Unit FP3: Further Pure Mathematics A2 3

Hyperbolic functions are combinations and ratios of $e^x$ and $e^{-x}$. The two basic functions are$$\sinh x=\frac{e^x-e^{-x}}2,\qquad \cosh x=\frac{e^x+e^{-x}}2.$$The other four are ratios or reciprocals of these, so their domains and asymptotes follow from where $\sinh x$ or $\cosh x$ can be zero.

The fundamental identity follows directly from the exponential definitions:$$\cosh^2x-\sinh^2x=\frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{4}=1.$$Dividing by $\cosh^2x$ gives $1-\tanh^2x=\operatorname{sech}^2x$; dividing by $\sinh^2x$ gives $\coth^2x-\operatorname{cosech}^2x=1$. Also, $\cosh^2x+\sinh^2x=\cosh2x$.

These addition formulae are proved by replacing every hyperbolic function with its exponential definition, expanding, and collecting $e^{A+B}$ with $e^{-(A+B)}$. They also allow a linear combination to be written in a shifted form. For example, $P\sinh x+Q\cosh x=R\sinh(x+\alpha)$ when $R\cosh\alpha=P$ and $R\sinh\alpha=Q$.

A universal route for $$a\cosh x+b\sinh x=c$$ is to set $u=e^x>0$. Multiplying the exponential form by $2u$ gives$$(a+b)u^2-2cu+(a-b)=0.$$Solve the quadratic, reject every root with $u\le0$, and use $x=\ln u$. For example, $5\cosh x+3\sinh x=7$ gives $4u^2-7u+1=0$, so$$x=\ln\frac{7+\sqrt{33}}8\quad\text{or}\quad x=\ln\frac{7-\sqrt{33}}8.$$

An inverse hyperbolic function reverses a one-to-one branch of a hyperbolic function. Its graph is the reflection of that restricted branch in $y=x$. The prefix $\operatorname{ar}$ means inverse function; it does not mean reciprocal.

To prove the official example, let $y=\operatorname{arsinh}x$, so $x=\sinh y$. With $u=e^y>0$,$$x=\frac{u-u^{-1}}2\quad\Longrightarrow\quad u^2-2xu-1=0.$$Thus $u=x\pm\sqrt{x^2+1}$. Since $\sqrt{x^2+1}>|x|$, only $u=x+\sqrt{x^2+1}$ is positive. Therefore$$y=\ln u=\ln\!\left(x+\sqrt{x^2+1}\right).$$

The same quadratic method gives the other forms. From $x=\cosh y$ with the principal restriction $y\ge0$, choose $e^y=x+\sqrt{x^2-1}$. From $x=\tanh y$, rearrangement gives $e^{2y}=(1+x)/(1-x)$, hence the factor $\tfrac12$ in $\operatorname{artanh}x$.

Branch information matters. If $\cosh y=x$ with $x\ge1$, then both$$y=\pm\operatorname{arcosh}x$$solve the unrestricted equation. A condition $y<0$ selects$$y=-\operatorname{arcosh}x=\ln\!\left(x-\sqrt{x^2-1}\right),$$because the two positive exponential roots are reciprocals.

Because $\sinh$ and $\tanh$ are odd, $\operatorname{arsinh}$ and $\operatorname{artanh}$ are odd. The principal $\operatorname{arcosh}$ graph begins at $(1,0)$ and is increasing. The logarithmic forms make exact equation solutions possible without treating inverse values as decimal approximations.

A standard ellipse or hyperbola is centred at the origin with its transverse or major axis on the $x$-axis. Assume $a>0$ and $b>0$. The signs in the Cartesian equation identify the curve: an ellipse has a sum equal to $1$, while a hyperbola has a difference equal to $1$. A parametric pair describes points on the curve using one parameter and is often the cleanest form for differentiation or moving-point problems.

Verify a parametrisation by substitution. For the ellipse,$$\frac{(a\cos t)^2}{a^2}+\frac{(b\sin t)^2}{b^2}=\cos^2t+\sin^2t=1.$$For the trigonometric hyperbola parametrisation, the corresponding identity is $\sec^2t-\tan^2t=1$; for the hyperbolic parametrisation it is $\cosh^2t-\sinh^2t=1$. This check also explains why the same parameter must appear in both coordinates.

To remove a parameter, first isolate functions of it and then use the matching identity. If $x=4\cos t$ and $y=3\sin t$, then $\cos t=x/4$ and $\sin t=y/3$, so$$\frac{x^2}{16}+\frac{y^2}{9}=1.$$The coordinate bounds $|x|\le4$ and $|y|\le3$ follow from the parametrisation as well as from the ellipse.

A Cartesian equation is usually better for intersections. Substitute the equation of the line into the conic to obtain a quadratic in one coordinate. Its two roots represent the two intersection coordinates; their sum and product can locate a midpoint or build a symmetric expression without solving for each point separately. Return to the line equation for the other coordinate, and check that every resulting point lies on both curves.

If a conic is translated, replace $x$ and $y$ by displacements from its centre. For example,$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$$has parametrisation $x=h+a\cos t$, $y=k+b\sin t$. Complete squares before reading a centre or semi-axis from a non-standard Cartesian equation.

For a fixed focus $S$ and directrix $\ell$, a conic is the locus of points $P$ satisfying$$PS=e\,d(P,\ell),$$where $e>0$ is the eccentricity and $d(P,\ell)$ is the perpendicular distance to the line. An ellipse has $0<e<1$ and a hyperbola has $e>1$. For the standard horizontal conics, the foci and directrices occur in symmetric pairs on the $x$-axis.

The focus is farther from the centre than the corresponding vertex for a hyperbola because $e>1$, but nearer for an ellipse because $e<1$. Thus the quantities $ae$ and $a/e$ switch their relative sizes: for an ellipse $ae<a<a/e$, whereas for a hyperbola $a/e<a<ae$. This is a useful geometric check on calculated coordinates and directrices.

The ellipse relation can be recovered from the locus definition. Use the right focus $S=(ae,0)$ and right directrix $x=a/e$. For $P=(x,y)$ on the appropriate side of that directrix,$$(x-ae)^2+y^2=e^2\left(x-\frac ae\right)^2.$$Expansion cancels the linear terms and gives$$(1-e^2)x^2+y^2=a^2(1-e^2).$$Dividing through and comparing with $x^2/a^2+y^2/b^2=1$ yields $b^2=a^2(1-e^2)$. The same distance principle, with $e>1$, gives the hyperbola relation.

Given any two compatible focus-directrix quantities, solve systematically. If a hyperbola has focus distance $ae=10$ and directrix distance $a/e=5/2$, then $a^2=(ae)(a/e)=25$, so $a=5$, $e=2$, and$$b^2=a^2(e^2-1)=75.$$Its standard equation is therefore $x^2/25-y^2/75=1$. The positive square root is used because eccentricity and semi-axis length are positive.

These displayed formulae assume a horizontal major or transverse axis. For a vertical conic, the roles of $x$ and $y$ swap: the foci lie at $(0,\pm ae)$ and the directrices are $y=\pm a/e$. Identify the axis from the equation before applying a memorised coordinate pattern.

A tangent has the curve's instantaneous gradient at the point of contact; a normal is perpendicular to it. There are two complementary routes. Use differentiation when the contact point is known or parametrised. Use a repeated-root condition when a line $y=mx+c$ is required to touch a conic at exactly one point.

For the ellipse $x^2/a^2+y^2/b^2=1$, implicit differentiation gives$$\frac{dy}{dx}=-\frac{b^2x}{a^2y}.$$At $(a\cos t,b\sin t)$ this is $-b\cos t/(a\sin t)$. Substitution in point-gradient form and simplification produces the tangent in the table. The normal gradient is the negative reciprocal, $a\sin t/(b\cos t)$, provided neither relevant gradient is vertical.

Parametric differentiation gives the same result without first eliminating $t$:$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}.$$For $x=a\sec t$, $y=b\tan t$, this gives $dy/dx=b\sec^2t/(a\sec t\tan t)$. Keep the contact point attached to the same parameter value when writing the line through it.

These conditions for $y=mx+c$ come from substituting the line into the Cartesian conic and setting the resulting quadratic discriminant to zero. For example, tangents of slope $3/4$ to $x^2/16+y^2/9=1$ satisfy$$c^2=16\left(\frac34\right)^2+9=18,$$so the two parallel tangents are $y=\tfrac34x\pm3\sqrt2$. A line through a specified external point supplies a second equation relating $m$ and $c$.

After finding a tangent, recover its contact point by solving the line and conic simultaneously. Because the intersection is a repeated root, the quadratic should have one repeated coordinate; substitution gives the other coordinate. This is also a check that the proposed line really touches rather than cuts the curve.

A locus records every position a moving point can occupy. In a coordinate problem, first express the moving point as $(x(t),y(t))$ from the geometry; then eliminate $t$ to obtain a Cartesian relation. The relation is not the whole answer until restrictions inherited from the parameter and construction have been checked.

Suppose a moving midpoint has coordinates$$x=2+3\cos t,\qquad y=-1+2\sin t.$$Then $\cos t=(x-2)/3$ and $\sin t=(y+1)/2$. Using $\cos^2t+\sin^2t=1$ gives the locus$$\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1.$$If $t$ ranges through a full $2\pi$ interval, the whole ellipse is traced; a smaller interval would select only part of it.

When coordinates are rational functions of a parameter, avoid solving unnecessarily complicated equations. If$$x=\frac{6u}{u^2+1},\qquad y=\frac{3(1-u^2)}{u^2+1},$$then direct calculation gives $x^2/9+y^2/9=1$. The missing or repeated points must still be checked from the original expressions; algebraic elimination alone does not record how the curve is traced.

Points created from intersections, midpoints or perpendicular lines should be built in layers. Find the defining lines or intersection coordinates first; use root sums for a midpoint when two intersections are roots of one quadratic; then eliminate the moving parameter. This keeps each geometric condition visible and prevents an unexplained leap to the final equation.

After elimination, complete squares and normalise to recognise the curve. For example,$$4x^2+9y^2+18y=27$$becomes$$\frac{x^2}{9}+\frac{(y+1)^2}{4}=1.$$A correct locus equation should be tested by substituting the original parametric coordinates back into it identically.

Hyperbolic derivatives follow from the exponential definitions and then obey the usual chain, product, quotient and reciprocal rules. The key pair mirrors differentiation of $e^x$:$$\frac{d}{dx}\sinh x=\cosh x,\qquad \frac{d}{dx}\cosh x=\sinh x.$$Unlike trigonometric differentiation, differentiating $\cosh x$ does not introduce a minus sign.

For example, the quotient rule and $\cosh^2u-\sinh^2u=1$ give$$\frac{d}{du}\tanh u=\frac{\cosh^2u-\sinh^2u}{\cosh^2u}=\operatorname{sech}^2u.$$The reciprocal rule then gives the negative sign in $d(\operatorname{sech}u)/du=-\operatorname{sech}u\tanh u$. These derivations are safer than importing trigonometric signs by memory.

Differentiate the outside function, keep its argument, then multiply by the derivative of that argument. Thus$$\frac{d}{dx}\tanh(5x-1)=5\operatorname{sech}^2(5x-1),$$and$$\frac{d}{dx}\operatorname{sech}(x^2)=-2x\operatorname{sech}(x^2)\tanh(x^2).$$Every nested layer contributes a factor.

For a product such as $y=x\sinh^2x$, keep both product-rule terms and apply a second chain rule to the square:$$\frac{dy}{dx}=\sinh^2x+2x\sinh x\cosh x=\sinh^2x+x\sinh2x.$$The identity $2\sinh x\cosh x=\sinh2x$ is an optional final simplification, not a replacement for the differentiation steps.

Write powers explicitly before differentiating a quotient. If$$y=\frac{\cosh2x}{\sqrt{x+1}}=\cosh(2x)(x+1)^{-1/2},\qquad x>-1,$$then$$\frac{dy}{dx}=2\sinh(2x)(x+1)^{-1/2}-\frac12\cosh(2x)(x+1)^{-3/2}.$$A common denominator may be formed only after both terms are correct.

If $y=f^{-1}(x)$ on a one-to-one branch of $f$, then $x=f(y)$ and implicit differentiation gives$$\frac{dy}{dx}=\frac{1}{f'(y)}.$$Rewrite the result in terms of $x$ using the defining identity and the principal range. The branch and domain are part of the derivative formula, especially when a square root appears.

To derive the inverse-sinh result, let $y=\operatorname{arsinh}x$, so $x=\sinh y$. Then$$1=\cosh y\frac{dy}{dx}\quad\Longrightarrow\quad\frac{dy}{dx}=\frac1{\cosh y}.$$Because $\cosh y>0$ and $\cosh^2y=1+\sinh^2y$, this becomes$$\frac{d}{dx}\operatorname{arsinh}x=\frac1{\sqrt{1+x^2}}.$$The positive square root is justified by the principal range, not chosen arbitrarily.

For a composite inverse function, differentiate the complete inner expression. For example, if $y=\operatorname{arcosh}(2x+3)$ and $x>-1$, then$$\frac{dy}{dx}=\frac{2}{\sqrt{(2x+3)^2-1}}.$$The stricter condition $2x+3>1$ ensures the derivative is finite; the function itself is defined at the endpoint where $2x+3=1$, but its derivative formula is singular there.

Useful simplification often appears only after applying the rule. For$$y=\arcsin x+x\sqrt{1-x^2},\qquad |x|<1,$$the product and chain rules give$$\frac{dy}{dx}=\frac1{\sqrt{1-x^2}}+\sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}=2\sqrt{1-x^2}.$$Keeping a common radical visible makes the cancellation reliable.

When simplifying radicals, use $\sqrt{x^2}=|x|$, not automatically $x$. For example,$$\frac{d}{dx}\operatorname{arsinh}\!\left(\sqrt{4x^2-1}\right)=\frac{2x}{|x|\sqrt{4x^2-1}},\qquad |x|>\frac12.$$This equals $2/\sqrt{4x^2-1}$ for $x>1/2$ but $-2/\sqrt{4x^2-1}$ for $x<-1/2$. A stated domain decides the sign.

After differentiating a combination, solve $f'(x)=0$ within the original domain. If clearing radicals requires squaring, the resulting equation is only a candidate equation: substitute each candidate back into the unsquared derivative equation and reject values outside the inverse-function or radical domain.

Hyperbolic antiderivatives reverse the derivative patterns from FP3.3. First recognise whether the integrand contains a function together with the derivative of its argument; then account for any constant chain factor. Always add $C$ to an indefinite integral.

Ratios lead to logarithms because $d(\ln|f|)/dx=f'/f$:$$\int\tanh x\,dx=\ln(\cosh x)+C,$$$$\int\coth x\,dx=\ln|\sinh x|+C.$$The absolute value is needed in the second result because $\sinh x$ may be negative; $\cosh x$ is always positive.

Two useful single-function forms follow by differentiating the inverse expression:$$\int\operatorname{sech}x\,dx=\arctan(\sinh x)+C,$$$$\int\operatorname{cosech}x\,dx=\ln\left|\tanh\frac{x}{2}\right|+C.$$These apply only on intervals where the integrand is defined.

For$$\int\left(3\cosh(3x)-2\operatorname{sech}^2(2x)\right)dx,$$each inner derivative is already present, so the result is$$\sinh(3x)-\tanh(2x)+C.$$Differentiate the result: $3\cosh(3x)-2\operatorname{sech}^2(2x)$ is recovered exactly.

Expressions involving powers may need an identity before integration. For example,$$\cosh^2x=\frac{1+\cosh2x}{2},\qquad \sinh^2x=\frac{\cosh2x-1}{2}.$$Thus $\int\cosh^2x\,dx=x/2+\sinh2x/4+C$. Choose an identity that turns the integrand into direct patterns rather than expanding exponentials unnecessarily.

An inverse trigonometric or inverse hyperbolic function is usually integrated by parts. Choose the inverse function as the part to differentiate and $1$ as the part to integrate:$$u=f^{-1}(x),\qquad dv=dx.$$Its derivative becomes an algebraic expression that can be simplified or integrated directly.

For example,$$\int\operatorname{arsinh}x\,dx=x\operatorname{arsinh}x-\int\frac{x}{\sqrt{1+x^2}}\,dx.$$With $w=1+x^2$, the remaining integral is $\sqrt{1+x^2}$, giving$$x\operatorname{arsinh}x-\sqrt{1+x^2}+C.$$Differentiating this result makes the two $x/\sqrt{1+x^2}$ terms cancel.

For a scaled argument, substitute first or apply integration by parts with the full argument. If $a>0$, then$$\int\arctan\left(\frac{x}{a}\right)dx=x\arctan\left(\frac{x}{a}\right)-\frac a2\ln(a^2+x^2)+C.$$Terms differing only by a constant inside a logarithm are absorbed into $C$.

For a definite integral, use one antiderivative consistently and retain exact principal inverse values. Check that the complete interval lies in the real domain: $[-1,1]$ for $\arcsin x$, $[1,\infty)$ for $\operatorname{arcosh}x$, and $(-1,1)$ for $\operatorname{artanh}x$.

A substitution is chosen so that a standard identity removes a quadratic expression. Assume $a>0$. Trigonometric substitutions are natural for $a^2-x^2$ and $a^2+x^2$; hyperbolic substitutions give equally direct forms for positive sums and differences. Transform the differential and limits as well as the integrand.

Use the same four moves each time: choose the identity-matching substitution; calculate $dx$; replace every occurrence of $x$; simplify the square root with the parameter range that fixes its sign. For a definite integral, convert both bounds immediately and evaluate in the new variable without back-substituting.

For $0<b<a$,$$\int_0^b\frac{dx}{(a^2-x^2)^{3/2}}$$uses $x=a\sin\theta$, $dx=a\cos\theta\,d\theta$, with $0\le\theta<\pi/2$. Since $(a^2-x^2)^{3/2}=a^3\cos^3\theta$, the integral becomes$$\frac1{a^2}\int_0^{\arcsin(b/a)}\sec^2\theta\,d\theta=\frac{b}{a^2\sqrt{a^2-b^2}}.$$The range justifies replacing $\sqrt{\cos^2\theta}$ by $\cos\theta$.

A hyperbolic substitution can be especially clean when $\sqrt{a^2+x^2}$ or $\sqrt{x^2-a^2}$ appears because the radical becomes $a\cosh t$ or $a\sinh t$. A trigonometric route may instead be preferable when exact angle values make definite limits simple. The valid method is the one whose identity matches the sign pattern and whose range is controlled.

For an integral involving $\sqrt{Ax^2+Bx+C}$, expose the quadratic's geometry before substituting. Complete the square, factor out any positive constant, and identify one of the patterns $a^2-u^2$, $a^2+u^2$ or $u^2-a^2$. In more complicated cases the required substitution may be supplied, but every algebraic change must still be shown.

Consider$$\int\frac{dx}{\sqrt{x^2+6x+13}}.$$Completing the square gives $(x+3)^2+4$. Put $x+3=2\sinh t$, so $dx=2\cosh t\,dt$ and the denominator is $2\cosh t$. The integral is $\int dt$, hence$$\operatorname{arsinh}\left(\frac{x+3}{2}\right)+C.$$Differentiation recovers the original integrand.

If the numerator contains the derivative $2Ax+B$ of the quadratic, try the direct substitution $u=Ax^2+Bx+C$ before a trigonometric or hyperbolic substitution. For example,$$\int\frac{2x+4}{\sqrt{x^2+4x+7}}\,dx=2\sqrt{x^2+4x+7}+C.$$This is shorter because the numerator already supplies $du$.

When a non-obvious substitution is given, derive its consequences rather than guessing the target. Calculate $dx$, rewrite the surd, state the parameter range needed for its sign, and simplify systematically. For definite integrals, translate endpoints before cancellation so that orientation and sign remain visible.

A reduction formula rewrites an integral indexed by $n$ in terms of the same family with a smaller index. It is derived from the definition of the indexed integral, usually by integration by parts and an identity. State the definition, restrictions and base cases before using the recurrence.

Let$$I_n=\int_0^{\pi/2}\sin^n x\,dx.$$For $n\ge2$, write $\sin^n x=\sin^{n-1}x\sin x$ and integrate by parts with $u=\sin^{n-1}x$, $dv=\sin x\,dx$. The boundary term $[-\sin^{n-1}x\cos x]_0^{\pi/2}$ is zero.

The remaining integral is$$(n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x\,dx=(n-1)(I_{n-2}-I_n).$$Therefore$$nI_n=(n-1)I_{n-2},\qquad n\ge2.$$Use $I_0=\pi/2$ for even powers and $I_1=1$ for odd powers. For instance, $I_4=(3/4)I_2=(3/4)(1/2)I_0=3\pi/16$.