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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 4.79 min answer

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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 uu be the inner function. With u=g(x)u = g(x) and du=gβ€²(x) dxdu = g'(x)\,dx,

∫f(g(x)) gβ€²(x) dx=∫f(u) du. \int f(g(x))\, g'(x)\, dx = \int f(u)\, du.

For a definite integral, change the limits to values of uu rather than substituting back.

Integration by parts

Integration by parts reverses the product rule:

∫u dvdx dx=uvβˆ’βˆ«v dudx dx. \int u\, \frac{dv}{dx}\, dx = uv - \int v\, \frac{du}{dx}\, dx.

Choose uu to be the factor that becomes simpler when differentiated. A useful priority for picking uu is logarithms, then inverse trigonometric functions, then algebraic, then trigonometric, then exponential.

Partial fractions

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

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

and each piece integrates to a logarithm.

Volumes of revolution

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

V=Ο€βˆ«aby2 dx=Ο€βˆ«ab(f(x))2 dx. V = \pi \int_a^b y^2 \, dx = \pi \int_a^b \big(f(x)\big)^2 \, dx.

Rotation about the yy axis gives V=Ο€βˆ«cdx2 dyV = \pi\int_c^d x^2 \, dy.

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.

2023 TASC5 marksUse the substitution u = sqrt(x) to evaluate the integral from 4 to 9 of dx/(1 + sqrt(x)).
Show worked answer β†’

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 the integral from u = 2 to u = 3 of (2u)/(1 + u) du.

Rewrite the integrand by division: u/(1 + u) = 1 - 1/(1 + u). So the integral = 2 times the integral from 2 to 3 of [1 - 1/(1 + u)] du = 2[u - ln(1 + u)] evaluated from 2 to 3.

Evaluate: 2[(3 - ln 4) - (2 - ln 3)] = 2[1 - ln 4 + ln 3] = 2 - 2 ln(4/3). Markers reward changing the limits with the substitution and the algebraic division before integrating.

2022 TASC6 marksDetermine the integral of e^(-x) sin(2x) dx.
Show worked answer β†’

Use integration by parts twice (or the standard reduction). Let I = integral of e^(-x) sin(2x) dx.

First parts: take u = sin(2x), dv = e^(-x) dx, so I = -e^(-x) sin(2x) + 2 integral of e^(-x) cos(2x) dx.

Second parts on the remaining integral: take u = cos(2x), dv = e^(-x) dx, giving integral of e^(-x) cos(2x) dx = -e^(-x) cos(2x) - 2 I.

Substitute back: I = -e^(-x) sin(2x) + 2[-e^(-x) cos(2x) - 2 I] = -e^(-x) sin(2x) - 2 e^(-x) cos(2x) - 4 I.

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

2022 TASC5 marksThe area bounded by the two quadratic functions 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 β†’

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 times the integral from 0 to 4 of [(2x(4 - x))^2 - (x(4 - x))^2] dx = pi times the integral of 3[x(4 - x)]^2 dx.

Expand: [x(4 - x)]^2 = x^2(16 - 8x + x^2) = 16x^2 - 8x^3 + x^4. Integrate from 0 to 4: 16(64/3) - 8(64) + 1024/5 = 1024/3 - 512 + 1024/5 = 512/15.

So V = 3 pi (512/15) = 512 pi / 5. Markers reward the difference of squares of the outer and inner radii, correct expansion, and the final volume.