Integration

1. Show that \[ \mathrm{sec}^2\left(\tfrac14\pi-\tfrac12 x\right)=\frac{2}{1+\sin x} \,. \] Hence integrate \(\dfrac{1}{1+\sin x}\) with respect to \(x\).
2. By means of the substitution \(y=\pi -x\), show that \[ \int_0^\pi x \f (\sin x)\, \d x = \frac \pi 2 \int_0^\pi \f(\sin x) \, \d x ,\] where \(\mathrm{f}\) is any function for which these integrals exist. Hence evaluate \[ \int_0^\pi \frac x {1+\sin x} \, \d x \,. \]
3. Evaluate \[ \int_0^\pi\frac{ 2x^3 -3\pi x^2}{(1+\sin x)^2}\, \d x .\]

1. Show that \(\int \ln (2-x) \d x = -(2-x)\ln (2-x) + (2-x) + c \,,\ \) where \(x<2\).
2. Sketch the curve \(A\) given by \(y= \ln \vert x^2-4\vert\).
3. Show that the area of the finite region enclosed by the positive \(x\)-axis, the \(y\)-axis and the curve \(A\) is \(4\ln(2+\sqrt3)-2\sqrt3\,\).
4. The curve \(B\) is given by \(y= \vert \ln \vert x^2-4\vert \vert\,\). Find the area between the curve \(B\) and the \(x\)-axis with \(| x| <2\). Note: you may assume that \(t \ln t \to 0\) as \(t\to 0\).

The numbers \(a\) and \(b\), where \(b > a\ge0\), are such that \[ \int_a^b x^2 \d x = \left ( \int_a^b x \d x\right)^{\!\!2}\,. \]

1. In the case \(a=0\) and \(b>0\), find the value of \(b\).
2. In the case \(a=1\), show that \(b\) satisfies \[ 3b^3 -b^2-7b -7 =0\,. \] Show further, with the help of a sketch, that there is only one (real) value of \(b\) that satisfies this equation and that it lies between \(2\) and \(3\).
3. Show that \(3p^2 + q^2 = 3p^2q\), where \(p=b+a\) and \(q=b-a\), and express \(p^2\) in terms of \(q\). Deduce that \(1< b-a\le\frac43\).

This question concerns the inequality \begin{equation} \int_0^\pi \bigl( f(x) \bigr)^2 \d x \le \int_0^\pi \bigl( f'(x)\bigr)^2 \d x\,.\tag{\(*\)} \end{equation}

1. Show that \((*)\) is satisfied in the case \(f(x)=\sin nx\), where \(n\) is a positive integer. Show by means of counterexamples that \((*)\) is not necessarily satisfied if either \(f(0) \ne 0\) or \(f(\pi)\ne0\).
2. You may now assume that \((*)\) is satisfied for any (differentiable) function \(f\) for which \(f(0)=f(\pi)=0\). By setting \(f(x) = ax^2 + bx +c\), where \(a\), \(b\) and \(c\) are suitably chosen, show that \(\pi^2\le 10\). By setting \(f(x) = p \sin \frac12 x + q\cos \frac12 x +r\), where \(p\), \(q\) and \(r\) are suitably chosen, obtain another inequality for \(\pi\). Which of these inequalities leads to a better estimate for \(\pi^2\,\)?

1. By using the substitution \(u=1/x\), show that for \(b>0\) \[ \int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1)} \d x =0 \,. \]
2. By using the substitution \(u=1/x\), show that for \(b>0\), \[ \int_{1/b}^b \frac{\arctan x}{x} \d x = \frac{\pi \ln b} 2\,. \]
3. By using the result \( \displaystyle \int_0^\infty \frac 1 {a^2+x^2} \d x = \frac {\pi}{2 a} \) (where \(a > 0\)),and a substitution of the form \(u=k/x\), for suitable \(k\), show that \[ \int_0^\infty \frac 1 {(a^2+x^2)^2} \d x = \frac {\pi}{4a^3 } \, \ \ \ \ \ \ (a > 0). \]

1. Let \[ I = \int_0^1 \bigl((y')^2 -y^2\bigr)\d x \qquad\text{and}\qquad I_1=\int_0^1 (y'+y\tan x)^2 \d x \,, \] where \(y\) is a given function of \(x\) satisfying \(y=0\) at \(x=1\). Show that \(I-I_1=0\) and deduce that \(I\ge0\). Show further that \(I=0\) only if \(y=0\) for all \(x\) (\(0\le x \le 1\)).
2. Let \[ J = \int_0^1 \bigl((y')^2 -a^2y^2\bigr)\d x \,, \] where \(a\) is a given positive constant and \(y\) is a given function of \(x\), not identically zero, satisfying \(y=0\) at \(x=1\). By considering an integral of the form \[ \int_0^1 (y'+ay\tan bx)^2 \d x \,, \] where \(b\) is suitably chosen, show that \(J\ge0\). You should state the range of values of \(a\), in the form \(a < k\), for which your proof is valid. In the case \(a=k\), find a function \(y\) (not everywhere zero) such that \(J=0\).

1. Show that, for \(n> 0\), \[ \int_0^{\frac14\pi} \tan^n x \,\sec^2 x \, \d x = \frac 1 {n+1} \; \text{ and } \int_0^{\frac14\pi} \!\! \sec ^n\! x \, \tan x \, \d x = \frac{(\sqrt 2)^n - 1}n \,. \]
2. Evaluate the following integrals: \[ \displaystyle \int_0^{\frac14\pi} \!\! x\, \sec ^4 \! x\, \tan x \, \d x \, \text{ and } \int_0^{\frac14\pi} \!\! x^2 \sec ^2 \! x\, \tan x \, \d x \,. \]

The function \(\f\) satisfies \(\f(x)>0\) for \(x\ge0\) and is strictly decreasing (which means that \(\f(b)<\f(a)\) for \(b>a\)).

1. For \(t\ge0\), let \(A_0(t)\) be the area of the largest rectangle with sides parallel to the coordinate axes that can fit in the region bounded by the curve \(y=\f(x)\), the \(y\)-axis and the line \(y=\f(t)\). Show that \(A_0(t)\) can be written in the form \[ A_0(t) =x_0\left( \f(x_0) -\f(t)\right), \] where \(x_0\) satisfies \(x_0 \f'(x_0) +\f(x_0) = \f(t)\,\).
2. The function g is defined, for \(t> 0\), by \[ \g(t) =\frac 1t \int_0^t \f(x) \d x\,. \] Show that \(t \g'(t) = \f(t) -\g(t)\,\). Making use of a sketch show that, for \(t>0\), \[ \int_0^t \left( \f(x) - \f(t)\right) \d x > A_0(t) \] and deduce that \(-t^2 \g'(t)> A_0(t)\).
3. In the case \(\f(x)= \dfrac 1 {1+x}\,\), use the above to establish the inequality \[ \ln \sqrt{1+t} > 1 - \frac 1 {\sqrt{1+t}} \,, \] for \(t>0\).

Given that \(t= \tan \frac12 x\), show that \(\dfrac {\d t}{\d x} = \frac12(1+t^2)\) and \( \sin x = \dfrac {2t}{1+t^2}\,\). Hence show that \[ \int_0^{\frac12\pi} \frac 1{1+a \sin x}\, \d x = \frac2 {\sqrt{1-a^2}} \arctan \frac{\sqrt{1-a}}{\sqrt{1+a}}\, \qquad \quad (0 < a < 1). \] Let \[ I_n = \int_0^{\frac12\pi} \frac{ \sin ^nx}{2+\sin x} \, \d x \qquad \quad (n\ge0). \] By considering \(I_{n+1}+2I_{n}\,\), or otherwise, evaluate \(I_3\).

1. Sketch the curve \(y=\sin x\) for \(0\le x \le \tfrac12 \pi\) and add to your diagram the tangent to the curve at the origin and the chord joining the origin to the point \((b, \sin b)\), where \(0 < b < \frac12\pi\). By considering areas, show that \[ 1-\tfrac12 b^2 <\cos b < 1-\tfrac 12 b \sin b\,. \]
2. By considering the curve \(y=a^x\), where \(a>1\), show that \[ \frac{2(a-1)}{a+1} < \ln a < -1 + \sqrt{2a-1\,}\,. \] [Hint: You may wish to write \(a^x\) as \(\e^{x\ln a}\).]

Show that \[ \int_0^{\frac14\pi} \sin (2x) \ln(\cos x)\, \d x = \frac14(\ln 2 -1)\,, \] and that \[ \int_0^{\frac14\pi} \cos (2x) \ln(\cos x)\, \d x = \frac18(\pi -\ln 4-2)\,. \] Hence evaluate \[ \int_{\frac14\pi}^{\frac12\pi} \big ( \cos(2x) + \sin (2x)\big) \, \ln \big( \cos x + \sin x\big)\, \d x\,. \]

Show that, for any function f (for which the integrals exist), \[ \int_0^\infty \f\big(x+\sqrt{1+x^2}\big) \,\d x = \frac12 \int_1^\infty \left(1+\frac 1 {t^2}\right) \f(t)\, \d t \,. \] Hence evaluate \[ \int_0^\infty \frac1 {2x^2 +1 + 2 x\sqrt{x^2+1} \ } \, \, \d x \,, \] and, using the substitution \(x=\tan\theta\), \[ \int_0^{\frac12\pi} \frac{1}{(1+\sin\theta)^3}\,\d \theta \,. \]

Given that \(0 < k < 1\), show with the help of a sketch that the equation \[ \sin x = k x \tag{\(*\)}\] has a unique solution in the range \(0 < x < \pi\). Let \[ I= \int_0^\pi \big\vert \sin x -kx\big\vert \, \d x\,. \] Show that \[ I= \frac{\pi^2 \sin\alpha }{2\alpha} -2\cos\alpha - \alpha \sin\alpha\,, \] where \(\alpha\) is the unique solution of \((*)\). Show that \(I\), regarded as a function of \(\alpha\), has a unique stationary value and that this stationary value is a minimum. Deduce that the smallest value of \(I\) is \[ -2 \cos \frac{\pi}{\sqrt2}\, .\]

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Note that the line \(y = x\) is the tangent \((0,0)\) and \(y = \sin x \) is always below it. For any other line through the origin with gradient \(0 < k < 1\) it must start below \(y = \sin x\), but finish above it at \(x = \pi\). It also can only cross once due the the convexity of \(\sin\) in this interval. \begin{align*} && I &= \int_0^\pi | \sin x -kx | \d x \\ &&&= \int_0^\alpha (\sin x -k x) \d x + \int_{\alpha}^\pi (kx - \sin x) \d x \\ &&&= \left [ -\cos x - \frac{kx^2}{2} \right]_0^{\alpha} + \left [ \cos x +\frac{kx^2}{2} \right]_{\alpha}^\pi \\ &&&= -\cos \alpha - \frac{k \alpha^2}{2} +1+(-1)+\frac{k\pi^2}{2} - \cos \alpha - \frac{k\alpha^2}{2} \\ &&&= -2\cos \alpha - k\left (\alpha^2 - \frac{\pi^2}{2} \right) \\ &&&= -2\cos \alpha - \frac{\sin \alpha}{\alpha}\left (\alpha^2 - \frac{\pi^2}{2} \right) \\ &&&= \frac{\pi^2 \sin \alpha}{\alpha} - \alpha \sin \alpha - 2\cos \alpha \end{align*} \begin{align*} && \frac{\d I}{\d \alpha} &= \frac{\pi^2(\alpha \cos \alpha - \sin \alpha)}{2\alpha^2} + 2 \sin \alpha - \sin \alpha - \alpha \cos \alpha \\ &&&= \frac{-2\alpha^3 \cos \alpha + 2\alpha \sin\alpha + \pi^2 \alpha \cos \alpha - \pi^2 \sin \alpha}{2\alpha^2} \\ \\ &&&= \left ( \alpha \cos \alpha - \sin \alpha\right) \left ( \frac{\pi^2}{2\alpha^2}-1 \right)\ \end{align*} Therefore \(I' = 0\) if \(\tan \alpha = \alpha\) or \(\alpha = \frac{\pi}{\sqrt{2}}\). Since \(\tan \alpha = \alpha\) only at \(\alpha = 0\) (between \(0 \leq \alpha < \pi\) (by considering the tangent), we must have a unique turning point when \(\alpha = \frac{\pi}{\sqrt{2}}\). Note that \(I(\frac{\pi}{\sqrt{2}}) = \frac{\pi^2 \sqrt{2} \sin \alpha}{2\pi} - \frac{\pi}{\sqrt{2}} \sin \alpha - 2\cos \frac{\pi}{\sqrt{2}}=- 2\cos \frac{\pi}{\sqrt{2}}\). Notice that \(I(0) = \frac{\pi^2}2 - 2 > 2\) and \(I(\pi) = 2\), but \(-2\cos \frac{\pi}{\sqrt{2}} < 2\) so we must be a at a minimum

For any given function \(\f\), let \[ I = \int [\f'(x)]^2 \,[\f(x)]^n \d x\,, \tag{\(*\)} \] where \(n\) is a positive integer. Show that, if \(\f(x)\) satisfies \(\f''(x) =k \f(x)\f'(x)\) for some constant \(k\), then (\(*\)) can be integrated to obtain an expression for \(I\) in terms of \(\f(x)\), \(\f'(x)\), \(k\) and \(n\).

1. Verify your result in the case \(\f(x) = \tan x\,\). Hence find \[ \displaystyle \int \frac{\sin^4x}{\cos^{8}x} \, \d x\;. \]
2. Find \[ \displaystyle \int \sec^2x\, (\sec x + \tan x)^6\,\d x\;. \]

Use the substitution \(x=\dfrac{1}{t^{2}-1}\; \), where \(t>1\), to show that, for \( x>0\), \[ \int \frac{1}{\sqrt{x\left(x+1\right) \; } \ }\; \d x =2 \ln \left(\sqrt x+ \sqrt{x +1} \; \right)+c \,. \] Note: You may use without proof the result \(\displaystyle \int \! \frac{1}{t^2-a^2} \, \d t = \frac{1}{2a} \ln \left| \frac{t-a}{t+a}\right| + \rm {constant}\). The section of the curve \[ y=\dfrac{1}{\sqrt{x}\; }-\dfrac{1}{\sqrt{x+1}\; } \] between \(x=\frac{1}{8}\) and \(x=\frac{9}{16}\) is rotated through \(360^{o}\) about the \(x\)-axis. Show that the volume enclosed is \(2\pi \ln \tfrac{5}{4}\,\). \(\phantom{\dfrac AB}\)

Prove that \[ \cos 3x = 4 \cos^3 x - 3 \cos x \,. \] Find and prove a similar result for \(\sin 3x\) in terms of \(\sin x\).

1. Let \[ {\rm I}(\alpha) = \int_0^\alpha \big(7\sin x - 8 \sin^3 x\big) \d x\,. \] Show that \[ {\rm I}(\alpha) = -\tfrac 8 3 c^3 + c +\tfrac5 3\,, \] where \(c = \cos \alpha\). Write down one value of \(c\) for which \({\rm I}(\alpha) =0\).
2. Useless Eustace believes that \[ \int \sin^n x \, \d x =\dfrac {\sin^{n+1}x}{n+1}\, \] for \(n=1, \ 2, \ 3, \ldots\, \). Show that Eustace would obtain the correct value of \({\rm I}(\beta)\,\), where \(\cos \beta= -\frac16\). Find all values of \(\alpha\) for which he would obtain the correct value of \({\rm I}(\alpha)\).

1. Let \[ I=\int_0^a \frac {\f(x)}{\f(x)+\f(a-x)} \, \d x\,. \] Use a substitution to show that \[ I = \int_0^a \frac {\f(a-x)}{\f(x)+\f(a-x)} \, \d x\, \] and hence evaluate \(I\) in terms of \(a\). Use this result to evaluate the integrals \[ \int_0^1 \frac{\ln (x+1)}{\ln (2+x-x^2)}\, \d x \ \ \ \ \ \ \text{ and }\ \ \ \ \ \int_0^{\frac\pi 2} \frac{\sin x } {\sin(x+\frac \pi 4 )} \, \d x \,. \]
2. Evaluate \[ \int_{\frac12}^2 \frac {\sin x}{x \big(\sin x + \sin \frac 1 x\big)} \, \d x\,. \]

The curves \(C_1\) and \(C_2\) are defined by \[ y= \e^{-x} \quad (x>0) \quad \text{ and } \quad y= \e^{-x}\sin x \quad (x>0), \] respectively. Sketch roughly \(C_1\) and \(C_2\) on the same diagram. Let \(x_n\) denote the \(x\)-coordinate of the \(n\)th point of contact between the two curves, where \(0 < x_1 < x_2 < \cdots\), and let \(A_n\) denote the area of the region enclosed by the two curves between \(x_n\) and \(x_{n+1}\). Show that \[ A_n = \tfrac12(\e^{2\pi}-1) \e^{-(4n+1)\pi/2} \] and hence find \(\displaystyle \sum_{n=1}^\infty A_n\).

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The curves touch when \(\sin x = 1\) ie \(x = \frac{4n+1}{2} \pi\). therefore \begin{align*} && I &= \int e^{-x} \sin x \d x \\ &&&= -e^{-x} \sin x + \int e^{-x} \cos x \d x \\ &&&= -e^{-x} \sin x -e^{-x} \cos x - I \\ && I &= -\frac{e^{-x}}2 ( \sin x + \cos x)\\ \\ && A_n &= \int_{\frac{4n+3}{2}\pi}^{\frac{4n-1}{2}\pi} e^{-x} (1-\sin x) \d x \\ &&&= \left [ -e^{-x} - \frac{e^{-x}}2(\sin x + \cos x) \right]_{\frac{4n+1}{2}\pi}^{\frac{4n+5}{2}\pi} \\ &&&= -\frac12\exp \left ( -\frac{4n+1}{2}\pi\right)+ \frac12\exp \left ( -\frac{4n-3}{2}\pi\right) \\ &&&= \frac12 (e^{2 \pi}-1) \left ( -\frac{4n+1}{2}\pi\right) \\ \\ && \sum_{n=1}^\infty A_n &= \frac12(e^{2\pi}-1)e^{-\pi/2} \sum_{n=1}^\infty e^{-2n\pi} \\ &&&= \frac12(e^{2\pi}-1)e^{-\pi/2} \frac{e^{-2\pi}}{1-e^{-2\pi}} \\ &&&= \frac12e^{-\pi/2} \end{align*}

1. Show that, for \(m>0\,\), \[ \int_{1/m}^m \frac{x^2}{x+1} \, \d x = \frac{(m-1)^3(m+1)}{2m^2}+ \ln m\,. \]
2. Show by means of a substitution that \[ \int_{1/m}^m \frac1 {x^n(x+1)}\,\d x = \int_{1/m}^m \frac {u^{n-1}}{u+1}\,\d u \,. \]
3. Evaluate:
   • \(\bf (a)\) \(\displaystyle \int_{1/2}^2 \frac {x^5+3}{x^3(x+1)}\,\d x \;;\)
   • \(\bf (b)\) \(\displaystyle \int_1^2 \frac{x^5+x^3 +1}{x^3(x+1)}\, \d x\;. \)

Show that, for any integer \(m\), \[ \int_0^{2\pi} \e^x \cos mx \, \d x = \frac {1}{m^2+1}\big(\e^{2\pi}-1\big)\,. \]

1. Expand \(\cos(A+B) +\cos(A-B)\). Hence show that \[\displaystyle \int_0^{2\pi} \e^x \cos x \cos 6x \, \d x\, = \tfrac{19}{650}\big( \e^{2\pi}-1\big)\,. \]
2. Evaluate $\displaystyle \int_0^{2\pi} \e^x \sin 2x \sin 4x \cos x \, \d x\,$.

Expand and simplify \((\sqrt{x-1}+1)^2\,\).

1. Evaluate \[ \int_{5}^{10} \frac{ \sqrt{x+2\sqrt{x-1} \;} + \sqrt{x-2\sqrt{x-1} \;} } {\sqrt{x-1}} \,\d x\;. \]
2. Find the total area between the curve \[ y= \frac{\sqrt{x-2\sqrt{x-1}\;}}{\sqrt{x-1}\;} \] and the \(x\)-axis between the points \(x=\frac54\) and \(x=10\).
3. Evaluate \[ \int_{\frac54}^{10} \frac{ \sqrt{x+2\sqrt{x-1}\;} + \sqrt{x-2\sqrt{x+1}+2 \;} } {\sqrt{x^2-1} } \;\d x\;. \]

Let \(y= (x-a)^n \e^{bx} \sqrt{1+x^2}\,\), where \(n\) and \(a\) are constants and \(b\) is a non-zero constant. Show that \[ \frac{\d y}{\d x} = \frac{(x-a)^{n-1} \e^{bx} \q(x)}{\sqrt{1+x^2}}\,, \] where \(\q(x)\) is a cubic polynomial. Using this result, determine:

1. $\displaystyle \int \frac {(x-4)^{14} \e^{4x}(4x^3-1)} {\sqrt{1+x^2\;}} \, \d x\,;\(
2. \)\displaystyle \int \frac{(x-1)^{21}\e^{12x}(12x^4-x^2-11)} {\sqrt{1+x^2\;}}\,\d x\,;\(
3. \)\displaystyle \int \frac{(x-2)^{6}\e^{4x}(4x^4+x^3-2)} {\sqrt{1+x^2\;}}\,\d x\,.$

For any given (suitable) function \(\f\), the Laplace transform of \(\f\) is the function \(\F\) defined by \[ \F(s) = \int_0^\infty \e^{-st}\f(t)\d t \quad \quad \, (s>0) \,. \]

1. Show that the Laplace transform of \(\e^{-bt}\f(t)\), where \(b>0\), is \(\F(s+b)\).
2. Show that the Laplace transform of \(\f(at)\), where \(a>0\), is \(a^{-1}\F(\frac s a)\,\).
3. Show that the Laplace transform of \(\f'(t)\) is \(s\F(s) -\f(0)\,\).
4. In the case \(\f(t)=\sin t\), show that \(\F(s)= \dfrac 1 {s^2+1}\,\).

The function \(\f\) is defined by \[ \f(x) = \frac{\e^x-1}{\e-1}, \ \ \ \ \ x\ge0, \] and the function \(\g\) is the inverse function to \(\f\), so that \(\g(\f(x))=x\). Sketch \(\f(x)\) and \(\g(x)\) on the same axes. Verify, by evaluating each integral, that \[ \int_0^\frac12 \f(x) \,\d x + \int_0^k \g(x) \,\d x = \frac1 {2(\sqrt \e +1)}\,, \] where \(\displaystyle k= \frac 1{\sqrt\e+1}\), and explain this result by means of a diagram.

Evaluate the integrals \[\int_0^{\frac{1}{2}\pi} \frac{\sin 2x}{1+\sin^2x} \d x \text{ and } \int_0^{\frac{1}{2}\pi} \frac{\sin x}{1+\sin^2x} \d x\] Show, using the binomial expansion, that \((1+\sqrt2\,)^5<99\). Show also that \(\sqrt 2 > 1.4\). Deduce that \(2^{\sqrt2} > 1+ \sqrt2\,\). Use this result to determine which of the above integrals is greater.