Prove formula by induction

8 It is given that \(I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sec ^ { n } x \mathrm {~d} x\) where \(n\) is a positive integer.

1. By writing \(\sec ^ { n } x = \sec ^ { n - 2 } x \sec ^ { 2 } x\), or otherwise, show that $$( n - 1 ) I _ { n } = ( \sqrt { 2 } ) ^ { n - 2 } + ( n - 2 ) I _ { n - 2 } \text { for } n > 1 .$$
2. Show that \(I _ { 8 } = \frac { 96 } { 35 }\).
3. Prove by induction that \(I _ { 2 n }\) is rational for all values of \(n > 1\). \section*{END OF QUESTION PAPER}

7 Let $$I _ { n } = \int _ { 0 } ^ { 1 } t ^ { n } \mathrm { e } ^ { - t } \mathrm {~d} t$$ where \(n \geqslant 0\). Show that, for all \(n \geqslant 1\), $$I _ { n } = n I _ { n - 1 } - \mathrm { e } ^ { - 1 }$$ Hence prove by induction that, for all positive integers \(n\), $$I _ { n } < n ! .$$

5 Let $$I _ { n } = \int _ { 1 } ^ { \mathrm { e } } x ( \ln x ) ^ { n } \mathrm {~d} x$$ where \(n \geqslant 1\). Show that $$I _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { 2 } - \frac { 1 } { 2 } ( n + 1 ) I _ { n }$$ Hence prove by induction that, for all positive integers \(n , I _ { n }\) is of the form \(A _ { n } \mathrm { e } ^ { 2 } + B _ { n }\), where \(A _ { n }\) and \(B _ { n }\) are rational numbers.

5
7 \end{array} \right) , \quad \mathbf { y } _ { 2 } = \left( \begin{array} { r } 2
- 3
4 \end{array} \right) , \quad \mathbf { y } _ { 3 } = \left( \begin{array} { r } 5
51
55 \end{array} \right)
\mathbf { P } = \left( \begin{array} { r r r } 1 & - 4 & 3
0 & 2 & 5
0 & 0 & - 7 \end{array} \right) . \end{gathered}$$

1. Show that \(\mathbf { y } _ { 1 } , \mathbf { y } _ { 2 } , \mathbf { y } _ { 3 }\) are linearly dependent.
2. Find a basis for the linear space spanned by the vectors \(\mathbf { P y } _ { 1 } , \mathbf { P y } _ { 2 } , \mathbf { P y } _ { 3 }\). 4 Given that \(y = x \sin x\), find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) and \(\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }\), simplifying your results as far as possible, and show that $$\frac { \mathrm { d } ^ { 6 } y } { \mathrm {~d} x ^ { 6 } } = - x \sin x + 6 \cos x$$ Use induction to establish an expression for \(\frac { \mathrm { d } ^ { 2 n } y } { \mathrm {~d} x ^ { 2 n } }\), where \(n\) is a positive integer. 5 The integral \(I _ { n }\) is defined by $$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sec ^ { n } x \mathrm {~d} x$$ By considering \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \tan x \sec ^ { n } x \right)\), or otherwise, show that $$( n + 1 ) I _ { n + 2 } = 2 ^ { \frac { 1 } { 2 } n } + n I _ { n }$$ Find the value of \(I _ { 6 }\).

5 Let \(I _ { n }\) denote \(\int _ { 0 } ^ { \infty } x ^ { n } \mathrm { e } ^ { - 2 x } \mathrm {~d} x\). Show that \(I _ { n } = \frac { 1 } { 2 } n I _ { n - 1 }\), for \(n \geqslant 1\). Prove by mathematical induction that, for all positive integers \(n , I _ { n } = \frac { n ! } { 2 ^ { n + 1 } }\).

1. Given that \(k\) is a real non-zero constant and that

$$y = x ^ { 3 } \sin k x$$ use Leibnitz's theorem to show that $$\frac { \mathrm { d } ^ { 5 } y } { \mathrm {~d} x ^ { 5 } } = \left( k ^ { 2 } x ^ { 2 } + A \right) k ^ { 3 } x \cos k x + B \left( k ^ { 2 } x ^ { 2 } + C \right) k ^ { 2 } \sin k x$$ where \(A\), \(B\) and \(C\) are integers to be determined.

5. $$y = \mathrm { e } ^ { 3 x } \sin x$$

1. Use Leibnitz's theorem to show that $$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = 28 \mathrm { e } ^ { 3 x } \sin x + 96 \mathrm { e } ^ { 3 x } \cos x$$
2. Hence express \(\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }\) in the form $$\operatorname { Re } ^ { 3 \mathrm { x } } \sin ( \mathrm { x } + \alpha )$$ where \(R\) and \(\alpha\) are constants to be determined, \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\)

1. Given \(k\) is a constant and that

$$y = x ^ { 3 } \mathrm { e } ^ { k x }$$ use Leibnitz theorem to show that $$\frac { \mathrm { d } ^ { n } y } { \mathrm {~d} x ^ { n } } = k ^ { n - 3 } \mathrm { e } ^ { k x } \left( k ^ { 3 } x ^ { 3 } + 3 n k ^ { 2 } x ^ { 2 } + 3 n ( n - 1 ) k x + n ( n - 1 ) ( n - 2 ) \right)$$