What is show the structural step for the mark?

Write the line that exposes the method: the peeled factor with the Pythagorean conversion (cos^3 x = cos x(1-sin^2 x)), the reserved sec^2 x or sec xtan x, or the product-to-sum line. That line is usually the first mark; jumping to a u-substitution without it can be marked down.

What is the wrong reserved factor for tan and sec?

Even power of sec reserves sec^2 x with u = tan x; odd power of tan reserves sec xtan x with u = sec x. Reserving the wrong one leaves a factor that is not the right du.

What is sign in the sin Asin B identity?

It is sin Asin B = 12[cos(A-B) - cos(A+B)], with a minus sign; the cosine product has a plus. Mixing them flips a sign through the whole answer.

NESA HSC Ext 2 (representative)-style practice: Find the exact value of _0^π/2 sin^3 x \, dx.

The power of sin is odd, so peel one factor of sin x and convert the rest with sin^2 x = 1 - cos^2 x: sin^3 x = sin x\,(1 - cos^2 x). Substitute u = cos x, so du = -sin x\,dx. The limits change: x = 0 u = 1, and x = π2 u = 0. Then _0^π/2sin^3 x\,dx = _1^0 (1 - u^2)(-\,du) = _0^1 (1 - u^2)\,du = [u - 13u^3]_0^1 = 1 - 13 = 23. Mark notes: 1 mark for peeling a sin x and using sin^2 x = 1 - cos^2 x, 1 mark for the substitution with corrected limits, 1 mark for the exact value 23.

NESA HSC Ext 2 (representative)-style practice: Find the exact value of _0^π/4 sec^2 x tan^3 x \, dx.

The power of sec is even (m = 2), so reserve the sec^2 x as the differential and substitute u = tan x, with du = sec^2 x\,dx. The limits change: x = 0 u = 0, and x = π4 u = tanπ4 = 1. Then _0^π/4sec^2 xtan^3 x\,dx = _0^1 u^3\,du = [14u^4]_0^1 = 14. Mark notes: 1 mark for reserving sec^2 x and choosing u = tan x, 1 mark for changing the limits, 1 mark for the value 14.

NESA HSC Ext 2 (representative)-style practice: By first converting the product to a sum, find cos 3x cos x \, dx.

Use cos Acos B = 12[cos(A - B) + cos(A + B)] with A = 3x, B = x: cos 3xcos x = 12[cos 2x + cos 4x]. Integrate each term: cos 3xcos x\,dx = 12[12sin 2x + 14sin 4x] + C = 14sin 2x + 18sin 4x + C. Mark notes: 1 mark for the correct product-to-sum conversion, 1 mark for integrating both terms with +C.

NSWMaths Extension 2Syllabus dot point

How do we integrate powers and products of trigonometric functions by choosing between the Pythagorean-identity substitution, the double-angle reduction, and the product-to-sum identities?

Integrate powers and products of trigonometric functions: powers of sin and cos (odd and even), powers of tan and sec, the integral of sec x, and products of sines and cosines via product-to-sum identities

A focused answer to the HSC Maths Extension 2 dot point on trigonometric integrals. The odd and even strategies for powers of sin and cos, powers of tan and sec, the standard integrals of sec x and sec cubed x, and products via product-to-sum identities, with every result verified by differentiation.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-21

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

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Quick answer

For $\int \sin^m x \cos^n x\,dx$ : if a power is odd, peel one factor off that function, convert the rest with $\sin^2 + \cos^2 = 1$ , and substitute; if both even, use the power-reduction identities $\cos^2 x = \tfrac{1}{2}(1+\cos 2x)$ and $\sin^2 x = \tfrac{1}{2}(1-\cos 2x)$ . For $\int \sec^m x \tan^n x\,dx$ : even $m$ , reserve $\sec^2 x$ and substitute $u=\tan x$ ; odd $n$ , reserve $\sec x\tan x$ and substitute $u=\sec x$ ; otherwise integrate by parts. Memorise $\int\tan x\,dx=-\ln|\cos x|$ , $\int\sec x\,dx=\ln|\sec x+\tan x|$ and $\int\sec^3 x\,dx=\tfrac{1}{2}(\sec x\tan x+\ln|\sec x+\tan x|)$ . For products of different multiples, convert to a sum with the product-to-sum identities, then integrate term by term.

Jump to a section

What this dot point is asking
Powers of sine and cosine
Powers of tangent and secant
Products of sines and cosines: product-to-sum
Choosing the method
How exam questions ask about trigonometric integrals

What this dot point is asking

NESA wants you to integrate powers and products of trigonometric functions without reaching for a general-purpose substitution every time. The integrand is built only from $\sin$ , $\cos$ , $\tan$ and $\sec$ , and the skill being examined is recognition: each shape has a matching tactic, and choosing the right one turns the integral into either a polynomial in a single trig function or a sum of simple terms. Four families are examined: powers $\int \sin^m x \cos^n x\,dx$ (the strategy splits on whether the exponents are odd or even); powers $\int \sec^m x \tan^n x\,dx$ (three sub-cases plus the special integrals of $\tan x$ and $\sec x$ ); the standout standard results $\int \sec x\,dx$ and $\int \sec^3 x\,dx$ ; and products such as $\sin mx\cos nx$ that surrender to the product-to-sum identities.

Powers of sine and cosine

Every integral $\int \sin^m x \cos^n x\,dx$ is decided by the parity of the exponents, and the reason is the derivative pairing $\tfrac{d}{dx}\sin x = \cos x$ and $\tfrac{d}{dx}\cos x = -\sin x$ . A substitution $u = \sin x$ needs a spare $\cos x$ to become $du$ ; a substitution $u = \cos x$ needs a spare $\sin x$ . An odd power always leaves exactly one such spare factor after you peel it off, and the Pythagorean identity converts the rest (now an even power) cleanly into the other function. When both powers are even there is no spare factor to peel, so substitution is dead and you drop down in degree with the double-angle identities instead.

Key fact

For $\displaystyle\int \sin^m x \cos^n x\,dx$ :

At least one exponent odd: peel one factor off the odd-power function, rewrite the remaining even power with $\sin^2 x + \cos^2 x = 1$ , then substitute ( $u=\cos x$ if you peeled a $\sin x$ , or $u=\sin x$ if you peeled a $\cos x$ ). The result is a polynomial integral.
Both exponents even: use the power-reduction identities $\cos^2 x = \tfrac{1}{2}(1+\cos 2x)$ and $\sin^2 x = \tfrac{1}{2}(1-\cos 2x)$ (and, for a product, $\sin x\cos x = \tfrac{1}{2}\sin 2x$ ) to halve the degree, then integrate; repeat if a squared cosine remains.

Why the odd case always works

The peel-and-substitute move is not a lucky trick, it is forced by the structure. Suppose the power of $\cos$ is odd, say $\cos^{2k+1}x$ . Detach one cosine: $\cos^{2k+1}x = (\cos^2 x)^k\cos x = (1-\sin^2 x)^k\cos x$ . Now every cosine except the detached one has been turned into sines, and the lone $\cos x$ is precisely the $du$ for $u = \sin x$ . The same logic mirrors for an odd power of $\sin$ : detach one sine (it becomes $-du$ for $u = \cos x$ ) and turn the remaining even power of $\sin$ into cosines. The companion power of the other function comes along untouched as part of the polynomial in $u$ , which is why the other exponent can be anything at all in the odd case, even, odd, or zero.

Why the even case needs double angles

If both exponents are even there is no leftover single factor to serve as $du$ , so no substitution rationalises the integrand. The way out is to lower the degree: $\cos^2 x = \tfrac{1}{2}(1+\cos 2x)$ replaces a square by a first power of a doubled angle, which integrates directly. Squaring out a fourth power produces a $\cos^2 2x$ , to which you apply the identity again. The mixed product $\sin^2 x\cos^2 x$ is best handled by first writing $\sin x\cos x = \tfrac{1}{2}\sin 2x$ , so $\sin^2 x\cos^2 x = \tfrac{1}{4}\sin^2 2x = \tfrac{1}{8}(1-\cos 4x)$ , a single application that lands a constant plus one cosine.

Powers of tangent and secant

The integral $\int \sec^m x \tan^n x\,dx$ is governed by the two derivative facts $\tfrac{d}{dx}\tan x = \sec^2 x$ and $\tfrac{d}{dx}\sec x = \sec x\tan x$ , together with the Pythagorean identity in its tangent form $1 + \tan^2 x = \sec^2 x$ . Each derivative fact is the $du$ of a substitution, and the question is which spare factor the integrand can afford to give up.

Key fact

For $\displaystyle\int \sec^m x \tan^n x\,dx$ :

$m$ even (and $m \ge 2$ ): reserve one $\sec^2 x$ as the differential, convert the rest with $\sec^2 x = 1 + \tan^2 x$ , then substitute $u = \tan x$ .
$n$ odd: reserve one $\sec x\tan x$ as the differential, convert the remaining (now even) power of $\tan$ with $\tan^2 x = \sec^2 x - 1$ , then substitute $u = \sec x$ .
$m$ odd and $n$ even (so neither reservation works): integrate by parts, which produces a reduction; this is exactly how $\int \sec^3 x\,dx$ is obtained.

Two special cases sit underneath: $\displaystyle\int \tan x\,dx = -\ln|\cos x| + C$ and $\displaystyle\int \sec x\,dx = \ln|\sec x + \tan x| + C$ .

The two special integrals you must know

For $\int \tan x\,dx$ , write $\tan x = \dfrac{\sin x}{\cos x}$ . The numerator is the negative derivative of the denominator, so the integral is a logarithm:

\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx = -\ln|\cos x| + C.

For $\int \sec x\,dx$ there is a famous device: multiply top and bottom by $\sec x + \tan x$ . The new numerator is then exactly the derivative of the new denominator, because $\tfrac{d}{dx}(\sec x + \tan x) = \sec x\tan x + \sec^2 x = \sec x(\sec x + \tan x)$ . So

\int \sec x\,dx = \int \frac{\sec x(\sec x + \tan x)}{\sec x + \tan x}\,dx = \int \frac{\sec x\tan x + \sec^2 x}{\sec x + \tan x}\,dx = \ln|\sec x + \tan x| + C.

Both of these are on the standard-integrals reference sheet for Extension 2, but you should be able to reproduce the $\sec x$ derivation, because the same multiply-by-the-conjugate idea is what unlocks $\int \sec^3 x\,dx$ .

Products of sines and cosines: product-to-sum

When the integrand is a product of trigonometric functions of different multiples of $x$ , such as $\sin 3x\cos 2x$ or $\cos 5x\cos 3x$ , there is no useful Pythagorean factor to peel. Instead, convert the product into a sum of single trig functions, each of which integrates immediately. The three identities follow from adding or subtracting the compound-angle formulas:

\sin A\cos B = \tfrac{1}{2}\big[\sin(A-B) + \sin(A+B)\big],

\cos A\cos B = \tfrac{1}{2}\big[\cos(A-B) + \cos(A+B)\big],

\sin A\sin B = \tfrac{1}{2}\big[\cos(A-B) - \cos(A+B)\big].

The only one easy to misremember is the last: $\sin A\sin B$ has a minus sign and the order is $\cos(A-B) - \cos(A+B)$ . After converting, each term is $\sin(kx)$ or $\cos(kx)$ , which integrates to $\mp\tfrac{1}{k}\cos(kx)$ or $\tfrac{1}{k}\sin(kx)$ . These integrals are the bridge to the harder Extension 2 material on summing trigonometric series and on roots of unity, where the same products appear.

Choosing the method

The whole topic is recognition. Read the integrand and ask, in order:

Exam technique

Decision order: (1) Is it a product of different angles like $\sin mx\cos nx$ ? Use product-to-sum. (2) Is it $\sin^m x\cos^n x$ ? Check parity: an odd power means peel-and-substitute, both even means power-reduction. (3) Is it $\sec^m x\tan^n x$ ? Even $\sec$ power reserves $\sec^2 x$ (sub $u=\tan x$ ); odd $\tan$ power reserves $\sec x\tan x$ (sub $u=\sec x$ ); the leftover odd- $\sec$ , even- $\tan$ case is integration by parts.
Show the structural step for the mark: Write the line that exposes the method: the peeled factor with the Pythagorean conversion ( $\cos^3 x = \cos x(1-\sin^2 x)$ ), the reserved $\sec^2 x$ or $\sec x\tan x$ , or the product-to-sum line. That line is usually the first mark; jumping to a $u$ -substitution without it can be marked down.
Definite integrals: when you substitute $u=\sin x$ , $u=\cos x$ , $u=\tan x$ or $u=\sec x$ , change the limits to $u$ -values and finish in $u$ , rather than back-substituting. Keep $+C$ on every indefinite answer, and keep the modulus inside every $\ln$ . "Find the exact value" means surds, $\pi$ and $\ln$ stay; no decimals.

How exam questions ask about trigonometric integrals

"Find $\int \cos^3 x\,dx$ " or " $\int \sin^5 x\,dx$ " (a single odd power). Peel one factor, convert the rest with the Pythagorean identity, substitute. The companion power can be absent (as here) or present.
"Evaluate $\int_0^{\pi/2}\cos^4 x\,dx$ " or any even power / even-by-even product. Power-reduction identities; apply twice for a fourth power, and remember the standard results $\int_0^{\pi/2}\cos^2 = \int_0^{\pi/2}\sin^2 = \tfrac{\pi}{4}$ .
"Find $\int \tan^4 x\,dx$ " / " $\int \sec^4 x\tan^2 x\,dx$ " (powers of tan and sec). Decide on the reserved factor from the parity rule, convert with $1+\tan^2 x = \sec^2 x$ , substitute.
"Find $\int \sec x\,dx$ " or " $\int \sec^3 x\,dx$ ." Quote (or, for $\sec^3 x$ , derive by parts) the standard result; these are flagged so you produce the log term and the $\tfrac{1}{2}(\sec x\tan x + \ln|\sec x+\tan x|)$ form exactly.
"By converting to a sum, find $\int \sin 3x\cos x\,dx$ " (a product of different multiples). Product-to-sum, then integrate term by term.
"Hence find ..." after a part that proved an identity. The earlier part hands you the identity to substitute; use it rather than starting over.

Integrate an odd power:

\displaystyle\int \sin^5 x \cos^2 x\,dx

Integrate the odd power $\displaystyle\int \sin^5 x \cos^2 x\,dx$

Here $\sin$ has the odd power ( $m = 5$ ) and $\cos$ the even power ( $n = 2$ ). The odd power decides everything: work with $\sin$ , peel one factor, and convert the rest to $\cos$ so that the substitution $u = \cos x$ takes over.

Peel one sine and convert the rest. Detach a single $\sin x$ and turn the remaining $\sin^4 x$ into cosines with $\sin^2 x = 1 - \cos^2 x$ :

\sin^5 x \cos^2 x = \sin x\,(\sin^2 x)^2\cos^2 x = \sin x\,(1 - \cos^2 x)^2\cos^2 x.

Substitute $u = \cos x$ . Then $du = -\sin x\,dx$ , so $\sin x\,dx = -\,du$ , and every remaining factor is a function of $u$ :

\int \sin^5 x \cos^2 x\,dx = \int (1 - u^2)^2 u^2\,(-\,du) = -\int (1 - 2u^2 + u^4)u^2\,du.

Expand to a polynomial. Multiplying through by $u^2$ ,

-\int (u^2 - 2u^4 + u^6)\,du.

Integrate term by term.

-\left(\frac{1}{3}u^3 - \frac{2}{5}u^5 + \frac{1}{7}u^7\right) + C = -\frac{1}{3}u^3 + \frac{2}{5}u^5 - \frac{1}{7}u^7 + C.

Back-substitute $u = \cos x$ .

\int \sin^5 x \cos^2 x\,dx = -\frac{1}{3}\cos^3 x + \frac{2}{5}\cos^5 x - \frac{1}{7}\cos^7 x + C.

Differentiating this back returns $\sin^5 x\cos^2 x$ exactly, which confirms the result.

Final answer: $\displaystyle\int \sin^5 x \cos^2 x\,dx = -\frac{1}{3}\cos^3 x + \frac{2}{5}\cos^5 x - \frac{1}{7}\cos^7 x + C$ .

Integrate an even power:

\displaystyle\int_0^{\pi/2} \cos^4 x\,dx

Evaluate the even power $\displaystyle\int_0^{\pi/2} \cos^4 x\,dx$

The power of $\cos$ is even and there is no $\sin$ factor to peel, so substitution is unavailable. Reduce the degree with the power-reduction identity, applied twice.

Reduce once. Using $\cos^2 x = \tfrac{1}{2}(1 + \cos 2x)$ and squaring,

\cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1}{4}\big(1 + 2\cos 2x + \cos^2 2x\big).

Reduce the squared cosine again. The term $\cos^2 2x$ is still even, so apply the identity once more with angle $2x$ : $\cos^2 2x = \tfrac{1}{2}(1 + \cos 4x)$ . Substituting,

\cos^4 x = \frac{1}{4}\left(1 + 2\cos 2x + \frac{1 + \cos 4x}{2}\right) = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x.

Integrate term by term.

\int \cos^4 x\,dx = \frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C.

Evaluate from $0$ to $\tfrac{\pi}{2}$ . At $x = \tfrac{\pi}{2}$ : $\sin 2x = \sin\pi = 0$ and $\sin 4x = \sin 2\pi = 0$ , so only the first term survives, giving $\tfrac{3}{8}\cdot\tfrac{\pi}{2} = \tfrac{3\pi}{16}$ . At $x = 0$ every term is $0$ .

Final answer: $\displaystyle\int_0^{\pi/2} \cos^4 x\,dx = \frac{3\pi}{16}$ (about $0.589$ ). As a sanity check, the average value of $\cos^4 x$ over a quarter period is $\tfrac{3}{8}$ , and $\tfrac{3}{8}\times\tfrac{\pi}{2} = \tfrac{3\pi}{16}$ as found.

Powers of tangent:

\displaystyle\int \tan^4 x\,dx

Integrate the power of tangent $\displaystyle\int \tan^4 x\,dx$

There is no $\sec$ factor here ( $m = 0$ ), and $n = 4$ is even, so neither standard reservation applies directly. The way in is the Pythagorean identity $\tan^2 x = \sec^2 x - 1$ , which trades two powers of $\tan$ for a $\sec^2 x$ (a ready differential) at a cost of a lower power of $\tan$ . This is a reduction in disguise.

Split off a $\tan^2 x$ . Write $\tan^4 x = \tan^2 x\,(\sec^2 x - 1) = \tan^2 x\sec^2 x - \tan^2 x$ , so

\int \tan^4 x\,dx = \int \tan^2 x\sec^2 x\,dx - \int \tan^2 x\,dx.

First integral by substitution. With $u = \tan x$ , $du = \sec^2 x\,dx$ :

\int \tan^2 x\sec^2 x\,dx = \int u^2\,du = \frac{1}{3}u^3 = \frac{1}{3}\tan^3 x.

Second integral, reduce again. Apply the identity once more: $\tan^2 x = \sec^2 x - 1$ , so

\int \tan^2 x\,dx = \int (\sec^2 x - 1)\,dx = \tan x - x.

Combine.

\int \tan^4 x\,dx = \frac{1}{3}\tan^3 x - (\tan x - x) + C = \frac{1}{3}\tan^3 x - \tan x + x + C.

Differentiating back gives $\sec^2 x\tan^2 x - \sec^2 x + 1 = \sec^2 x(\tan^2 x - 1) + 1$ ; using $\sec^2 x = 1 + \tan^2 x$ this simplifies to $\tan^4 x$ , confirming the answer.

Final answer: $\displaystyle\int \tan^4 x\,dx = \frac{1}{3}\tan^3 x - \tan x + x + C$ .

The standard result

\displaystyle\int \sec^3 x\,dx

by parts

Derive the standard integral $\displaystyle\int \sec^3 x\,dx$

This is the odd- $\sec$ , even- $\tan$ case ( $m = 3$ , $n = 0$ ): no reserved factor works, so integrate by parts. The integral reappears on the right, which we solve for, exactly the boomerang technique. It relies on already knowing $\int \sec x\,dx = \ln|\sec x + \tan x|$ .

Set up parts. Write $\sec^3 x = \sec x\cdot\sec^2 x$ and let $u = \sec x$ , $dv = \sec^2 x\,dx$ . Then $du = \sec x\tan x\,dx$ and $v = \tan x$ . Integration by parts $\int u\,dv = uv - \int v\,du$ gives

\int \sec^3 x\,dx = \sec x\tan x - \int \tan x\,(\sec x\tan x)\,dx = \sec x\tan x - \int \sec x\tan^2 x\,dx.

Convert the leftover $\tan^2 x$ . Using $\tan^2 x = \sec^2 x - 1$ ,

\int \sec x\tan^2 x\,dx = \int \sec x(\sec^2 x - 1)\,dx = \int \sec^3 x\,dx - \int \sec x\,dx.

Spot the boomerang. Substituting back, and writing $I = \int \sec^3 x\,dx$ ,

I = \sec x\tan x - \left(I - \int \sec x\,dx\right) = \sec x\tan x - I + \ln|\sec x + \tan x|.

Solve for $I$ . Adding $I$ to both sides, $2I = \sec x\tan x + \ln|\sec x + \tan x|$ , hence

\int \sec^3 x\,dx = \frac{1}{2}\Big(\sec x\tan x + \ln|\sec x + \tan x|\Big) + C.

Differentiating the right side returns $\sec^3 x$ , confirming the result.

Final answer: $\displaystyle\int \sec^3 x\,dx = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln|\sec x + \tan x| + C$ .

A product via product-to-sum:

\displaystyle\int \cos 5x\cos 3x\,dx

Integrate the product $\displaystyle\int \cos 5x\cos 3x\,dx$

The integrand is a product of cosines of different multiples of $x$ . There is no Pythagorean factor to exploit, so convert the product to a sum and integrate each piece.

Apply the product-to-sum identity. With $\cos A\cos B = \tfrac{1}{2}\big[\cos(A-B) + \cos(A+B)\big]$ and $A = 5x$ , $B = 3x$ :

\cos 5x\cos 3x = \frac{1}{2}\big[\cos 2x + \cos 8x\big].

Integrate term by term. Each cosine integrates to a sine over its multiple:

\int \cos 5x\cos 3x\,dx = \frac{1}{2}\left[\frac{1}{2}\sin 2x + \frac{1}{8}\sin 8x\right] + C = \frac{1}{4}\sin 2x + \frac{1}{16}\sin 8x + C.

Differentiating back gives $\tfrac{1}{2}\cos 2x + \tfrac{1}{2}\cos 8x = \cos 5x\cos 3x$ , confirming the answer.

Final answer: $\displaystyle\int \cos 5x\cos 3x\,dx = \frac{1}{4}\sin 2x + \frac{1}{16}\sin 8x + C$ .

As a definite-integral variant, $\displaystyle\int_0^{\pi/4}\cos 5x\cos 3x\,dx = \left[\tfrac{1}{4}\sin 2x + \tfrac{1}{16}\sin 8x\right]_0^{\pi/4} = \tfrac{1}{4}\sin\tfrac{\pi}{2} + \tfrac{1}{16}\sin 2\pi = \tfrac{1}{4}$ .

Common traps

Treating an even power like an odd one: $\int \cos^4 x\,dx$ has no spare factor to peel, so $u = \sin x$ leaves an unmanaged $\cos^3 x$ . Both-even means power-reduction, not substitution.
Forgetting to convert before substituting: After peeling a $\sin x$ from $\sin^5 x\cos^2 x$ you must turn the remaining $\sin^4 x$ into $(1-\cos^2 x)^2$ before $u = \cos x$ works; leaving any $\sin$ behind (other than the one in $du$ ) breaks the substitution.
The wrong reserved factor for tan and sec: Even power of $\sec$ reserves $\sec^2 x$ with $u = \tan x$ ; odd power of $\tan$ reserves $\sec x\tan x$ with $u = \sec x$ . Reserving the wrong one leaves a factor that is not the right $du$ .
Sign in the $\sin A\sin B$ identity: It is $\sin A\sin B = \tfrac{1}{2}[\cos(A-B) - \cos(A+B)]$ , with a minus sign; the cosine product has a plus. Mixing them flips a sign through the whole answer.
Dropping the modulus or the boomerang halving: Keep $\ln|\cos x|$ and $\ln|\sec x + \tan x|$ with the bars, and remember $\int \sec^3 x\,dx$ ends in a factor of $\tfrac{1}{2}$ after solving $2I = \ldots$ ; forgetting it doubles the answer.

Exam-style practice questions

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

HSC Ext 2 (representative)3 marksFind the exact value of

\displaystyle\int_0^{\pi/2} \sin^3 x \, dx

Show worked answer →

The power of $\sin$ is odd, so peel one factor of $\sin x$ and convert the rest with $\sin^2 x = 1 - \cos^2 x$ :

$\sin^3 x = \sin x\,(1 - \cos^2 x)$ .

Substitute $u = \cos x$ , so $du = -\sin x\,dx$ . The limits change: $x = 0 \Rightarrow u = 1$ , and $x = \tfrac{\pi}{2} \Rightarrow u = 0$ . Then

$\displaystyle\int_0^{\pi/2}\sin^3 x\,dx = \int_1^0 (1 - u^2)(-\,du) = \int_0^1 (1 - u^2)\,du = \left[u - \tfrac{1}{3}u^3\right]_0^1 = 1 - \tfrac{1}{3} = \tfrac{2}{3}.$

Mark notes: 1 mark for peeling a $\sin x$ and using $\sin^2 x = 1 - \cos^2 x$ , 1 mark for the substitution with corrected limits, 1 mark for the exact value $\tfrac{2}{3}$ .

HSC Ext 2 (representative)3 marksFind the exact value of

\displaystyle\int_0^{\pi/4} \sec^2 x \tan^3 x \, dx

Show worked answer →

The power of $\sec$ is even ( $m = 2$ ), so reserve the $\sec^2 x$ as the differential and substitute $u = \tan x$ , with $du = \sec^2 x\,dx$ . The limits change: $x = 0 \Rightarrow u = 0$ , and $x = \tfrac{\pi}{4} \Rightarrow u = \tan\tfrac{\pi}{4} = 1$ . Then

$\displaystyle\int_0^{\pi/4}\sec^2 x\tan^3 x\,dx = \int_0^1 u^3\,du = \left[\tfrac{1}{4}u^4\right]_0^1 = \tfrac{1}{4}.$

Mark notes: 1 mark for reserving $\sec^2 x$ and choosing $u = \tan x$ , 1 mark for changing the limits, 1 mark for the value $\tfrac{1}{4}$ .

HSC Ext 2 (representative)2 marksBy first converting the product to a sum, find

\displaystyle\int \cos 3x \cos x \, dx

Show worked answer →

Use $\cos A\cos B = \tfrac{1}{2}\big[\cos(A - B) + \cos(A + B)\big]$ with $A = 3x$ , $B = x$ :

$\cos 3x\cos x = \tfrac{1}{2}\big[\cos 2x + \cos 4x\big].$

Integrate each term:

$\displaystyle\int \cos 3x\cos x\,dx = \tfrac{1}{2}\left[\tfrac{1}{2}\sin 2x + \tfrac{1}{4}\sin 4x\right] + C = \tfrac{1}{4}\sin 2x + \tfrac{1}{8}\sin 4x + C.$

Mark notes: 1 mark for the correct product-to-sum conversion, 1 mark for integrating both terms with $+C$ .

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksFind the exact value of

\displaystyle\int_0^{\pi/2} \sin^2 x \, dx

Show worked solution →

Recognise the type. This is an even power of $\sin$ , so the odd-power substitution is not available. Use the power-reduction (double-angle) identity instead.

Apply the identity. Since $\sin^2 x = \frac{1}{2}(1 - \cos 2x)$ ,

\int_0^{\pi/2} \sin^2 x \, dx = \frac{1}{2}\int_0^{\pi/2} (1 - \cos 2x)\, dx = \frac{1}{2}\left[\, x - \frac{1}{2}\sin 2x \,\right]_0^{\pi/2}.

Evaluate at the limits. At $x = \frac{\pi}{2}$ : $x = \frac{\pi}{2}$ and $\sin 2x = \sin\pi = 0$ . At $x = 0$ both terms vanish. So the value is $\frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4}$ .

Answer: $\displaystyle\int_0^{\pi/2} \sin^2 x \, dx = \frac{\pi}{4}$ (about $0.785$ ).

Mark notes: 1 mark for the power-reduction identity, 1 mark for the correct exact value.

foundation2 marksFind

\displaystyle\int \cos^3 x \, dx

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Recognise the type. The power of $\cos$ is odd, so peel off one factor of $\cos x$ and convert the rest to $\sin$ .

Peel and convert. Write $\cos^3 x = \cos x \,(1 - \sin^2 x)$ using $\cos^2 x = 1 - \sin^2 x$ .

Substitute $u = \sin x$ , so $du = \cos x \, dx$ :

\int \cos^3 x \, dx = \int (1 - u^2)\, du = u - \frac{1}{3}u^3 + C.

Back-substitute $u = \sin x$ :

\int \cos^3 x \, dx = \sin x - \frac{1}{3}\sin^3 x + C.

Mark notes: 1 mark for peeling a $\cos x$ and using the Pythagorean identity, 1 mark for the integrated answer with $+C$ .

core3 marksFind

\displaystyle\int \sin^3 x \cos^2 x \, dx

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Recognise the type. $\sin$ appears to an odd power, so work with $\sin$ : peel one $\sin x$ and turn the remaining $\sin^2 x$ into $1 - \cos^2 x$ .

Peel and convert. $\sin^3 x \cos^2 x = \sin x\,(1 - \cos^2 x)\cos^2 x$ .

Substitute $u = \cos x$ , so $du = -\sin x \, dx$ , i.e. $\sin x \, dx = -\,du$ :

\int \sin^3 x \cos^2 x \, dx = \int (1 - u^2)u^2 \,(-\,du) = -\int (u^2 - u^4)\, du.

Integrate the polynomial.

-\int (u^2 - u^4)\, du = -\frac{1}{3}u^3 + \frac{1}{5}u^5 + C.

Back-substitute $u = \cos x$ :

\int \sin^3 x \cos^2 x \, dx = -\frac{1}{3}\cos^3 x + \frac{1}{5}\cos^5 x + C.

Mark notes: 1 mark for choosing to work with $\sin$ (odd power) and peeling, 1 mark for the substitution and polynomial, 1 mark for the back-substituted answer.

core3 marksFind the exact value of

\displaystyle\int_0^{\pi/4} \tan^3 x \, dx

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Recognise the type. A pure power of $\tan$ . Split off $\tan^2 x = \sec^2 x - 1$ to expose a $\sec^2 x$ factor (whose integral pairs with $\tan x$ ).

Split with the Pythagorean identity.

\tan^3 x = \tan x\,(\sec^2 x - 1) = \tan x \sec^2 x - \tan x.

Integrate each piece. For the first, $u = \tan x$ gives $\int \tan x \sec^2 x \, dx = \frac{1}{2}\tan^2 x$ . For the second, $\int \tan x \, dx = -\ln|\cos x|$ . So

\int \tan^3 x \, dx = \frac{1}{2}\tan^2 x + \ln|\cos x| + C.

Evaluate from $0$ to $\frac{\pi}{4}$ . At $x = \frac{\pi}{4}$ : $\tan\frac{\pi}{4} = 1$ and $\cos\frac{\pi}{4} = \frac{1}{\sqrt 2}$ , so the bracket is $\frac{1}{2} + \ln\frac{1}{\sqrt 2} = \frac{1}{2} - \frac{1}{2}\ln 2$ . At $x = 0$ : $\frac{1}{2}(0) + \ln 1 = 0$ .

Answer: $\displaystyle\int_0^{\pi/4} \tan^3 x \, dx = \frac{1}{2} - \frac{1}{2}\ln 2$ (about $0.153$ ).

Mark notes: 1 mark for the $\tan^2 x = \sec^2 x - 1$ split, 1 mark for the antiderivative, 1 mark for the exact value.

core3 marksUsing a product-to-sum identity, find

\displaystyle\int \sin 4x \cos 2x \, dx

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Recognise the type. A product of a sine and a cosine of different multiples of $x$ . Convert it to a sum with $\sin A \cos B = \frac{1}{2}\big[\sin(A - B) + \sin(A + B)\big]$ .

Convert to a sum. With $A = 4x$ , $B = 2x$ :

\sin 4x \cos 2x = \frac{1}{2}\big[\sin 2x + \sin 6x\big].

Integrate term by term.

\int \sin 4x \cos 2x \, dx = \frac{1}{2}\left[-\frac{1}{2}\cos 2x - \frac{1}{6}\cos 6x\right] + C = -\frac{1}{4}\cos 2x - \frac{1}{12}\cos 6x + C.

Mark notes: 1 mark for the correct product-to-sum identity, 1 mark for the converted integrand, 1 mark for integrating each term correctly with $+C$ .

exam4 marksShow that

\displaystyle\int_0^{\pi/4} \sec^4 x \tan^2 x \, dx = \frac{8}{15}

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Recognise the type. Powers of $\sec$ and $\tan$ with an even power of $\sec$ ( $m = 4$ ). Save one $\sec^2 x$ for the differential and turn the rest into $\tan$ using $\sec^2 x = 1 + \tan^2 x$ .

Reserve a $\sec^2 x$ . Write $\sec^4 x = \sec^2 x \cdot \sec^2 x = (1 + \tan^2 x)\sec^2 x$ , so

\sec^4 x \tan^2 x = (1 + \tan^2 x)\tan^2 x \, \sec^2 x.

Substitute $u = \tan x$ , so $du = \sec^2 x \, dx$ . The limits change: $x = 0 \Rightarrow u = 0$ , and $x = \frac{\pi}{4} \Rightarrow u = \tan\frac{\pi}{4} = 1$ . Then

\int_0^{\pi/4} \sec^4 x \tan^2 x \, dx = \int_0^1 (1 + u^2)u^2 \, du = \int_0^1 (u^2 + u^4)\, du.

Integrate and evaluate.

\int_0^1 (u^2 + u^4)\, du = \left[\frac{1}{3}u^3 + \frac{1}{5}u^5\right]_0^1 = \frac{1}{3} + \frac{1}{5} = \frac{5 + 3}{15} = \frac{8}{15}.

So the integral equals $\frac{8}{15}$ as required (about $0.533$ ).

Mark notes: 1 mark for reserving a $\sec^2 x$ factor, 1 mark for $\sec^2 x = 1 + \tan^2 x$ and the substitution, 1 mark for changing the limits, 1 mark for the evaluation to $\frac{8}{15}$ .

exam3 marksFind

\displaystyle\int \sec^5 x \tan x \, dx

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Recognise the type. Powers of $\sec$ and $\tan$ where the power of $\tan$ is odd ( $n = 1$ ). Reserve the factor $\sec x \tan x$ for the differential and write everything else in terms of $\sec x$ .

Reserve $\sec x \tan x$ . Split off one $\sec x$ and the single $\tan x$ :

\sec^5 x \tan x = \sec^4 x \cdot (\sec x \tan x).

Substitute $u = \sec x$ , so $du = \sec x \tan x \, dx$ :

\int \sec^5 x \tan x \, dx = \int u^4 \, du = \frac{1}{5}u^5 + C.

Back-substitute $u = \sec x$ :

\int \sec^5 x \tan x \, dx = \frac{1}{5}\sec^5 x + C.

Mark notes: 1 mark for reserving the $\sec x \tan x$ factor, 1 mark for the $u = \sec x$ substitution, 1 mark for the back-substituted answer with $+C$ .

What this dot point is asking

Powers of sine and cosine

Why the odd case always works

Why the even case needs double angles

Powers of tangent and secant

The two special integrals you must know

Products of sines and cosines: product-to-sum

Choosing the method

How exam questions ask about trigonometric integrals

Exam-style practice questions

Practice questions

Related dot points