Trigonometric substitution (sin, cos, tan)

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\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 \(A\).

6

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

6

1. Use the substitution \(x = 2 \tan \theta\) to show that $$\int _ { 0 } ^ { 2 } \frac { 8 } { \left( 4 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \cos ^ { 2 } \theta \mathrm {~d} \theta$$
2. Hence find the exact value of $$\int _ { 0 } ^ { 2 } \frac { 8 } { \left( 4 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x$$

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

11. \begin{figure}[h] \end{figure}

1. By writing \(\sec \theta\) as \(\frac { 1 } { \cos \theta }\), show that when \(x = 3 \sec \theta\), $$\frac { \mathrm { d } x } { \mathrm {~d} \theta } = 3 \sec \theta \tan \theta$$ Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \frac { \sqrt { x ^ { 2 } - 9 } } { x } \quad x \geqslant 3$$ The finite region \(R\), shown shaded in Figure 3, is bounded by the curve \(C\), the \(x\)-axis and the line with equation \(x = 6\)
2. Use the substitution \(x = 3 \sec \theta\) to find the exact value of the area of \(R\). [Solutions based entirely on graphical or numerical methods are not acceptable.]

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

6. \begin{figure}[h] \end{figure} Figure 2 Figure 2 shows a sketch of the curve with equation \(y = \sqrt { ( 3 - x ) ( x + 1 ) } , 0 \leqslant x \leqslant 3\)
The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis, and the \(y\)-axis.

1. Use the substitution \(x = 1 + 2 \sin \theta\) to show that $$\int _ { 0 } ^ { 3 } \sqrt { ( 3 - x ) ( x + 1 ) } d x = k \int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } \theta d \theta$$ where \(k\) is a constant to be determined.
2. Hence find, by integration, the exact area of \(R\).

6. (i) Given that \(y > 0\), find $$\int \frac { 3 y - 4 } { y ( 3 y + 2 ) } d y$$ (ii) (a) Use the substitution \(x = 4 \sin ^ { 2 } \theta\) to show that $$\int _ { 0 } ^ { 3 } \sqrt { \left( \frac { x } { 4 - x } \right) } \mathrm { d } x = \lambda \int _ { 0 } ^ { \frac { \pi } { 3 } } \sin ^ { 2 } \theta \mathrm {~d} \theta$$ where \(\lambda\) is a constant to be determined.
(b) Hence use integration to find $$\int _ { 0 } ^ { 3 } \sqrt { \left( \frac { x } { 4 - x } \right) } d x$$ giving your answer in the form \(a \pi + b\), where \(a\) and \(b\) are exact constants.

3. Use the substitution \(x = \tan \theta\) to show that $$\int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x = \frac { \pi } { 8 } + \frac { 1 } { 4 }$$ (8)

1. In this question you must show all stages of your working.

\section*{Solutions relying on calculator technology are not acceptable.}

1. Use the substitution \(x = 2 \sin u\) to show that $$\int _ { 0 } ^ { 1 } \frac { 3 x + 2 } { \left( 4 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } d x = \int _ { 0 } ^ { p } \left( \frac { 3 } { 2 } \operatorname { secutanu } + \frac { 1 } { 2 } \sec ^ { 2 } u \right) d u$$ where \(p\) is a constant to be found.
2. Hence find the exact value of $$\int _ { 0 } ^ { 1 } \frac { 3 x + 2 } { \left( 4 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } d x$$

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 \(\int _ { 0 } ^ { 1 } \sqrt { \frac { x } { 1 - x } } \mathrm {~d} x\).

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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

4

1. Show that the substitution \(x = \tan \theta\) transforms \(\int \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x\) to \(\int \cos ^ { 2 } \theta \mathrm {~d} \theta\).
2. Hence find the exact value of \(\int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x\).
   \(5 A B C D\) is a parallelogram. The position vectors of \(A , B\) and \(C\) are given respectively by $$\mathbf { a } = 2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k } , \quad \mathbf { b } = 3 \mathbf { i } - 2 \mathbf { j } , \quad \mathbf { c } = \mathbf { i } - \mathbf { j } - 2 \mathbf { k } .$$
3. Find the position vector of \(D\).
4. Determine, to the nearest degree, the angle \(A B C\).

7. (i) Use the substitution \(x = 2 \sin u\) to evaluate $$\int _ { 0 } ^ { \sqrt { 3 } } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x$$ (ii) Evaluate $$\int _ { 0 } ^ { \frac { \pi } { 2 } } x \cos x \mathrm {~d} x$$

2 Use the substitution \(x = \tan \theta\) to find the exact value of $$\int _ { 1 } ^ { \sqrt { 3 } } \frac { 1 - x ^ { 2 } } { 1 + x ^ { 2 } } \mathrm {~d} x$$

4 Use the substitution \(x = \frac { 1 } { 3 } \sin \theta\) to find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 6 } } \frac { 1 } { \left( 1 - 9 x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x$$

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1. The curve \(y = \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { 2 } }\) is shown in Fig. 8. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{a13f7a05-e2d3-4354-a0c7-ef7283eff514-08_495_1058_1105_315} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
   1. Show that \(\frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 20 x ^ { 2 } - 4 } { \left( 1 + x ^ { 2 } \right) ^ { 4 } }\).
   2. In this question you must show detailed reasoning. Find the set of values of \(x\) for which the curve is concave downwards.
2. Use the substitution \(x = \tan \theta\) to find the exact value of \(\int _ { - 1 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } d x\). Answer all the questions.
   Section B (15 marks) The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.

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1. Given that \(x = \frac { 1 } { \sin \theta }\), use the quotient rule to show that \(\frac { \mathrm { d } x } { \mathrm {~d} \theta } = - \operatorname { cosec } \theta \cot \theta\).
   (3 marks)
2. Use the substitution \(x = \operatorname { cosec } \theta\) to find \(\int _ { \sqrt { 2 } } ^ { 2 } \frac { 1 } { x ^ { 2 } \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x\), giving your answer to three significant figures.
   (9 marks)

6. (a) Use the substitution \(x = 2 \sin u\) to evaluate $$\int _ { 0 } ^ { \sqrt { 3 } } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x$$ (b) Use integration by parts to evaluate $$\int _ { 0 } ^ { \frac { \pi } { 2 } } x \cos x d x$$ 6. continued

8 Use a suitable trigonometric substitution to find \(\int \frac { x ^ { 2 } } { \sqrt { 1 - x ^ { 2 } } } \mathrm {~d} x\).

4

1. Show that the substitution \(x = \tan \theta\) transforms \(\int \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x\) to \(\int \cos ^ { 2 } \theta \mathrm {~d} \theta\).
2. Hence find the exact value of \(\int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x\).
   \(5 A B C D\) is a parallelogram. The position vectors of \(A , B\) and \(C\) are given respectively by $$\mathbf { a } = 2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k } , \quad \mathbf { b } = 3 \mathbf { i } - 2 \mathbf { j } , \quad \mathbf { c } = \mathbf { i } - \mathbf { j } - 2 \mathbf { k } .$$
3. Find the position vector of \(D\).
4. Determine, to the nearest degree, the angle \(A B C\). 6 The equation of a curve is \(x y ^ { 2 } = 2 x + 3 y\).
5. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 - y ^ { 2 } } { 2 x y - 3 }\).
6. Show that the curve has no tangents which are parallel to the \(y\)-axis. 7 A curve is given parametrically by the equations $$x = t ^ { 2 } , \quad y = \frac { 1 } { t }$$
7. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), giving your answer in its simplest form.
8. Show that the equation of the tangent at the point \(P \left( 4 , - \frac { 1 } { 2 } \right)\) is $$x - 16 y = 12$$
9. Find the value of the parameter at the point where the tangent at \(P\) meets the curve again. June 2005
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10. Given that \(\frac { 3 x + 4 } { ( 1 + x ) ( 2 + x ) ^ { 2 } } \equiv \frac { A } { 1 + x } + \frac { B } { 2 + x } + \frac { C } { ( 2 + x ) ^ { 2 } }\), find \(A , B\) and \(C\).
11. Hence or otherwise expand \(\frac { 3 x + 4 } { ( 1 + x ) ( 2 + x ) ^ { 2 } }\) in ascending powers of \(x\), up to and including the term in \(x ^ { 2 }\).
12. State the set of values of \(x\) for which the expansion in part (ii) is valid. 9 Newton's law of cooling states that the rate at which the temperature of an object is falling at any instant is proportional to the difference between the temperature of the object and the temperature of its surroundings at that instant. A container of hot liquid is placed in a room which has a constant temperature of \(20 ^ { \circ } \mathrm { C }\). At time \(t\) minutes later, the temperature of the liquid is \(\theta ^ { \circ } \mathrm { C }\).
13. Explain how the information above leads to the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - 20 )$$ where \(k\) is a positive constant.
14. The liquid is initially at a temperature of \(100 ^ { \circ } \mathrm { C }\). It takes 5 minutes for the liquid to cool from \(100 ^ { \circ } \mathrm { C }\) to \(68 ^ { \circ } \mathrm { C }\). Show that $$\theta = 20 + 80 \mathrm { e } ^ { - \left( \frac { 1 } { 5 } \ln \frac { 5 } { 3 } \right) t }$$
15. Calculate how much longer it takes for the liquid to cool by a further \(32 ^ { \circ } \mathrm { C }\). 1 Simplify \(\frac { x ^ { 3 } - 3 x ^ { 2 } } { x ^ { 2 } - 9 }\). 2 Given that \(\sin y = x y + x ^ { 2 }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). 3
16. Find the quotient and the remainder when \(3 x ^ { 3 } - 2 x ^ { 2 } + x + 7\) is divided by \(x ^ { 2 } - 2 x + 5\).
17. Hence, or otherwise, determine the values of the constants \(a\) and \(b\) such that, when \(3 x ^ { 3 } - 2 x ^ { 2 } + a x + b\) is divided by \(x ^ { 2 } - 2 x + 5\), there is no remainder. 4
18. Use integration by parts to find \(\int x \sec ^ { 2 } x \mathrm {~d} x\).
19. Hence find \(\int x \tan ^ { 2 } x \mathrm {~d} x\).