 } 2 x \right) \mathrm { d } x\).
2. Find the exact value of \(\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } \frac { \sin \left( x + \frac { 1 } { 6 } \pi \right) } { \sin x } \mathrm {~d} x\).

6 Let \(I = \int _ { 0 } ^ { 1 } \frac { \sqrt { } x } { 2 - \sqrt { } x } \mathrm {~d} x\).

1. Using the substitution \(u = 2 - \sqrt { } x\), show that \(I = \int _ { 1 } ^ { 2 } \frac { 2 ( 2 - u ) ^ { 2 } } { u } \mathrm {~d} u\).
2. Hence show that \(I = 8 \ln 2 - 5\).