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

13 Let \(I _ { n } = \int _ { 1 } ^ { \mathrm { e } } ( \ln x ) ^ { n } \mathrm {~d} x\), where \(n\) is a positive integer.

1. By considering \(\frac { \mathrm { d } } { \mathrm { d } x } \left( x ( \ln x ) ^ { n } \right)\), or otherwise, show that \(I _ { n } = \mathrm { e } - n I _ { n - 1 }\).
2. Let \(J _ { n } = \frac { I _ { n } } { n ! }\). Prove by induction that $$\sum _ { r = 2 } ^ { n } \frac { ( - 1 ) ^ { r } } { r ! } = \frac { 1 } { \mathrm { e } } \left( 1 + ( - 1 ) ^ { n } J _ { n } \right)$$ for all positive integers \(n \geqslant 2\).