1.08h Integration by substitution

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

Solutions relying entirely on calculator technology are not acceptable. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve \(C\) with equation \(y ^ { 2 } = 8 x\) and part of the line \(l\) with equation \(x = 18\) The region \(R\), shown shaded in Figure 2, is bounded by \(C\) and \(l\)

1. Show that the perimeter of \(R\) is given by $$\alpha + 2 \int _ { 0 } ^ { \beta } \sqrt { 1 + \frac { y ^ { 2 } } { 16 } } d y$$ where \(\alpha\) and \(\beta\) are positive constants to be determined.
2. Use the substitution \(y = 4 \sinh u\) and algebraic integration to determine the exact perimeter of \(R\), giving your answer in simplest form.

6. The hyperbola \(H\) has equation $$\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 4 } = 1$$ The line \(l\) is a tangent to \(H\) at the point \(P ( 4 \cosh \alpha , 2 \sinh \alpha )\), where \(\alpha\) is a constant, \(\alpha \neq 0\)

1. Using calculus, show that an equation for \(l\) is $$2 y \sinh \alpha - x \cosh \alpha + 4 = 0$$ The line \(l\) cuts the \(y\)-axis at the point \(A\).
2. Find the coordinates of \(A\) in terms of \(\alpha\). The point \(B\) has coordinates ( \(0,10 \sinh \alpha\) ) and the point \(S\) is the focus of \(H\) for which \(x > 0\)
3. Show that the line segment \(A S\) is perpendicular to the line segment \(B S\).

1. (a) Show that, under the substitution \(x = \frac { 3 } { 4 } \sinh u\),

$$\int \frac { x ^ { 2 } } { \sqrt { 16 x ^ { 2 } + 9 } } \mathrm {~d} x = k \int ( \cosh 2 u - 1 ) \mathrm { d } u$$ where \(k\) is a constant to be determined.
(b) Hence show that $$\int _ { 0 } ^ { 1 } \frac { 64 x ^ { 2 } } { \sqrt { 16 x ^ { 2 } + 9 } } \mathrm {~d} x = p + q \ln 3$$ where \(p\) and \(q\) are rational numbers to be found.

7. The curve \(C\) has parametric equations $$x = 3 t ^ { 4 } , \quad y = 4 t ^ { 3 } , \quad 0 \leqslant t \leqslant 1$$ The curve \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis. The area of the curved surface generated is \(S\).

1. Show that $$S = k \pi \int _ { 0 } ^ { 1 } t ^ { 5 } \left( t ^ { 2 } + 1 \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} t$$ where \(k\) is a constant to be found.
2. Use the substitution \(u ^ { 2 } = t ^ { 2 } + 1\) to find the value of \(S\), giving your answer in the form \(p \pi ( 11 \sqrt { 2 } - 4 )\) where \(p\) is a rational number to be found.

1. (a) Find

$$\int \frac { 5 + x } { \sqrt { 4 - 3 x ^ { 2 } } } \mathrm {~d} x$$ (b) Hence find the exact value of $$\int _ { 0 } ^ { 1 } \frac { 5 + x } { \sqrt { 4 - 3 x ^ { 2 } } } d x$$ giving your answer in the form \(p \pi \sqrt { 3 } + q\), where \(p\) and \(q\) are rational numbers to be found.

3. Using the substitution \(\mathrm { x } = \frac { 3 } { \sinh \theta }\), or otherwise, find the exact value of $$\int _ { 4 } ^ { 3 \sqrt { } 3 } \frac { 1 } { x \sqrt { } \left( x ^ { 2 } + 9 \right) } d x$$ giving your answer in the form a ln b , where a and b are rational numbers.
(Total 8 marks)

5. A particle moves along the \(x\)-axis. At time \(t\) seconds, \(t \geqslant 0\), the velocity of the particle is \(v \mathrm {~ms} ^ { - 1 }\) in the direction of \(x\) increasing, where \(v = 2 t ^ { \frac { 3 } { 2 } } - 6 t + 2\) At time \(t = 0\) the particle passes through the origin \(O\). At the instant when the acceleration of the particle is zero, the particle is at the point \(A\). Find the distance \(O A\).
(8)

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

Solutions relying entirely on calculator technology are not acceptable.

1. Determine $$\int \frac { 1 } { \sqrt { 5 + 4 x - x ^ { 2 } } } d x$$
2. Use the substitution \(x = 3 \sec \theta\) to determine the exact value of $$\int _ { 2 \sqrt { 3 } } ^ { 6 } \frac { 18 } { \left( x ^ { 2 } - 9 \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x$$ Give your answer in the form \(A + B \sqrt { 3 }\) where \(A\) and \(B\) are constants to be found.

7. $$I _ { n } = \int \frac { x ^ { n } } { \sqrt { 10 - x ^ { 2 } } } \mathrm {~d} x \quad n \in \mathbb { N } \quad | x | < \sqrt { 10 }$$

1. Show that $$n I _ { n } = 10 ( n - 1 ) I _ { n - 2 } - x ^ { n - 1 } \left( 10 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \quad n \geqslant 2$$
2. Hence find the exact value of $$\int _ { 0 } ^ { 1 } \frac { x ^ { 5 } } { \sqrt { 10 - x ^ { 2 } } } \mathrm {~d} x$$ giving your answer in the form \(\frac { 1 } { 15 } ( p \sqrt { 10 } + q )\) where \(p\) and \(q\) are integers to be determined.

6. $$I _ { n } = \int _ { 0 } ^ { \sqrt { \frac { \pi } { 2 } } } x ^ { n } \cos \left( x ^ { 2 } \right) \mathrm { d } x \quad n \geqslant 1$$

1. Prove that, for \(n \geqslant 5\) $$I _ { n } = \frac { 1 } { 2 } \left( \frac { \pi } { 2 } \right) ^ { \frac { n - 1 } { 2 } } - \frac { 1 } { 4 } ( n - 1 ) ( n - 3 ) I _ { n - 4 }$$
2. Hence, determine the exact value of \(I _ { 5 }\), giving your answer in its simplest form.

8. $$y = \arccos ( 2 \sqrt { x } )$$

1. Determine \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
2. Show that $$\int y \mathrm {~d} x = x \arccos ( 2 \sqrt { x } ) + \int \frac { \sqrt { x } } { \sqrt { 1 - 4 x } } \mathrm {~d} x$$
3. Use the substitution \(\sqrt { x } = \frac { 1 } { 2 } \cos \theta\) to show that $$\int _ { 0 } ^ { \frac { 1 } { 8 } } \frac { \sqrt { x } } { \sqrt { 1 - 4 x } } \mathrm {~d} x = \frac { 1 } { 4 } \int _ { a } ^ { b } \cos ^ { 2 } \theta \mathrm {~d} \theta$$ where \(a\) and \(b\) are limits to be determined.
4. Hence, determine the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 8 } } \arccos ( 2 \sqrt { x } ) d x$$

5. $$I _ { n } = \int _ { 0 } ^ { 5 } \frac { x ^ { n } } { \sqrt { } \left( 25 - x ^ { 2 } \right) } d x , \quad n \geqslant 0$$

1. Find an expression for \(\int \frac { x } { \sqrt { } \left( 25 - x ^ { 2 } \right) } \mathrm { d } x , \quad 0 \leqslant x \leqslant 5\).
2. Using your answer to part (a), or otherwise, show that $$I _ { n } = \frac { 25 ( n - 1 ) } { n } I _ { n - 2 } \quad n \geqslant 2$$
3. Find \(I _ { 4 }\) in the form \(k \pi\), where \(k\) is a fraction.

1. A curve, which is part of an ellipse, has parametric equations

$$x = 3 \cos \theta , \quad y = 5 \sin \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 } .$$ The curve is rotated through \(2 \pi\) radians about the \(x\)-axis.

1. Show that the area of the surface generated is given by the integral $$k \pi \int _ { 0 } ^ { \alpha } \sqrt { } \left( 16 c ^ { 2 } + 9 \right) \mathrm { d } c , \quad \text { where } c = \cos \theta$$ and where \(k\) and \(\alpha\) are constants to be found.
2. Using the substitution \(c = \frac { 3 } { 4 } \sinh u\), or otherwise, evaluate the integral, showing all of your working and giving the final answer to 3 significant figures.

2. Use calculus to find the exact value of \(\int _ { - 2 } ^ { 1 } \frac { 1 } { x ^ { 2 } + 4 x + 13 } \mathrm {~d} x\).

7. $$\mathrm { f } ( x ) = 5 \cosh x - 4 \sinh x , \quad x \in \mathbb { R }$$

1. Show that \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + 9 \mathrm { e } ^ { - x } \right)\) Hence
2. solve \(\mathrm { f } ( x ) = 5\)
3. show that \(\int _ { \frac { 1 } { 2 } \ln 3 } ^ { \ln 3 } \frac { 1 } { 5 \cosh x - 4 \sinh x } \mathrm {~d} x = \frac { \pi } { 18 }\)

8. \begin{figure}[h] \end{figure} The curve \(C\), shown in Figure 2, has equation $$y = 2 x ^ { \frac { 1 } { 2 } } , \quad 1 \leqslant x \leqslant 8$$

1. Show that the length \(s\) of curve \(C\) is given by the equation $$s = \int _ { 1 } ^ { 8 } \sqrt { } \left( 1 + \frac { 1 } { x } \right) \mathrm { d } x$$
2. Using the substitution \(x = \sinh ^ { 2 } u\), or otherwise, find an exact value for \(s\). Give your answer in the form \(a \sqrt { } 2 + \ln ( b + c \sqrt { } 2 )\) where \(a , b\) and \(c\) are integers.

7. The curve \(C\) has equation $$y = \mathrm { e } ^ { - x } , \quad x \in \mathbb { R }$$ The part of the curve \(C\) between \(x = 0\) and \(x = \ln 3\) is rotated through \(2 \pi\) radians about the \(x\)-axis.

1. Show that the area \(S\) of the curved surface generated is given by $$S = 2 \pi \int _ { 0 } ^ { \ln 3 } \mathrm { e } ^ { - x } \sqrt { 1 + \mathrm { e } ^ { - 2 x } } \mathrm {~d} x$$
2. Use the substitution \(\mathrm { e } ^ { - x } = \sinh u\) to show that $$S = 2 \pi \int _ { \operatorname { arsinh } \alpha } ^ { \operatorname { arsinh } \beta } \cosh ^ { 2 } u \mathrm {~d} u$$ where \(\alpha\) and \(\beta\) are constants to be determined.
3. Show that $$2 \int \cosh ^ { 2 } u \mathrm {~d} u = \frac { 1 } { 2 } \sinh 2 u + u + k$$ where \(k\) is an arbitrary constant.
4. Hence find the value of \(S\), giving your answer to 3 decimal places.

4. (i) Find, without using a calculator, $$\int _ { 3 } ^ { 5 } \frac { 1 } { \sqrt { 15 + 2 x - x ^ { 2 } } } d x$$ giving your answer as a multiple of \(\pi\).
(ii)

1. Show that $$5 \cosh x - 4 \sinh x = \frac { \mathrm { e } ^ { 2 x } + 9 } { 2 \mathrm { e } ^ { x } }$$
2. Hence, using the substitution \(u = e ^ { x }\) or otherwise, find $$\int \frac { 1 } { 5 \cosh x - 4 \sinh x } d x$$

4. Use the substitution \(x + 2 = u ^ { 2 }\), where \(u > 0\), to show that $$\int _ { - 1 } ^ { 7 } \frac { ( x + 2 ) ^ { \frac { 1 } { 2 } } } { x + 5 } \mathrm {~d} x = a + b \pi \sqrt { 3 }$$ where \(a\) and \(b\) are rational numbers to be found. \includegraphics[max width=\textwidth, alt={}, center]{image-not-found}
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4. The curve \(C\) has equation $$y = \operatorname { arsinh } x + x \sqrt { x ^ { 2 } + 1 } , \quad 0 \leqslant x \leqslant 1$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \sqrt { x ^ { 2 } + 1 }\)
2. Hence show that the length of the curve \(C\) is given by $$\int _ { 0 } ^ { 1 } \sqrt { 4 x ^ { 2 } + 5 } d x$$
3. Using the substitution \(x = \frac { \sqrt { 5 } } { 2 } \sinh u\), find the exact length of the curve \(C\), giving your answer in the form \(a + b \ln c\), where \(a , b\) and \(c\) are constants to be found.

7 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 \(\int _ { 0 } ^ { 1 } 2 x \ln ( 1 + x ) \mathrm { d } x\).

5 Using the substitution \(u = 2 x + 1\), show that \(\int _ { 0 } ^ { 1 } \frac { x } { 2 x + 1 } \mathrm {~d} x = \frac { 1 } { 4 } ( 2 - \ln 3 )\).

1. Evaluate

$$\int _ { 2 } ^ { 15 } \frac { 1 } { \sqrt [ 3 ] { 2 x - 3 } } d x$$

7 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 $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( u ^ { \frac { 3 } { 2 } } - u ^ { \frac { 1 } { 2 } } \right) \mathrm { d } u$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.