Further Calculus Skills

Further calculus skills was worth 8 marks on NESA's sample paper for the 2024 syllabus and about 7–16 marks in recent HSC papers. Substitution and the inverse trig integrals turn up almost every year, often inside a bigger question on areas, volumes or differential equations.

• Parametric derivatives: $\frac{dy}{dx}$ for a curve given by $x(t)$ and $y(t)$, and problems such as tangents.
• Inverse functions: the derivative of $f^{-1}$ is the reciprocal of $f^{\prime }$ at the matching point.
• Inverse trig derivatives: examine the proofs, use the chain rule forms, and combine with the product, quotient and chain rules.
• Inverse trig integrals: integrals of $\frac{1}{\sqrt{a^2-x^2}}$ and $\frac{a}{a^2+x^2}$.
• Substitution: definite and indefinite integrals, given the substitution $u=g(x)$ or $x=g(u)$.
• ${\sin }^2$ and ${\cos }^2$: prove the identities ${\sin }^{2}nx=\frac{1}{2}(1-\cos 2nx)$ and ${\cos }^{2}nx=\frac{1}{2}(1+\cos 2nx)$ and use them to integrate.

In this course the substitution is always given. Integration by parts and partial fractions in general are Extension 2 (partial fractions appear here only for the logistic equation in ME12-CA).

Parametric curves

If $x$ and $y$ are both functions of a parameter $t$, the chain rule gives

Worked example 1: tangent to a parametric curve

A curve has $x=3\cos t$, $y=2\sin t$. Find the gradient and the equation of the tangent at $t=\frac{\pi }{4}$.

At $t=\frac{\pi }{4}$, $\cot t=1$, so the gradient is $-\frac{2}{3}$. The point is $\left(\frac{3}{\sqrt{2}},\frac{2}{\sqrt{2}}\right)=\left(\frac{3\sqrt{2}}{2},\sqrt{2}\right)$. The tangent is

(This curve is the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$, but you did not need its Cartesian equation.)

Inverse functions

Differentiating $f(f^{-1}(x))=x$ with the chain rule gives

In words: find the point on $f$ that matches, then take the reciprocal of the gradient there. For example, $f(x)=e^{2x}+x$ has $f(0)=1$, so $f^{-1}(1)=0$. Since $f^{\prime }(x)=2e^{2x}+1$ and $f^{\prime }(0)=3$, ${\left(f^{-1}\right)}^{\prime }(1)=\frac{1}{3}$. You never need a formula for $f^{-1}$.

The syllabus asks you to examine the proofs. For $y={\sin }^{-1}x$: $x=\sin y$, so $\frac{dx}{dy}=\cos y$, and for $-1<x<1$, $y$ is in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ so $\cos y>0$, giving $\cos y=\sqrt{1-x^2}$. Taking the reciprocal gives the result below. The chain rule versions:

Note that $\frac{d}{dx}{\cos }^{-1}x=-\frac{d}{dx}{\sin }^{-1}x$, which matches ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ having derivative 0.

Worked example 2: product rule with an inverse trig function

Show that $\frac{d}{dx}\left(x{\sin }^{-1}x+\sqrt{1-x^2}\right)={\sin }^{-1}x$, and hence find ${\int }_0^{1/2}{\sin }^{-1}xdx$.

So the primitive of ${\sin }^{-1}x$ is $x{\sin }^{-1}x+\sqrt{1-x^2}$:

Reading the derivatives backwards gives two standard integrals ($a>0$):

The first needs $|x|<a$. If $x$ has a coefficient, take it out first or use a substitution: for example $\int \frac{dx}{\sqrt{4-9x^2}}=\frac{1}{3}{\sin }^{-1}\frac{3x}{2}+C$. With $a^2=3$, the second gives

A substitution changes an integral in $x$ into a simpler one in $u$. The steps:

1. Write the substitution and differentiate it to link $dx$ and $du$.
2. Replace every $x$ in the integrand, including any left over, by an expression in $u$.
3. For a definite integral, change the limits to $u$-values (then there is no need to go back to $x$).
4. Integrate and evaluate.

The substitution may be given as $u=g(x)$, or with $x$ as the subject, $x=g(u)$. In the second case, $dx=g^{\prime }(u)du$.

Worked example 3: substitution with a leftover x

Use the substitution $u=2x-1$ to evaluate ${\int }_1^5\frac{x}{\sqrt{2x-1}}dx$.

From $u=2x-1$: $du=2dx$ and $x=\frac{u+1}{2}$. When $x=1$, $u=1$; when $x=5$, $u=9$.

From $\cos 2A=1-2{\sin }^{2}A=2{\cos }^{2}A-1$, with $A=nx$:

Rewrite, then integrate term by term. For example:

Notice the angle doubles: ${\cos }^{2}2x$ becomes $\cos 4x$.

• Forgetting to change the limits in a substitution, or changing them and then substituting back to $x$ as well.
• Leaving an $x$ in the integrand after substituting.
• Missing the chain rule factor: $\frac{d}{dx}{\tan }^{-1}(3x)=\frac{3}{1+9x^2}$, not $\frac{1}{1+9x^2}$.
• Using $\frac{1}{\sqrt{a^2-x^2}}$ for an integrand like $\frac{1}{\sqrt{x^2-a^2}}$ or $\frac{1}{a^2-x^2}$, which are different.
• Writing ${\sin }^{2}x=\frac{1}{2}(1-\cos x)$ (the angle must double) or getting the signs of the two identities swapped.
• For $(f^{-1}{)}^{\prime }(a)$, evaluating $f^{\prime }(a)$ instead of $f^{\prime }$ at $f^{-1}(a)$.
• Using degrees: ${\sin }^{-1}$ and ${\tan }^{-1}$ in calculus are in radians.

• Check the reference sheet for the derivatives of the inverse trig functions and the standard integrals, but practise using them with a chain rule factor and a coefficient on $x$.
• Show that a derivative equals something: differentiate fully and simplify; then "hence" usually means integrate it, as in worked example 2.
• Write the substitution working in full ($du=\dots dx$, new limits). Markers award a mark for the correctly transformed integral even if the final arithmetic slips.
• Give exact answers (with $\pi$, surds and fractions) unless a decimal is requested.
• If a question asks for the gradient of a parametric curve at a point given in $(x,y)$ form, first find the value of $t$ at that point.

• Inverse trigonometric functions (ME12-IT): the definitions and ranges behind the derivatives.
• Further trigonometry (ME11-TR): the double angle formulas give the ${\sin }^2$ and ${\cos }^2$ identities.
• Further applications of calculus (ME12-CA): these integrals are used for volumes of revolution and solving differential equations.
• Introduction to vectors (ME12-VE): a path $\underset{\sim }{r}(t)$ is a parametric curve.
• Advanced, Integral Calculus (MA12-IC): the reverse chain rule is a substitution done mentally.