Reduction Formulae

1. Let \(\displaystyle I_n = \int_0^{\beta} (\sec x + \tan x)^n \, dx\), where \(n\) is a non-negative integer and \(0 < \beta < \dfrac{\pi}{2}\). For \(n \geqslant 1\), show that \[ \tfrac{1}{2}(I_{n+1} + I_{n-1}) = \frac{1}{n}\bigl[(\sec\beta + \tan\beta)^n - 1\bigr]. \] Show also that \[ I_n < \frac{1}{n}\bigl[(\sec\beta + \tan\beta)^n - 1\bigr]. \]
2. Let \(\displaystyle J_n = \int_0^{\beta} (\sec x \cos\beta + \tan x)^n \, dx\), where \(n\) is a non-negative integer and \(0 < \beta < \dfrac{\pi}{2}\). For \(n \geqslant 1\), show that \[ J_n < \frac{1}{n}\bigl[(1 + \tan\beta)^n - \cos^n\beta\bigr]. \]

For non-negative integers \(a\) and \(b\), let \[ \mathrm{I}(a,b) = \int_0^{\frac{\pi}{2}} \cos^a x \cos bx \; \mathrm{d}x. \]

1. Show that for positive integers \(a\) and \(b\), \[ \mathrm{I}(a,b) = \frac{a}{a+b} \, \mathrm{I}(a-1, b-1). \]
2. Prove by induction on \(n\) that for non-negative integers \(n\) and \(m\), \[ \int_0^{\frac{\pi}{2}} \cos^n x \cos(n+2m+1)x \; \mathrm{d}x = (-1)^m \frac{2^n \, n! \, (2m)! \, (n+m)!}{m! \, (2n+2m+1)!}. \]

Let \[ \displaystyle I_n= \int_{-\infty}^\infty \frac 1 {(x^2+2ax+b)^n} \, \d x \] where \(a\) and \(b\) are constants with \(b > a^2\), and \(n\) is a positive integer.

1. By using the substitution \(x + a = \sqrt{b- a^2} \, \tan u\,\), or otherwise, show that \[ I_1 = \dfrac \pi {\sqrt{b-a^2}}\, . \]
2. Show that \(2n(b - a^2)\, I_{n+1} =(2n - 1) \, I_n\,\).
3. Hence prove by induction that \[ I_n =\frac{\pi}{2^{2n-2}( b - a^2)^{n-\frac12}} \, \binom {2n-2}{n-1} \]

1. Let \[ I_n= \int_0^\infty \frac 1 {(1+u^2)^n}\, \d u \,, \] where \(n\) is a positive integer. Show that \[ I_n - I_{n+1} = \frac 1 {2n} I_n \] and deduce that \[ I_{n+1} = \frac{(2n)!\, \pi}{2^{2n+1}(n!)^2} \,. \]
2. Let \[ J = \int_0^\infty \f\big( (x- x^{-1})^2\big ) \, \d x \,, \] where \(\f\) is any function for which the integral exists. Show that \[ J = \int_0^\infty x^{-2} \f\big( (x- x^{-1})^2\big) \, \d x \, = \frac12 \int_0^\infty (1 + x^{-2}) \f\big( (x- x^{-1})^2\big ) \, \d x \, = \int_0^\infty \f\big(u^2\big) \,\d u \,. \]
3. Hence evaluate \[ \int_0^\infty \frac {x^{2n-2}}{(x^4-x^2+1)^n} \, \d x \,, \] where \(n\) is a positive integer.

For \(n\ge 0\), let \[ I_n = \int_0^1 x^n(1-x)^n\d x\,. \]

1. For \(n\ge 1\), show by means of a substitution that \[ \int_0^1 x^{n-1}(1-x)^n\d x = \int_0^1 x^n(1-x)^{n-1}\d x\, \] and deduce that \[ 2 \int_0^1 x^{n-1}(1-x)^n\d x = I_{n-1}\,. \] Show also, for \(n\ge1\), that \[ I_n = \frac n {n+1} \int_0^1 x^{n-1} (1-x)^{n+1} \d x \] and hence that \(I_n = \dfrac{n}{2(2n+1)} I_{n-1}\,.\)
2. When \(n\) is a positive integer, show that \[ I_n = \frac{(n!)^2}{(2n+1)!}\,. \]
3. Use the substitution \(x= \sin^2 \theta\) to show that \(I_{\frac12}= \frac \pi 8\), and evaluate \(I_{\frac32}\).

Let \(m\) be a positive integer and let \(n\) be a non-negative integer.

1. Use the result \(\displaystyle \lim_{t\to\infty}\e^{-mt} t^n=0\) to show that \[ \lim_{x\to0} x^m (\ln x)^n =0\,. \] By writing \(x^x\) as \(\e^{x\ln x}\) show that \[ \lim _{x\to0} x^x=1\,. \]
2. Let \(\displaystyle I_{n} = \int_0^1 x^m (\ln x)^n \d x\,\). Show that \[ I_{n+1} = - \frac {n+1}{m+1} I_{n} \] and hence evaluate \(I_{n}\).
3. Show that \[ \int_0^1 x^x \d x = 1 -\left(\tfrac12\right)^2 +\left(\tfrac13\right)^3 -\left(\tfrac14\right)^4 + \cdots \,. \]

For \(n=1\), \(2\), \(3\), \(\ldots\,\), let \[ I_n = \int_0^1 {t^{n-1} \over \l t+1 \r^n} \, \mathrm{d} t \, . \] By considering the greatest value taken by \(\displaystyle {t \over t+1}\) for \(0 \le t \le 1\) show that \(I_{n+1} < {1 \over 2} I_{n}\,\). Show also that \(\; \displaystyle I_{n+1}= - \frac 1{\; n\, 2^n} + I_{n}\,\). Deduce that \(\; \displaystyle I_n < \frac1 {\; n \, 2^{n-1}}\,\). Prove that \[ \ln 2 = \sum_{r=1}^n {1 \over \; r\, 2^r} + I_{n+1} \] and hence show that \({2 \over 3} < \ln 2 < {17 \over 24}\,\).

Let \[ u_{n}=\int_{0}^{\frac{1}{2}\pi}\sin^{n}t\,\mathrm{d}t \] for each integer \(n\geqslant0\). By integrating \[ \int_{0}^{\frac{1}{2}\pi}\sin t\sin^{n-1}t\,\mathrm{d}t \] by parts, or otherwise, obtain a formula connecting \(u_{n}\) and \(u_{n-2}\) when \(n\geqslant2\) and deduce that \[ nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2} \] for all \(n\geqslant2\). Deduce that \[ nu_{n}u_{n-1}=\tfrac{1}{2}\pi. \] Sketch graphs of \(\sin^{n}t\) and \(\sin^{n-1}t\), for \(0\leqslant t\leqslant\frac{1}{2}\pi,\) on the same diagram and explain why \(0 < u_{n} < u_{n-1}.\) By using the result of the previous paragraph show that \[ nu_{n}^{2} < \tfrac{1}{2}\pi < nu_{n-1}^{2} \] for all \(n\geqslant1\). Hence show that \[ \left(\frac{n}{n+1}\right)\tfrac{1}{2}\pi < nu_{n}^{2} < \tfrac{1}{2}\pi \] and deduce that \(nu_{n}^{2}\rightarrow\tfrac{1}{2}\pi\) as \(n\rightarrow\infty\).

Show Solution

\begin{align*} && u_n &= \int_0^{\tfrac12 \pi} \sin^{n} t \, \d t \\ && &= \int_0^{\tfrac12 \pi} \sin t \sin^{n-1} t \, \d t \\ && &= \left [ -\cos t \sin^{n-1} t \right]_0^{\tfrac12 \pi} + \int_0^{\tfrac12 \pi} \cos t (n-1) \sin^{n-2} t \cos t \d t \\ && &= 0 + (n-1)\int_0^{\tfrac12 \pi} \cos^2 t \sin^{n-2} t \d t \\ && &= (n-1) \int_0^{\tfrac12 \pi}(1-\sin^2 t) \sin^{n-2} t \d t \\ && &= (n-1)u_{n-2} - (n-1)u_n \\ \Rightarrow && n u_n &= (n-1)u_{n-2} \\ \end{align*} Mutplying both sides by \(u_{n-1}\) we obtain \(nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2}\). Therefore \(nu_nu_{n-1}\) is constant, ie is equal to \(\displaystyle u_1u_0 = \int_0^{\tfrac12 \pi} \sin^{1} t \, \d t \int_0^{\tfrac12 \pi} \sin^{0} t \, \d t = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2}\)

+ - ↺

Since \(0 < \sin t < 1\) for \(t \in (0, \tfrac{\pi}{2})\) we must have \(0 < \sin^n t < \sin^{n-1} t\), in particular \(0 < u_n < u_{n-1}\) Therefore \begin{align*} && nu_{n}u_{n-1} &= \tfrac{1}{2}\pi \\ \Rightarrow && nu_n u_n &< \tfrac{1}{2}\pi \tag{\(u_n < u_{n-1}\)} \\ \Rightarrow && nu_{n-1} u_{n-1} &> \tfrac{1}{2}\pi \tag{\(u_n < u_{n-1}\)} \\ \Rightarrow && nu_n^2 &< \tfrac12 \pi < n u_{n-1}^2 \end{align*} However we also have \(\tfrac12 \pi < (n+1)u_n^2\) (by considering the next inequality), so \(\left ( \frac{n}{n+1}\right) \tfrac12 \pi < n u_n^2 < \tfrac12 \pi\) but since as \(n \to \infty\) the right hand bound is constant and the left hand bound tends to \(\tfrac12 \pi\) therefore \(n u_n^2 \to \tfrac12 \pi\)

If \[ \mathrm{I}_{n}=\int_{0}^{a}x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,\mathrm{d}x, \] show that \(\mathrm{I}_{0}=\pi a^{2}/8.\) Show that \((2n+4)\mathrm{I}_{n}=(2n+1)a\mathrm{I}_{n-1}\) and hence evaluate \(\mathrm{I}_{n}\).

Calculate \[ \int_{0}^{x}\mathrm{sech}\, t\,\mathrm{d}t. \] Find the reduction formula involving \(I_{n}\) and \(I_{n-2}\), where \[ I_{n}=\int_{0}^{x}\mathrm{sech}^{n}t\,\mathrm{d}t \] and, hence or otherwise, find \(I_{5}\) and \(I_{6}.\)

Given that \({\displaystyle I_{n}=\int_{0}^{\pi}\frac{x\sin^{2}(nx)}{\sin^{2}x}\,\mathrm{d}x,}\) where \(n\) is a positive integer, show that \(I_{n}-I_{n-1}=J_{n},\) where \[ J_{n}=\int_{0}^{\pi}\frac{x\sin(2n-1)x}{\sin x}\,\mathrm{d}x. \] Obtain also a reduction formula for \(J_{n}.\) The curve \(C\) is given by the cartesian equation \[ y=\dfrac{x\sin^{2}(nx)}{\sin^{2}x}, \] where \(n\) is a positive integer and \(0\leqslant x\leqslant\pi.\) Show that the area under the curve \(C\) is \(\frac{1}{2}n\pi^{2}.\)

Give a rough sketch of the function \(\tan^{k}\theta\) for \(0\leqslant\theta\leqslant\frac{1}{4}\pi\) in the two cases \(k=1\) and \(k\gg1\) (i.e. \(k\) is much greater than 1). Show that for any positive integer \(n\) \[ \int_{0}^{\frac{1}{4}\pi}\tan^{2n+1}\theta\,\mathrm{d}\theta=(-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right), \] and deduce that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m}=\tfrac{1}{2}\ln2. \] Show similarly that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m-1}=\frac{\pi}{4}. \]

Show Solution

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Let \(\displaystyle I_n = \int_0^{\pi/4} \tan^{n} \theta \, \d \theta\), then \begin{align*} I_0 &= \int_0^{\pi/4} \tan \theta \d \theta \\ &= \left [ -\ln \cos \theta \right]_0^{\pi/4} \\ &= -\ln \frac{1}{\sqrt{2}} - 0 \\ &= \frac12 \ln 2 \\ \\ \\ I_{2n+1} &= \int_0^{\pi/4} \tan^{2n+1} \theta \, \d \theta \\&= \int_0^{\pi/4} \tan^{2n-1} \theta \tan ^2 \theta \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta (\sec^2 \theta - 1) \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta \sec^2 \theta - \tan^{2n-1} \theta \, \d \theta \\ &= \left[ \frac{1}{2n} \tan^{2n} \theta \right]_0^{\pi/4} - I_{2n-1} \\ &= \frac1{2n} - I_{2n-1} \end{align*} Therefore we can see that \(\displaystyle I_{2n+1} = (-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right)\). As we can see as \(n \to \infty\), \(I_n \to 0\) Therefore \begin{align*} && 0 &= \tfrac{1}{2}\ln2+\lim_{n \to \infty} \sum_{m=1}^{n}\frac{(-1)^{m}}{2m} \\ \Rightarrow && \tfrac{1}{2}\ln2 &= \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m} \end{align*} \begin{align*} && I_{-1} &= \int_0^{\pi/4} 1 \d \theta \\ &&&= \frac{\pi}{4} \end{align*} Therefore \(\displaystyle I_{2n} = (-1)^n \left ( \frac{\pi}{4} + \sum_{m=1}^n \frac{(-1)^m}{2m-1} \right)\) and since \(I_{2m} \to 0\) the same result follows.

1. The integral \(I_{k}\) is defined by \[ I_{k}=\int_{0}^{\theta}\cos^{k}x\,\cos kx\,\mathrm{d}x. \] Prove that \(2kI_{k}=kI_{k-1}+\cos^{k}\theta\sin k\theta.\)
2. Prove that \[ 1+m\cos2\theta+\binom{m}{2}\cos4\theta+\cdots+\binom{m}{r}\cos2r\theta+\cdots+\cos2m\theta=2^{m}\cos^{m}\theta\cos m\theta. \]
3. Using the results of (i) and (ii), show that \[ m\frac{\sin2\theta}{2}+\binom{m}{2}\frac{\sin4\theta}{4}+\cdots+\binom{m}{r}\frac{\sin2r\theta}{2r}+\cdots+\frac{\sin2m\theta}{2m} \] is equal to \[ \cos\theta\sin\theta+\cos^{2}\theta\sin2\theta+\cdots+\frac{1}{r}2^{r-1}\cos^{r}\theta\sin r\theta+\cdots+\frac{1}{m}2^{m-1}\cos^{m}\theta\sin m\theta. \]

Let \({\displaystyle I_{m,n}=\int\cos^{m}x\sin nx\,\mathrm{d}x,}\) where \(m\) and \(n\) are non-negative integers. Prove that for \(m,n\geqslant1,\) \[ (m+n)I_{m,n}=-\cos^{m}x\cos nx+mI_{m-1,n-1}. \]

1. Show that \({\displaystyle \int_{0}^{\pi}\cos^{m}x\sin nx\,\mathrm{d}x=0}\) whenever \(m,n\) are both even or both odd.
2. Evaluate \({\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2}x\sin3x\,\mathrm{d}x.}\)