Further calculus, polar coordinates & hyperbolic functions

The topic on one screen

• Volumes of revolution come from thin discs or shells. About the $x$-axis, $V=\pi\int y^2\,dx$; about the $y$-axis, $V=\pi\int x^2\,dy$. Worked example: rotating $y=\sqrt{x}$, $0\le x\le4$, about the $x$-axis gives $V=\pi\int_0^4x\,dx=8\pi$ (CP-5.1).
• An improper integral is a limit, not an ordinary substitution. Replace an infinite endpoint by $R$, or split at a singularity, integrate, then take the limit. For example, $\int_1^\infty x^{-2}\,dx=\lim_{R\to\infty}[-x^{-1}]_1^R=1$ (CP-5.2).
• The mean value of $f$ on $[a,b]$ is $\dfrac1{b-a}\int_a^bf(x)\,dx$. It is the height of a rectangle with the same signed area; for $\sin x$ on $[0,\pi]$ it is $2/\pi$ (CP-5.3).
• Partial fractions are useful only after checking that the fraction is proper. Distinct linear factors give simple terms; repeated factors need every power. Integrate logarithmic terms with their coefficient and use absolute values where required (CP-5.4).
• Know $\dfrac{d}{dx}\sin^{-1}x=(1-x^2)^{-1/2}$, $\dfrac{d}{dx}\cos^{-1}x=-(1-x^2)^{-1/2}$ and $\dfrac{d}{dx}\tan^{-1}x=(1+x^2)^{-1}$, including scaled forms (CP-5.5).
• Choose $x=a\sin\theta$ for $\sqrt{a^2-x^2}$ because $1-\sin^2\theta=\cos^2\theta$. Also $\int\dfrac{dx}{a^2-x^2}=\dfrac1{2a}\ln\left|\dfrac{a+x}{a-x}\right|+C$, obtained by partial fractions (CP-5.6).
• Polar and Cartesian coordinates satisfy $x=r\cos\theta$, $y=r\sin\theta$ and $r^2=x^2+y^2$. Negative $r$ places the point in the opposite direction, so a sketch should be built from symmetry, zeros, maxima and a small value table (CP-7.1 to CP-7.2).
• Polar area is $A=\dfrac12\int_{\alpha}^{\beta}r^2\,d\theta$. Find the correct tracing interval first. Worked example: $r=3\sin\theta$ is $x^2+(y-3/2)^2=(3/2)^2$ and is traced once for $0\le\theta\le\pi$, giving area $9\pi/4$ (CP-7.3).
• From exponentials, $\sinh x=\dfrac{e^x-e^{-x}}2$, $\cosh x=\dfrac{e^x+e^{-x}}2$ and $\tanh x=\sinh x/\cosh x$. All have domain $\mathbb R$; their ranges are $\mathbb R$, $[1,\infty)$ and $(-1,1)$ respectively. The $\sinh$ and $\tanh$ graphs are odd and increasing; $\cosh$ is even with minimum $1$; $\tanh$ has asymptotes $y=\pm1$ (CP-8.1).
• Differentiate as if hyperbolic functions were a paired system: $(\sinh x)'=\cosh x$, $(\cosh x)'=\sinh x$, and $(\tanh x)'=\operatorname{sech}^2x$. The sign difference from circular trig is deliberate (CP-8.2).
• Inverse notation reverses the restricted functions: $y=\operatorname{arsinh}x$ means $x=\sinh y$; similarly for $\operatorname{arcosh}$ and $\operatorname{artanh}$. Their domain-range pairs are $\mathbb R\to\mathbb R$, $[1,\infty)\to[0,\infty)$ and $(-1,1)\to\mathbb R$ (CP-8.3).
• Solve the exponential definitions to obtain $\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1})$, $\operatorname{arcosh}x=\ln(x+\sqrt{x^2-1})$ for $x\ge1$, and $\operatorname{artanh}x=\dfrac12\ln\left(\dfrac{1+x}{1-x}\right)$ for $|x|<1$ (CP-8.4).
• For $\sqrt{x^2+a^2}$ use $x=a\sinh u$; for $\sqrt{x^2-a^2}$ with $x\ge a$ use $x=a\cosh u$. Thus $\int dx/\sqrt{x^2-a^2}=\operatorname{arcosh}(x/a)+C=\ln(x+\sqrt{x^2-a^2})+C$ up to the constant $-\ln a$ (CP-8.5).
• Common errors: using $\pi\int y\,dx$ for a volume, evaluating an improper endpoint directly, missing the factor $1/2$ in polar area, treating negative $r$ as impossible, or copying circular-trig derivative signs into hyperbolic work.
• Exam technique: state the limiting process, tracing interval or substitution before calculating. Those setup lines are the method marks and make a calculator decimal check meaningful.

Where students actually lose marks

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Questions are written in the style of past Edexcel papers (source shown on each) — never copied from them.

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