JC2 H2 Maths Bukit Timah | Sin² Cos² Tan² Integrals

Did you know? The integral of sin²x is not −cos²x and is not simply −cos x. The square changes the problem: to integrate sin²x cleanly, first rewrite it as (1−cos 2x)/2. That one identity turns the difficult-looking expression into a constant and an ordinary cosine. In JC2 H2 Mathematics, recognising the identity is often more important than performing the final integration.

This JC2 H2 Mathematics Tutor Bukit Timah lesson answers one A-Level 9758 question: how can you integrate sin²x, cos²x and tan²x correctly, including scaled angles, exact definite limits and domain restrictions? The quick formulae are followed by derivative checks, worked applications and 32 original exam-style clinics. The explanations separate the squared trigonometric functions from simple products such as sin x cos x, because different algebraic structures call for different methods.

The 2027 SEAB H2 Mathematics syllabus 9758 lists integration of sin²x, cos²x and tan²x within Integration Techniques §5.3. It excludes reduction formulae. This guide uses the double-angle identities and tan²x=sec²x−1 rather than methods beyond the specified syllabus. Each worked calculation is a fictional mathematical problem, not an official examination item or an actual tuition student’s answer.

Quick answer: rewrite the square before integrating

The relevant identities are sin²x=(1−cos 2x)/2, cos²x=(1+cos 2x)/2 and tan²x=sec²x−1 wherever tangent is defined. Consequently the antiderivatives are ∫sin²x dx=x/2−sin 2x/4+C, ∫cos²x dx=x/2+sin 2x/4+C, and ∫tan²x dx=tan x−x+C on any interval containing no tangent singularity.

sin²x: split into 1/2−cos 2x/2; the oscillating term integrates with a minus sign.
cos²x: split into 1/2+cos 2x/2; the oscillating term integrates with a plus sign.
tan²x: use sec²x−1; the antiderivative is tan x−x.
Scaled angle kx: ∫sin²(kx)dx=x/2−sin(2kx)/(4k)+C for k≠0.
Scaled cosine: ∫cos²(kx)dx=x/2+sin(2kx)/(4k)+C.
Scaled tangent: ∫tan²(kx)dx=tan(kx)/k−x+C on any interval with no poles.
Before a definite integral: check endpoints and the function’s continuity, particularly for tan².

Worked example one: integrate sin²x from 0 to π

Evaluate ∫₀^π sin²x dx. Replace sin²x by (1−cos 2x)/2. Integrate term by term: the antiderivative is x/2−sin 2x/4. At x=π, its value is π/2−sin 2π/4=π/2. At x=0, its value is zero. Therefore ∫₀^π sin²x dx=π/2.

A useful mental check is that sin²x ranges between zero and one and has average value 1/2 over a full period. The interval length π covers one full period of sin²x, so the accumulated area is π×1/2=π/2. The result is positive, as it must be because the integrand is non-negative.

Worked example two: integrate cos²x from 0 to π/2

Evaluate ∫₀^(π/2) cos²x dx. The half-angle identity is cos²x=(1+cos 2x)/2. An antiderivative is x/2+sin 2x/4. At x=π/2 the sine term is sin π=0, so the result is π/4. The derivative of x/2+sin 2x/4 equals 1/2+cos 2x/2=cos²x, providing an independent check.

It is common to remember both sine-squared and cosine-squared antiderivatives but swap their signs. One quick audit is to differentiate the proposed answer. If a student writes x/2−sin 2x/4 for cos²x, its derivative is sin²x instead. A derivative check identifies the wrong function immediately.

Worked example three: tan²x is sec²x minus one

Evaluate ∫₀^(π/4) tan²x dx. Rewrite tan²x as sec²x−1. Then the antiderivative is tan x−x. Evaluating gives tan(π/4)−π/4−(tan0−0), or 1−π/4. Its value is approximately 0.214602, positive but less than the interval length, as expected because 0≤tan²x≤1 on this interval.

Do not use ∫tan²x dx=tan³x/3. That would be a valid pattern only if an appropriate differential of tan x appeared, such as sec²x dx. In this problem it does not. The identity sec²x=1+tan²x supplies the correct antiderivative without introducing an incorrect substitution.

Differentiation is the fastest formula check

Check the sine-squared antiderivative F(x)=x/2−sin 2x/4. Differentiating gives F′(x)=1/2−[2cos 2x]/4=(1−cos 2x)/2=sin²x. For G(x)=x/2+sin 2x/4, the derivative is (1+cos 2x)/2=cos²x. For H(x)=tan x−x, the derivative is sec²x−1=tan²x where defined.

These checks are exact identities, not approximations or calculator outputs. They are particularly useful in a timed H2 examination because a sign mistake in an antiderivative is easy to make but also easy to detect before evaluating limits. A twenty-second derivative check can prevent multiple later steps being built on the wrong expression.

Scaled angles: the factor k belongs in the denominator

If the integrand is sin²(3x), write (1−cos 6x)/2. Integrating gives x/2−sin 6x/12+C. The denominator of the oscillating term is 4×3=12. To check, differentiate −sin 6x/12: it contributes −(6/12)cos 6x=−cos 6x/2. The factor three in the original angle requires a chain-rule adjustment during integration.

Similarly, ∫cos²(2x)dx=x/2+sin 4x/8+C. The coefficient 1/8 is not copied from the unscaled expression; it reflects the doubled angle 4x inside the sine. More generally, the sin(2kx) term carries denominator 4k. A student who writes 4 in every case is forgetting the chain rule in reverse.

Sine squared and cosine squared complement each other

Since sin²x+cos²x=1, the two antiderivatives should add to x+C. Our formulae do exactly that: the −sin 2x/4 and +sin 2x/4 terms cancel, leaving x. This gives a second check independent of differentiating each one. If the oscillating terms fail to cancel, one sign or coefficient has been copied incorrectly.

For definite integrals over the same interval, ∫sin²x dx+∫cos²x dx equals the interval length. For example, on 0≤x≤π/4, the two values are π/8−1/4 and π/8+1/4; their sum is π/4, the interval length. When a problem has both functions, use this identity as an audit rather than solving them twice from scratch.

Half an interval and a nonzero sine endpoint

Consider ∫₀^(π/4) sin²x dx. Substitute the antiderivative: [x/2−sin 2x/4]₀^(π/4)=π/8−1/4. For cosine squared over the same interval, the answer is π/8+1/4. Both are positive, but cosine accumulates more area on [0,π/4] because cos x≥sin x there.

If you assume the sine term always disappears at a trigonometric endpoint, you will miss the ±1/4 contribution here. It vanishes at endpoints where sin 2x=0, such as 0 and π/2, but not at x=π/4. Calculate the transformed angle inside the sine before claiming an endpoint term equals zero.

A tangent-squared integral can fail at an endpoint

The expression tan²x is not defined at x=π/2. Therefore ∫₀^(π/2) tan²x dx is not an ordinary proper definite integral over a continuous bounded integrand. As the upper bound approaches π/2 from below, tan x−x grows without bound. The integral does not have a finite value. It is not correct to substitute tan(π/2)=0 or treat its undefined value as a cancellation.

This is a domain check, not a request to study improper integration beyond the 2027 syllabus. For a routine assessed integral of tan²x, work on an interval that avoids its poles, such as 0≤x≤π/4. The same caution applies to tan²(kx): points where cos(kx)=0 separate the natural intervals on which tan(kx)/k−x is a valid antiderivative.

Do not confuse sin²x with sin(x²)

sin²x means [sin x]²: square the output of the sine function. By contrast, sin(x²) means first square x, then take its sine. These are different functions with different derivatives and integration techniques. The double-angle identity applies to [sin x]², not to sin(x²). A learner who overlooks the location of the square will choose the wrong method before any calculus begins.

Similarly, cos²(3x) means [cos(3x)]², so its identity uses cos(6x), whereas cos(9x²) has a completely different argument. Parentheses are an efficient way to check: rewrite sin²(3x) as (sin(3x))² before transforming it. Mathematical notation needs to be read as an instruction sequence.

Using a graphing calculator as an audit

For ∫₀^(π/4) tan²x dx, a graphing calculator should give approximately 0.2146. Your exact result 1−π/4 produces the same decimal. If the calculator reports a negative value, check the signs and the radian mode. A graph can also show that tan²x rises sharply as x approaches π/2, warning that the endpoint is a singularity rather than a harmless value.

Calculator validation does not replace the identity-based working when an examination question requires it. The 2027 H2 syllabus expects students to know the limitations of their approved graphing calculators. Combining an exact antiderivative with a decimal check is usually safer than using an unsupported screen output alone.

One-minute diagnostic before practice

Ask a student to complete three identities: sin²x=(1−cos 2x)/2; cos²x=(1+cos 2x)/2; tan²x=sec²x−1. Next ask for the antiderivatives and the derivative check. Finally change sin²x to sin²(3x), then ask which denominator appears under sin 6x: 12. This sequence identifies whether the difficulty lies in the trig identity, the chain-rule scaling or the evaluation.

Thirty-two exact trigonometric-integration clinics

The next 32 original cases use a positive integer k and safe finite integration ranges. The exact answers contain π and simple rational coefficients, not numerical approximations. The problems vary whether the integrand is sine squared, cosine squared or tangent squared, and whether the integration interval is a quarter, half or complete period. Each case names the identity, applies its antiderivative and checks the sign or bound.

1. Sine squared, half-wave, k=1

Question. Evaluate ∫ from x=0 to x=π/2 of sin²(1x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(1x)=[1−cos(2x)]/2. An antiderivative is F(x)=x/2−sin(2x)/(4)+C. Evaluating it between the exact bounds gives π/4≈0.785398. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

Interpret and transfer. The double-angle identity removes the square before integration. Change k to another positive integer while shortening both endpoints in the same proportion. The area scales by the reciprocal factor because the curve oscillates more quickly. If the question changes sin² to cos² on the same interval, use the identity sin²+cos²=1 as an independent check. Do not treat the two antiderivative signs as interchangeable.

2. Cosine squared, half-wave, k=1

Question. Evaluate ∫ from x=0 to x=π/2 of cos²(1x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(1x)=[1+cos(2x)]/2. An antiderivative is F(x)=x/2+sin(2x)/(4)+C. Evaluating it between the exact bounds gives π/4≈0.785398. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

Interpret and transfer. A positive integer k changes the oscillation speed and scales the exact integral by 1/k on the proportionally shortened interval. Change k to another positive integer while shortening both endpoints in the same proportion. The area scales by the reciprocal factor because the curve oscillates more quickly. If the question changes sin² to cos² on the same interval, use the identity sin²+cos²=1 as an independent check. Do not treat the two antiderivative signs as interchangeable.

3. Sine squared, quarter-wave, k=1

Question. Evaluate ∫ from x=0 to x=π/4 of sin²(1x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(1x)=[1−cos(2x)]/2. An antiderivative is F(x)=x/2−sin(2x)/(4)+C. Evaluating it between the exact bounds gives (π−2)/8≈0.142699. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

Interpret and transfer. The sin² and cos² answers over the same range add to the length of that range. Change k to another positive integer while shortening both endpoints in the same proportion. The area scales by the reciprocal factor because the curve oscillates more quickly. If the question changes sin² to cos² on the same interval, use the identity sin²+cos²=1 as an independent check. Do not treat the two antiderivative signs as interchangeable.

4. Cosine squared, quarter-wave, k=1

Question. Evaluate ∫ from x=0 to x=π/4 of cos²(1x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(1x)=[1+cos(2x)]/2. An antiderivative is F(x)=x/2+sin(2x)/(4)+C. Evaluating it between the exact bounds gives (π+2)/8≈0.642699. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

Interpret and transfer. The oscillating endpoint term must be evaluated; it is not always zero. Change k to another positive integer while shortening both endpoints in the same proportion. The area scales by the reciprocal factor because the curve oscillates more quickly. If the question changes sin² to cos² on the same interval, use the identity sin²+cos²=1 as an independent check. Do not treat the two antiderivative signs as interchangeable.

5. Tangent squared before a pole, k=1

Question. Evaluate ∫ from x=0 to x=π/4 of tan²(1x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use tan²(1x)=sec²(1x)−1. An antiderivative is F(x)=tan(1x)/1−x+C. Evaluating it between the exact bounds gives (4−π)/4≈0.214602. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

Interpret and transfer. The tangent-squared case is valid here because the chosen interval stays away from tan’s poles. Change k to another positive integer while shortening both endpoints in the same proportion. The area scales by the reciprocal factor because the curve oscillates more quickly. If the question changes sin² to cos² on the same interval, use the identity sin²+cos²=1 as an independent check. Do not treat the two antiderivative signs as interchangeable.

6. Sine squared, full period, k=1

Question. Evaluate ∫ from x=0 to x=π/1 of sin²(1x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(1x)=[1−cos(2x)]/2. An antiderivative is F(x)=x/2−sin(2x)/(4)+C. Evaluating it between the exact bounds gives π/2≈1.570796. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

Interpret and transfer. An antiderivative should be differentiated to check the sign in front of sin(2kx). Change k to another positive integer while shortening both endpoints in the same proportion. The area scales by the reciprocal factor because the curve oscillates more quickly. If the question changes sin² to cos² on the same interval, use the identity sin²+cos²=1 as an independent check. Do not treat the two antiderivative signs as interchangeable.

7. Cosine squared, full period, k=1

Question. Evaluate ∫ from x=0 to x=π/1 of cos²(1x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(1x)=[1+cos(2x)]/2. An antiderivative is F(x)=x/2+sin(2x)/(4)+C. Evaluating it between the exact bounds gives π/2≈1.570796. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

Interpret and transfer. The correct answer is nonnegative because the squared trigonometric function is nonnegative throughout the interval. Change k to another positive integer while shortening both endpoints in the same proportion. The area scales by the reciprocal factor because the curve oscillates more quickly. If the question changes sin² to cos² on the same interval, use the identity sin²+cos²=1 as an independent check. Do not treat the two antiderivative signs as interchangeable.

8. Sine squared, final quarter, k=1

Question. Evaluate ∫ from x=π/4 to x=π/2 of sin²(1x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(1x)=[1−cos(2x)]/2. An antiderivative is F(x)=x/2−sin(2x)/(4)+C. Evaluating it between the exact bounds gives (π+2)/8≈0.642699. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

Interpret and transfer. Leaving an exact expression in terms of π avoids unnecessary rounding and makes symmetry checks easier. Change k to another positive integer while shortening both endpoints in the same proportion. The area scales by the reciprocal factor because the curve oscillates more quickly. If the question changes sin² to cos² on the same interval, use the identity sin²+cos²=1 as an independent check. Do not treat the two antiderivative signs as interchangeable.

9. Sine squared, half-wave, k=3

Question. Evaluate ∫ from x=0 to x=π/6 of sin²(3x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(3x)=[1−cos(6x)]/2. An antiderivative is F(x)=x/2−sin(6x)/(12)+C. Evaluating it between the exact bounds gives π/12≈0.261799. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

10. Cosine squared, half-wave, k=3

Question. Evaluate ∫ from x=0 to x=π/6 of cos²(3x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(3x)=[1+cos(6x)]/2. An antiderivative is F(x)=x/2+sin(6x)/(12)+C. Evaluating it between the exact bounds gives π/12≈0.261799. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

11. Sine squared, quarter-wave, k=3

Question. Evaluate ∫ from x=0 to x=π/12 of sin²(3x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(3x)=[1−cos(6x)]/2. An antiderivative is F(x)=x/2−sin(6x)/(12)+C. Evaluating it between the exact bounds gives (π−2)/24≈0.047566. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

12. Cosine squared, quarter-wave, k=3

Question. Evaluate ∫ from x=0 to x=π/12 of cos²(3x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(3x)=[1+cos(6x)]/2. An antiderivative is F(x)=x/2+sin(6x)/(12)+C. Evaluating it between the exact bounds gives (π+2)/24≈0.214233. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

13. Tangent squared before a pole, k=3

Question. Evaluate ∫ from x=0 to x=π/12 of tan²(3x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use tan²(3x)=sec²(3x)−1. An antiderivative is F(x)=tan(3x)/3−x+C. Evaluating it between the exact bounds gives (4−π)/12≈0.071534. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

14. Sine squared, full period, k=3

Question. Evaluate ∫ from x=0 to x=π/3 of sin²(3x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(3x)=[1−cos(6x)]/2. An antiderivative is F(x)=x/2−sin(6x)/(12)+C. Evaluating it between the exact bounds gives π/6≈0.523599. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

15. Cosine squared, full period, k=3

Question. Evaluate ∫ from x=0 to x=π/3 of cos²(3x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(3x)=[1+cos(6x)]/2. An antiderivative is F(x)=x/2+sin(6x)/(12)+C. Evaluating it between the exact bounds gives π/6≈0.523599. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

16. Sine squared, final quarter, k=3

Question. Evaluate ∫ from x=π/12 to x=π/6 of sin²(3x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(3x)=[1−cos(6x)]/2. An antiderivative is F(x)=x/2−sin(6x)/(12)+C. Evaluating it between the exact bounds gives (π+2)/24≈0.214233. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

17. Sine squared, half-wave, k=5

Question. Evaluate ∫ from x=0 to x=π/10 of sin²(5x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(5x)=[1−cos(10x)]/2. An antiderivative is F(x)=x/2−sin(10x)/(20)+C. Evaluating it between the exact bounds gives π/20≈0.15708. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

18. Cosine squared, half-wave, k=5

Question. Evaluate ∫ from x=0 to x=π/10 of cos²(5x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(5x)=[1+cos(10x)]/2. An antiderivative is F(x)=x/2+sin(10x)/(20)+C. Evaluating it between the exact bounds gives π/20≈0.15708. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

19. Sine squared, quarter-wave, k=5

Question. Evaluate ∫ from x=0 to x=π/20 of sin²(5x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(5x)=[1−cos(10x)]/2. An antiderivative is F(x)=x/2−sin(10x)/(20)+C. Evaluating it between the exact bounds gives (π−2)/40≈0.02854. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

20. Cosine squared, quarter-wave, k=5

Question. Evaluate ∫ from x=0 to x=π/20 of cos²(5x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(5x)=[1+cos(10x)]/2. An antiderivative is F(x)=x/2+sin(10x)/(20)+C. Evaluating it between the exact bounds gives (π+2)/40≈0.12854. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

21. Tangent squared before a pole, k=5

Question. Evaluate ∫ from x=0 to x=π/20 of tan²(5x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use tan²(5x)=sec²(5x)−1. An antiderivative is F(x)=tan(5x)/5−x+C. Evaluating it between the exact bounds gives (4−π)/20≈0.04292. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

22. Sine squared, full period, k=5

Question. Evaluate ∫ from x=0 to x=π/5 of sin²(5x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(5x)=[1−cos(10x)]/2. An antiderivative is F(x)=x/2−sin(10x)/(20)+C. Evaluating it between the exact bounds gives π/10≈0.314159. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

23. Cosine squared, full period, k=5

Question. Evaluate ∫ from x=0 to x=π/5 of cos²(5x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(5x)=[1+cos(10x)]/2. An antiderivative is F(x)=x/2+sin(10x)/(20)+C. Evaluating it between the exact bounds gives π/10≈0.314159. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

24. Sine squared, final quarter, k=5

Question. Evaluate ∫ from x=π/20 to x=π/10 of sin²(5x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(5x)=[1−cos(10x)]/2. An antiderivative is F(x)=x/2−sin(10x)/(20)+C. Evaluating it between the exact bounds gives (π+2)/40≈0.12854. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

25. Sine squared, half-wave, k=7

Question. Evaluate ∫ from x=0 to x=π/14 of sin²(7x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(7x)=[1−cos(14x)]/2. An antiderivative is F(x)=x/2−sin(14x)/(28)+C. Evaluating it between the exact bounds gives π/28≈0.1122. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

26. Cosine squared, half-wave, k=7

Question. Evaluate ∫ from x=0 to x=π/14 of cos²(7x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(7x)=[1+cos(14x)]/2. An antiderivative is F(x)=x/2+sin(14x)/(28)+C. Evaluating it between the exact bounds gives π/28≈0.1122. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

27. Sine squared, quarter-wave, k=7

Question. Evaluate ∫ from x=0 to x=π/28 of sin²(7x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(7x)=[1−cos(14x)]/2. An antiderivative is F(x)=x/2−sin(14x)/(28)+C. Evaluating it between the exact bounds gives (π−2)/56≈0.020386. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

28. Cosine squared, quarter-wave, k=7

Question. Evaluate ∫ from x=0 to x=π/28 of cos²(7x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(7x)=[1+cos(14x)]/2. An antiderivative is F(x)=x/2+sin(14x)/(28)+C. Evaluating it between the exact bounds gives (π+2)/56≈0.091814. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

29. Tangent squared before a pole, k=7

Question. Evaluate ∫ from x=0 to x=π/28 of tan²(7x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use tan²(7x)=sec²(7x)−1. An antiderivative is F(x)=tan(7x)/7−x+C. Evaluating it between the exact bounds gives (4−π)/28≈0.030657. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

30. Sine squared, full period, k=7

Question. Evaluate ∫ from x=0 to x=π/7 of sin²(7x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(7x)=[1−cos(14x)]/2. An antiderivative is F(x)=x/2−sin(14x)/(28)+C. Evaluating it between the exact bounds gives π/14≈0.224399. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

31. Cosine squared, full period, k=7

Question. Evaluate ∫ from x=0 to x=π/7 of cos²(7x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use cos²(7x)=[1+cos(14x)]/2. An antiderivative is F(x)=x/2+sin(14x)/(28)+C. Evaluating it between the exact bounds gives π/14≈0.224399. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

32. Sine squared, final quarter, k=7

Question. Evaluate ∫ from x=π/28 to x=π/14 of sin²(7x) dx. Start by identifying whether the square applies to the trigonometric output or to its input. The stated interval stays in a region where the integrand is continuous; for tangent squared, the upper bound deliberately avoids a pole. Predict whether the result should be positive before integrating.

Identity and antiderivative. Use sin²(7x)=[1−cos(14x)]/2. An antiderivative is F(x)=x/2−sin(14x)/(28)+C. Evaluating it between the exact bounds gives (π+2)/56≈0.091814. The sine or cosine endpoint contribution has been included, not discarded automatically. Differentiating F(x) recovers the original squared trigonometric function and checks the scaled-angle denominator.

Ten exact results for quick recall

∫₀^π sin²x dx=π/2.
∫₀^(π/2) cos²x dx=π/4.
∫₀^(π/4) sin²x dx=π/8−1/4.
∫₀^(π/4) cos²x dx=π/8+1/4.
∫₀^(π/4) tan²x dx=1−π/4.
∫₀^(π/2) sin²(2x) dx=π/4.
∫₀^(π/2) cos²(2x) dx=π/4.
∫sin²(3x)dx=x/2−sin6x/12+C.
∫cos²(2x)dx=x/2+sin4x/8+C.
∫tan²(2x)dx=tan(2x)/2−x+C, within an interval free of poles.

Five-day practice plan

Day one: write the three squared-trigonometric identities from memory and verify each using basic double-angle results. Day two: differentiate the three proposed antiderivatives, correcting sign errors before working on definite integrals. Day three: vary the scale k in sin²(kx) and cos²(kx), checking the chain-rule coefficient. Day four: evaluate exact limits including x=π/4, where endpoint sine terms do not vanish. Day five: mix sine, cosine and tangent squares with products such as sin x cos x, asking the learner to choose the appropriate identity or substitution before calculating.

For a three-learner tutorial, one student can identify the identity, another can integrate the transformed expression, and a third can differentiate the final answer as an independent audit. Rotate roles and change both the argument multiplier and upper limit. The purpose is to build decision-making accuracy rather than merely memorising three formulae without knowing when each applies.

Frequently asked questions

Can I integrate sin²x as though it were sin x?

No. Squaring a function changes its derivative and antiderivative. Use the half-angle identity first.

Why do sin²x and cos²x have different signs?

Their identities are (1−cos 2x)/2 and (1+cos 2x)/2, so the integrated oscillating term carries a minus or plus sign.

What is the antiderivative of tan²x?

It is tan x−x+C on any interval where tangent is defined, because tan²x=sec²x−1.

Why is there a 4k in the denominator?

The double-angle identity introduces cos(2kx); integrating it produces a 1/(2k) factor, combined with the identity’s 1/2 factor.

Can I integrate across a pole of tan²x?

Not as an ordinary proper definite integral. Check the endpoints and continuity first; an integral approaching a tangent pole may not converge.

Is a squared sine integral always nonnegative?

Over an interval with lower bound less than upper bound, its integrand is nonnegative, so the definite integral is nonnegative.

Does the 2027 H2 syllabus include these squared functions?

Yes. Integration of sin²x, cos²x and tan²x is explicitly listed in §5.3.

Is this the same as integrating sin x cos x?

No. The product sin x cos x is often handled directly by substitution or a double-angle identity; the squared functions require their own identities.

Related JC2 H2 guides and official curriculum

Parametric Tangents: Horizontal or Vertical?
Definite Integration by Substitution: Change the Limits
Integrating Sin² x, Cos² x and Tan² x: current guide
Connected Rates: Differentiate with Respect to Time

For the wider integration owner, see Bukit Timah H2 Integration Techniques, Areas and Volumes. The neighbouring Definite Integration by Substitution article teaches changed bounds, while the JC2 Bukit Timah tuition hub places the lesson in its local learning route. The immutable Clementi three-pupil Mathematics tutoring reference illustrates the teaching approach at another year level.

Official syllabus and editorial note

Source: SEAB H2 Mathematics 9758 syllabus for examination in 2027, §5.3 Integration Techniques. The worked examples and exact answers are original mathematics exercises; the guide does not reproduce actual assessment scripts or guarantee examination outcomes.

JC2 H2 Mathematics Tutor Bukit Timah | Integrating Sin² x, Cos² x and Tan² x