Integration using inverse trig and hyperbolic functions

Let \(f(x) = \sqrt{x^2 + 1} - x\).

1. Using a binomial series, or otherwise, show that, for large \(|x|\), \(\sqrt{x^2 + 1} \approx |x| + \frac{1}{2|x|}\). Sketch the graph \(y = f(x)\).
2. Let \(g(x) = \tan^{-1} f(x)\) and, for \(x \neq 0\), let \(k(x) = \frac{1}{2}\tan^{-1}\frac{1}{x}\).
   1. Show that \(g(x) + g(-x) = \frac{1}{2}\pi\).
   2. Show that \(k(x) + k(-x) = 0\).
   3. Show that \(\tan k(x) = \tan g(x)\) for \(x > 0\).
   4. Sketch the graphs \(y = g(x)\) and \(y = k(x)\) on the same axes.
   5. Evaluate \(\int_0^1 k(x) \, dx\) and hence write down the value of \(\int_{-1}^0 g(x) \, dx\).

1. Show, by means of the substitution \(u=\cosh x\,\), that \[ \int \frac{\sinh x}{\cosh 2x} \d x = \frac 1{2\sqrt2} \ln \left\vert \frac{\sqrt2 \cosh x - 1}{\sqrt2 \cosh x + 1 } \right\vert + C \,.\]
2. Use a similar substitution to find an expression for \[ \int \frac{\cosh x}{\cosh 2x} \d x \,.\]
3. Using parts (i) and (ii) above, show that \[ \int_0^1 \frac 1{1+u^4} \d u = \frac{\pi + 2\ln(\sqrt2 +1)}{4\sqrt2}\,. \]

The following result applies to any function \(\f\) which is continuous, has positive gradient and satisfies \(\f(0)=0\,\): \[ ab\le \int_0^a\f(x)\, \d x + \int_0^b \f^{-1}(y)\, \d y\,, \tag{\(*\)}\] where \(\f^{-1}\) denotes the inverse function of \(\f\), and \(a\ge 0\) and \(b\ge 0\).

1. By considering the graph of \(y=\f(x)\), explain briefly why the inequality \((*)\) holds. In the case \(a>0\) and \(b>0\), state a condition on \(a\) and \(b\) under which equality holds.
2. By taking \(\f(x) = x^{p-1}\) in \((*)\), where \(p>1\), show that if \(\displaystyle \frac 1p + \frac 1q =1\) then \[ ab \le \frac{a^p}p + \frac{b^q}q\,. \] Verify that equality holds under the condition you stated above.
3. Show that, for \(0\le a \le \frac12 \pi\) and \(0\le b \le 1\), \[ ab \le b\arcsin b + \sqrt{1-b^2} \, - \cos a\,. \] Deduce that, for \(t\ge1\), \[ \arcsin (t^{-1}) \ge t - \sqrt{t^2-1}\,. \]

The definite integrals \(T\), \(U\), \(V\) and \(X\) are defined by \begin{align*} T&= \int_{\frac13}^{\frac12} \frac{{\rm artanh}\, t}t \,\d t\,, & U&= \int _{\ln 2 }^{\ln 3 } \frac{u}{2\sinh u}\, \d u \,, \\[3mm] V&= - \int_{\frac13}^{\frac12} \frac{\ln v}{1-v^2} \,\d v \,, & X&= \int _{\frac12\ln2}^{\frac12\ln3} \ln ({\coth x})\, \d x\,. \end{align*} Show, without evaluating any of them, that \(T\), \(U\), \(V\) and \(X\) are all equal.

In this question, \(a\) is a positive constant.

1. Express \(\cosh a\) in terms of exponentials. By using partial fractions, prove that \[ \int_0^1 \frac 1{ x^2 +2x\cosh a +1} \, \d x = \frac a {2\sinh a}\,. \]
2. Find, expressing your answers in terms of hyperbolic functions, \[ \int_1^\infty \frac 1 {x^2 +2x \sinh a -1} \,\d x \, \] and \[ \int_0^\infty \frac 1 {x^4 +2x^2\cosh a +1} \,\d x \,.\]

1. Show, with the aid of a sketch, that \(y> \tanh (y/2)\) for \(y>0\) and deduce that \begin{equation} \textrm{arcosh} x > \dfrac{x-1}{\sqrt{x^2-1}} \text{ for } x>1. \tag{\(*\)} \end{equation}
2. By integrating \((*)\), show that $\textrm{arcosh} x > 2 \dfrac{{x-1}}{\sqrt{x^2-1}} \( for \)x>1$.
3. Show that $\textrm{arcosh} x >3 \dfrac{\sqrt{x^2-1}}{{x+2}} \( for \)x>1$.

Show that if \(\displaystyle \int\frac1{u \, \f(u)}\; \d u = \F(u) + c\;\), then \(\displaystyle \int\frac{m}{x \, \f(x^m)} \;\d x = \F(x^m) + c\;\), where \(m\ne0\). Find:

1. \(\displaystyle\int\frac1{x^n-x} \, \d x\,\);
2. \(\displaystyle\int\frac1 {\sqrt{x^n+x^2}}\, \d x\,\).

Show that \[ \int_0^a \frac{\sinh x}{2\cosh^2 x -1} \, \mathrm{d} x = \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}\cosh a -1}{\sqrt{2}\cosh a +1}\r + \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \] and find \[ \int_0^a \frac{\cosh x}{1+2\sinh^2 x} \, \mathrm{d} x \, . \] Hence show that \[ \int_0^\infty \frac{\cosh x - \sinh x}{1+2\sinh^2 x} \, \mathrm{d} x = \frac{\pi}{2\sqrt{2}} - \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \, . \] By substituting \(u = \e^x\) in this result, or otherwise, find \[ \int_1^\infty \frac{1}{1+u^4} \, \mathrm{d} u \, . \]

Given that \(x+a>0\) and \(x+b>0\,\), and that \(b>a\,\), show that \[ \frac{\mathrm{d} \ }{\mathrm{d} x} \arcsin \left ( \frac{x + a }{ \ x + b} \right) = \frac{ \sqrt{\;b - a\;}} {( x + b ) \sqrt{ a + b + 2x} \ \ } \] and find $\displaystyle \frac{\mathrm{d} \ }{ \mathrm{d} x} \; \mathrm{arcosh} \left ( \frac{x + b }{ \ x + a} \right)$. Hence, or otherwise, integrate, for \(x > -1\,\),

1. \[ \int \frac{1}{ ( x + 1) \sqrt{x + 3} } \mathrm{d} x \]
2. \[ \int \frac{1} {( x + 3 ) \sqrt{x + 1} } \mathrm{d} x \] .

Show that \( \cosh^{-1} x = \ln ( x + \sqrt{x^2-1})\). Show that the area of the region defined by the inequalities \(\displaystyle y^2 \ge x^2-8\) and \(\displaystyle x^2\ge 25y^2 -16 \) is \((72/5) \ln 2\).

Show Solution

\begin{align*} && x &= \cosh y \\ \Rightarrow && x &= \tfrac12 (e^y + e^{-y} ) \\ \Rightarrow && 0 &= e^{2y} - 2xe^y + 1 \\ \Rightarrow && e^y &= \frac{2x \pm \sqrt{4x^2-4}}{2} \\ &&&= x \pm \sqrt{x^2-1} \\ \Rightarrow &&e^y &= x + \sqrt{x^2-1} \tag{by convention \(\cosh^{-1} > 0\)} \\ \Rightarrow && y &= \ln (x + \sqrt{x^2-1}) \end{align*}

+ - ↺

\begin{align*} && A &= 4 \left ( \int_0^3 \frac15\sqrt{16+x^2} \d x - \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x \right) \\ \\ x = 4\sinh u: && \int_0^3 \sqrt{4^2+x^2} \d x &= \int_{u=0}^{u=\sinh^{-1}(3/4)} \sqrt{4^2 (1+\sinh^2 u)} 4 \cosh u \d u \\ &&&= \int_0^{\sinh^{-1}(3/4)} 16 \cosh^{2}u \d u \\ &&&= 8\int_0^{\sinh^{-1}(3/4)} (1+\cosh 2u) \d u \\ &&&= 8 \left[u + \frac12 \sinh 2u\right]_0^{\sinh^{-1}(3/4)} \\ &&&= 8 \left (\sinh^{-1}(3/4) + \frac12 \sinh \left ( 2 \sinh^{-1}(3/4) \right) \right) \\ \\ && \sinh^{-1}(3/4) &= \ln\left ( \frac34 + \sqrt{\left ( \frac{3}{4} \right)^2 + 1} \right) \\ &&&= \ln \left ( \frac34 +\frac{5}{4} \right) \\ &&&= \ln 2 \\ \\ \Rightarrow && \int_0^3 \sqrt{4^2+x^2} \d x &= 8 \ln 2 + 4 \left ( \frac{e^{2 \ln 2} - e^{-2\ln2}}{2} \right) \\ &&&= 8 \ln 2 + 2 \cdot 4 - 2\cdot \frac{1}{4} \\ &&&= 8 \ln 2 + \frac{15}{2} \end{align*} \begin{align*} x = 2\sqrt{2} \cosh u: && \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x &= \int_{u=0}^{u = \cosh^{-1} \frac{3}{2\sqrt{2}}} \sqrt{8(\cosh^2 u - 1)} 2 \sqrt{2} \sinh u \d u \\ &&&= \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} 8\sinh^2 u \d u \\ &&&= 4 \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} 2\sinh^2 u \d u \\ &&&= 4 \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} \cosh 2 u -1 \d u \\ &&&= 4 \left [\frac12 \sinh 2u - u \right]_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} \\ \\ && \cosh^{-1} \frac{3}{2\sqrt{2}} &= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\left ( \frac{3}{2\sqrt{2}} \right)^2-1} \right) \\ &&&= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\frac{9}{8} - 1} \right) \\ &&&= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\frac{1}{8} } \right) \\ &&&= \ln \frac{4}{2\sqrt{2}} \\ &&&= \frac12 \ln 2 \\ \\ && \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x &= 4 \left ( \frac12 \frac{e^{\ln2} - e^{-\ln2}}{2} - \frac12 \ln 2\right) \\ &&&= 2 - \frac12 -2 \ln 2 \\ &&&= \frac32 - 2 \ln 2 \end{align*} \begin{align*} A &= 4 \left (\frac15\left(8\ln 2 + \frac{15}2 \right)- \left ( \frac32 - 2 \ln 2\right)\right) \\ &=4\cdot \left( \frac{8}{5} + 2 \right) \ln 2 \\ &= \frac{72}{5} \ln 2 \end{align*}

Calculate the following integrals

1. \({\displaystyle \int\frac{x}{(x-1)(x^{2}-1)}\,\mathrm{d}x}\);
2. \({\displaystyle \int\frac{1}{3\cos x+4\sin x}\,\mathrm{d}x}\);
3. \({\displaystyle \int\frac{1}{\sinh x}\,\mathrm{d}x}.\)