What is volumes of revolution?

When the region under y = f(x) between x = a and x = b is rotated about the x axis, the solid formed has volume $V = π _a^b y^2 \, dx = π _a^b (f(x))^2 \, dx.Rotation about theyaxis givesV = \pi\int_c^d x^2 \, dy$.

What are integration using trigonometric identities?

Powers of sin and cos are integrated by first reducing the power with an identity. For even powers use the double-angle forms sin^2θ = 1 - cos 2θ2 and cos^2θ = 1 + cos 2θ2; for odd powers peel off one factor and convert the rest with sin^2θ + cos^2θ = 1, then substitute.

TASC TCE 2023-style practice: Use the substitution u = sqrt(x) to evaluate _4^9 dx1 + sqrt(x).

Let u = sqrt(x), so x = u^2 and dx = 2u\,du. Change the limits: when x = 4, u = 2; when x = 9, u = 3. The integral becomes _2^3 2u1 + u\,du. Rewrite the integrand by division: u1 + u = 1 - 11 + u. So the integral is 2_2^3[1 - 11 + u]du = 2[u - ln(1 + u)]_2^3. Evaluate: 2[(3 - ln 4) - (2 - ln 3)] = 2[1 - ln 4 + ln 3] = 2 - 2ln43. Markers reward changing the limits with the substitution and the algebraic division before integrating.

TASC TCE 2022-style practice: Determine e^-xsin(2x)\,dx.

Use integration by parts twice. Let I = e^-xsin(2x)\,dx. First parts: take u = sin(2x), dv = e^-x\,dx, so I = -e^-xsin(2x) + 2 e^-xcos(2x)\,dx. Second parts on the remaining integral: take u = cos(2x), dv = e^-x\,dx, giving e^-xcos(2x)\,dx = -e^-xcos(2x) - 2I. Substitute back: I = -e^-xsin(2x) + 2[-e^-xcos(2x) - 2I] = -e^-xsin(2x) - 2e^-xcos(2x) - 4I. So 5I = -e^-x(sin(2x) + 2cos(2x)), giving I = -15e^-x(sin(2x) + 2cos(2x)) + C. Markers reward applying parts twice and solving the resulting equation for I.

§-Syllabus dot point

TASSpecialist MathematicsSyllabus dot point

Which techniques let us integrate functions that the basic rules cannot handle?

Integrate using substitution, integration by parts and partial fractions, and find volumes of revolution.

Integration by substitution, by parts and by partial fractions, plus volumes of revolution, with fully worked examples and pitfalls for TCE Mathematics Specialised.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Basic antiderivatives only cover a handful of forms. Specialised integration is about recognising structure and applying the matching technique.

Substitution

Substitution reverses the chain rule. If part of the integrand is the derivative of another part, let $u$ be the inner function. With $u = g(x)$ and $du = g'(x)\,dx$ ,

\int f(g(x))\, g'(x)\, dx = \int f(u)\, du.

For a definite integral, change the limits to values of

u

rather than substituting back.

Integration by parts

Integration by parts reverses the product rule:

\int u\, \frac{dv}{dx}\, dx = uv - \int v\, \frac{du}{dx}\, dx.

Choose

u

to be the factor that becomes simpler when differentiated. A useful priority for picking

u

is logarithms, then inverse trigonometric functions, then algebraic, then trigonometric, then exponential.

Integration by parts

Evaluate $\displaystyle \int x e^{x}\, dx$ .

Step 1: Choose $u$ and $\dfrac{dv}{dx}$

Integration by parts requires choosing which factor to call $u$ and which to call $\dfrac{dv}{dx}$ . We want $u$ to become simpler when differentiated. Here $u = x$ is algebraic and differentiates to $1$ , while $e^x$ is easy to integrate. This is the natural split:

u = x, \quad \frac{du}{dx} = 1, \qquad \frac{dv}{dx} = e^{x}, \quad v = e^{x}.

Step 2: Apply the integration by parts formula

The formula $\displaystyle\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$ gives:

\int x e^{x}\, dx = x e^{x} - \int e^{x}\cdot 1\, dx = x e^{x} - e^{x} + C = e^{x}(x - 1) + C.

Step 3: Verify by differentiating

Differentiating the result using the product rule confirms correctness: $\tfrac{d}{dx}\big(e^{x}(x-1)\big) = e^{x}(x-1) + e^{x} = x e^{x}$ .

Final answer: $\displaystyle\int x e^{x}\, dx = e^{x}(x - 1) + C$ .

Partial fractions

A proper rational function with a factorised denominator can be split into simpler fractions. For distinct linear factors,

\frac{P(x)}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b},

and each piece integrates to a logarithm.

Partial fractions

Evaluate $\displaystyle \int \frac{5x - 1}{(x - 1)(x + 2)}\, dx$ .

Step 1: Set up the partial fraction decomposition

Because the denominator has two distinct linear factors, we can write the integrand as a sum of two simpler fractions. We assume:

\frac{5x - 1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}.

Multiplying both sides by $(x-1)(x+2)$ clears the denominators:

5x - 1 = A(x + 2) + B(x - 1).

Step 2: Find $A$ and $B$ by substituting convenient values

Choosing values of $x$ that zero out one factor at a time is the quickest method. Set $x = 1$ : $5(1) - 1 = A(3)$ , so $A = \tfrac{4}{3}$ . Set $x = -2$ : $5(-2) - 1 = B(-3)$ , so $B = \tfrac{11}{3}$ .

Step 3: Integrate each partial fraction

Each piece integrates to a logarithm:

\int \frac{5x - 1}{(x-1)(x+2)}\, dx = \frac{4}{3}\ln|x - 1| + \frac{11}{3}\ln|x + 2| + C.

Final answer: $\dfrac{4}{3}\ln|x - 1| + \dfrac{11}{3}\ln|x + 2| + C$ .

Volumes of revolution

When the region under $y = f(x)$ between $x = a$ and $x = b$ is rotated about the $x$ axis, the solid formed has volume

V = \pi \int_a^b y^2 \, dx = \pi \int_a^b \big(f(x)\big)^2 \, dx.

Rotation about the

y

axis gives

V = \pi\int_c^d x^2 \, dy

Volume of revolution

The region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated about the $x$ -axis. Find the volume.

Step 1: Write down the volume formula

When a region is rotated about the $x$ -axis, each thin vertical strip sweeps out a disc of radius $y = f(x)$ and thickness $dx$ . Summing these discs gives:

V = \pi \int_a^b \big(f(x)\big)^2 \, dx.

Step 2: Substitute and simplify the integrand

With $f(x) = \sqrt{x}$ , squaring removes the square root:

V = \pi \int_0^4 \big(\sqrt{x}\big)^2 \, dx = \pi \int_0^4 x \, dx.

Step 3: Evaluate the definite integral

\pi \int_0^4 x \, dx = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi\cdot\frac{16}{2} = 8\pi.

Final answer: $V = 8\pi$ cubic units.

Partial fractions with repeated and quadratic factors

The split depends on the denominator's factor types. A repeated linear factor needs one term per power, and an irreducible quadratic factor needs a linear numerator:

\frac{P(x)}{(x - a)^2} = \frac{A}{x - a} + \frac{B}{(x - a)^2}, \qquad \frac{P(x)}{(x - a)(x^2 + c)} = \frac{A}{x - a} + \frac{Bx + C}{x^2 + c}.

Always reduce an improper fraction (numerator degree at least the denominator degree) by polynomial division first; partial fractions only apply to a proper fraction.

Integration using trigonometric identities

Powers of $\sin$ and $\cos$ are integrated by first reducing the power with an identity. For even powers use the double-angle forms $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}$ and $\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}$ ; for odd powers peel off one factor and convert the rest with $\sin^2\theta + \cos^2\theta = 1$ , then substitute.

Integrating an even power of cosine

Evaluate $\displaystyle\int_0^{\pi/2} \cos^2 x\,dx$ .

Step 1: Use the double-angle identity to reduce the power

$\cos^2 x$ cannot be integrated directly in its current form. The double-angle identity $\cos^2 x = \dfrac{1 + \cos 2x}{2}$ rewrites the integrand as a sum of an easy constant and a cosine with a doubled argument:

\int_0^{\pi/2}\cos^2 x\,dx = \frac{1}{2}\int_0^{\pi/2}(1 + \cos 2x)\,dx.

Step 2: Integrate term by term

Both terms are straightforward: $1$ integrates to $x$ , and $\cos 2x$ integrates to $\dfrac{\sin 2x}{2}$ :

= \frac{1}{2}\left[x + \frac{\sin 2x}{2}\right]_0^{\pi/2}.

Step 3: Evaluate at the limits

At $x = \dfrac{\pi}{2}$ : the bracket gives $\dfrac{\pi}{2} + \dfrac{\sin\pi}{2} = \dfrac{\pi}{2} + 0 = \dfrac{\pi}{2}$ . At $x = 0$ : the bracket gives $0 + 0 = 0$ . So the integral is $\dfrac{1}{2}\cdot\dfrac{\pi}{2} = \dfrac{\pi}{4}$ .

Final answer: $\displaystyle\int_0^{\pi/2} \cos^2 x\,dx = \dfrac{\pi}{4}$ .

Inverse-trigonometric standard forms

Two integrals appear constantly and should be recognised on sight:

\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C, \qquad \int \frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan\frac{x}{a} + C.

Spotting an $a^2 - x^2$ under a square root, or an $a^2 + x^2$ in a denominator, signals one of these forms (or a trigonometric substitution $x = a\sin\theta$ or $x = a\tan\theta$ ).

Why this matters

The integration toolkit (substitution, parts, partial fractions and the trigonometric and inverse-trigonometric standard forms) is the engine behind volumes of revolution and the solution of differential equations. Recognising which technique a given integrand calls for is the skill TASC tests most often, so practise classifying an integral before you start computing.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20235 marksUse the substitution

u = \sqrt{x}

to evaluate

\displaystyle\int_4^9 \dfrac{dx}{1 + \sqrt{x}}

Show worked answer →

Let $u = \sqrt{x}$ , so $x = u^2$ and $dx = 2u\,du$ . Change the limits: when $x = 4$ , $u = 2$ ; when $x = 9$ , $u = 3$ .

The integral becomes $\displaystyle\int_2^3 \dfrac{2u}{1 + u}\,du$ .

Rewrite the integrand by division: $\dfrac{u}{1 + u} = 1 - \dfrac{1}{1 + u}$ . So the integral is $2\displaystyle\int_2^3\left[1 - \dfrac{1}{1 + u}\right]du = 2\Big[u - \ln(1 + u)\Big]_2^3$ .

Evaluate: $2\big[(3 - \ln 4) - (2 - \ln 3)\big] = 2\big[1 - \ln 4 + \ln 3\big] = 2 - 2\ln\dfrac{4}{3}$ . Markers reward changing the limits with the substitution and the algebraic division before integrating.

TCE 20226 marksDetermine

\displaystyle\int e^{-x}\sin(2x)\,dx

Show worked answer →

Use integration by parts twice. Let $I = \displaystyle\int e^{-x}\sin(2x)\,dx$ .

First parts: take $u = \sin(2x)$ , $dv = e^{-x}\,dx$ , so $I = -e^{-x}\sin(2x) + 2\displaystyle\int e^{-x}\cos(2x)\,dx$ .

Second parts on the remaining integral: take $u = \cos(2x)$ , $dv = e^{-x}\,dx$ , giving $\displaystyle\int e^{-x}\cos(2x)\,dx = -e^{-x}\cos(2x) - 2I$ .

Substitute back: $I = -e^{-x}\sin(2x) + 2\big[-e^{-x}\cos(2x) - 2I\big] = -e^{-x}\sin(2x) - 2e^{-x}\cos(2x) - 4I$ .

So $5I = -e^{-x}\big(\sin(2x) + 2\cos(2x)\big)$ , giving $I = -\dfrac{1}{5}e^{-x}\big(\sin(2x) + 2\cos(2x)\big) + C$ . Markers reward applying parts twice and solving the resulting equation for $I$ .

TCE 20225 marksThe area bounded by the curves

y = x(4 - x)

and

y = 2x(4 - x)

is rotated completely about the

x

-axis. Find the volume of the solid of revolution formed.

Show worked answer →

Both curves pass through $x = 0$ and $x = 4$ , and for $0 < x < 4$ the outer curve is $y = 2x(4 - x)$ and the inner is $y = x(4 - x)$ .

The volume by the washer method is $V = \pi\displaystyle\int_0^4\Big[\big(2x(4 - x)\big)^2 - \big(x(4 - x)\big)^2\Big]dx = \pi\displaystyle\int_0^4 3\big[x(4 - x)\big]^2\,dx$ .

Expand: $\big[x(4 - x)\big]^2 = x^2(16 - 8x + x^2) = 16x^2 - 8x^3 + x^4$ . Integrate from $0$ to $4$ : $16\cdot\dfrac{64}{3} - 8\cdot 64 + \dfrac{1024}{5} = \dfrac{1024}{3} - 512 + \dfrac{1024}{5} = \dfrac{512}{15}$ .

So $V = 3\pi\cdot\dfrac{512}{15} = \dfrac{512\pi}{5}$ . Markers reward the difference of squares of the outer and inner radii, correct expansion, and the final volume.