01 · Cosine squared
Find ∫cos²x dx.
Hint
Use the version with a plus sign.
Worked solution
x/2 + sin2x/4 + C.
Understand · explore · practise
Integration using trigonometric identities
Integrate sin squared, cos squared, tan squared, trigonometric products and fourth powers using identities. Original worked examples, interactive graphs and exact-answer practice.
Before you startDouble-angle and compound-angle identities; standard and linear-argument integrals.
01 / Rewrite the integrand first
An equal expression can be easier to integrate.
The integral of sin²x is not −cos²x. Instead, a double-angle identity turns the square into a constant and a cosine. Integrate those simpler terms, keeping the chain-rule factor.
The two forms in the model have exactly the same values. Switch between them or overlay their graphs; rewriting changes the calculation, not the function.
One integrand, two formsExplore
At x = 0: original = 0; rewritten = 0.
A primitive is x/2 − sin(2x)/4, plus C.
Inputs are radians. Small rounding differences in the numerical readout come from floating-point calculation; the identities are exact. The plotted interval is 0 ≤ x ≤ π.
02 / Integrate sine and cosine squares
Use a double-angle identity in reverse.
sin²x = (1 − cos2x)/2
cos²x = (1 + cos2x)/2
Find ∫sin²x dx.Worked example
∫sin²x dx = (1/2)∫(1 − cos2x) dx
Replace the square using the identity.
∫1 dx = x and ∫cos2x dx = (1/2)sin2x
The double angle produces another factor 1/2.
x/2 − sin2x/4 + C
Differentiate: 1/2 − cos2x/2 = sin²x.
Watch: transform the double-angle curve
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Keep the inner-angle factor
Square identities and linear integration work together.
Find ∫sin²(3x) dx.Worked example
sin²(3x) = (1 − cos6x)/2
Double the whole angle 3x.
The cosine primitive is sin6x/6
Its inner slope is 6.
x/2 − sin6x/12 + C
Keep the outer factor 1/2.
Find ∫cos²(2x + 1) dx.Worked example
cos²(2x + 1) = [1 + cos(4x + 2)]/2
Double both terms inside the angle.
x/2 + sin(4x + 2)/8 + C
The linear argument has derivative 4.
02 · A multiplier outside
Find ∫6sin²(2x) dx.
Hint
Rewrite as 3(1 − cos4x).
Worked solution
3x − (3/4)sin4x + C.
03 · Double the whole angle
Rewrite cos²(x − 2) before integrating.
Hint
2(x − 2) = 2x − 4.
Worked solution
[1 + cos(2x − 4)]/2, giving x/2 + sin(2x − 4)/4 + C.
04 / Integrate tan² and cot²
Use the reciprocal Pythagorean identities.
tan²x = sec²x − 1
cot²x = cosec²x − 1
Find ∫tan²(4x) dx.Worked example
tan²(4x) = sec²(4x) − 1
Both new terms have standard integrals.
(1/4)tan4x − x + C
Work on an interval without tan4x poles.
04 · Cotangent squared
Find ∫cot²(2x) dx.
Hint
The integral of cosec²(2x) has a minus sign.
Worked solution
−(1/2)cot2x − x + C, on an interval avoiding sin2x = 0.
05 · A constant shift
Find ∫(2tan²x + 3) dx.
Hint
Rewrite the integrand as 2sec²x + 1.
Worked solution
2tan x + x + C, away from the poles.
05 / Turn products into sums
Add or subtract compound-angle formulas.
2sin A cos B = sin(A + B) + sin(A − B)
2cos A cos B = cos(A + B) + cos(A − B)
2sin A sin B = cos(A − B) − cos(A + B)
Find ∫sin3x cos x dx.Worked example
sin3x cos x = (sin4x + sin2x)/2
Use A = 3x and B = x.
−cos4x/8 − cos2x/4 + C
Integrate the two frequencies separately.
06 · Cosine product
Find ∫cos3x cos x dx.
Hint
Use (cos4x + cos2x)/2.
Worked solution
sin4x/8 + sin2x/4 + C.
07 · Sine product
Find ∫sin3x sin x dx.
Hint
The difference is (cos2x − cos4x)/2.
Worked solution
sin2x/4 − sin4x/8 + C.
06 / Reduce fourth powers in two stages
After the first identity, another square remains.
Find ∫sin⁴x dx.Worked example
sin⁴x = [(1 − cos2x)/2]²
Square the entire right-hand side.
= [1 − 2cos2x + cos²2x]/4
Expand, keeping the cross term.
cos²2x = (1 + cos4x)/2
Apply the square identity again.
sin⁴x = (3 − 4cos2x + cos4x)/8
Collect the constant and cosine terms.
∫sin⁴x dx = 3x/8 − sin2x/4 + sin4x/32 + C
Each frequency supplies its own integration factor.
08 · Cosine to the fourth
Find ∫cos⁴x dx.
Hint
cos⁴x = (3 + 4cos2x + cos4x)/8.
Worked solution
3x/8 + sin2x/4 + sin4x/32 + C.
09 · Avoid a missing cross term
Why is sin⁴x not (1 − cos²2x)/4?
Hint
Expand (1 − cos2x)².
Worked solution
The correct expansion is (1 − 2cos2x + cos²2x)/4. Squaring a difference does not give a difference of squares.
07 / Handle sin²x cos²x efficiently
A double-angle product can reduce the work.
sin²x cos²x = (1/4)sin²2x = (1 − cos4x)/8
Find ∫sin²x cos²x dx.Worked example
Replace the product by (1 − cos4x)/8
One double-angle identity removes the product.
x/8 − sin4x/32 + C
Integrate the constant and cosine.
10 · A scaled product
Find ∫8sin²(2x)cos²(2x) dx.
Hint
The integrand simplifies to 1 − cos8x.
Worked solution
x − sin8x/8 + C.
08 / Expand before choosing an identity
Mixed terms may need different standard methods.
Find ∫(2 + sin x)² dx.Worked example
(2 + sin x)² = 4 + 4sin x + sin²x
Keep the middle term.
= 9/2 + 4sin x − (1/2)cos2x
Use the half-angle identity only on the squared term.
(9/2)x − 4cos x − sin2x/4 + C
Integrate term by term.
11 · An expanded cosine expression
Find ∫(1 − cos x)² dx.
Hint
It becomes 3/2 − 2cos x + (1/2)cos2x.
Worked solution
(3/2)x − 2sin x + sin2x/4 + C.
09 / Keep exact radian values
Evaluate the new primitive at the original limits.
Evaluate ∫₀^(π/2) sin⁴x dx.Worked example
Use F(x) = 3x/8 − sin2x/4 + sin4x/32
The indefinite work is already done.
sinπ = sin2π = sin0 = 0
All sine terms vanish at these endpoints.
The exact integral is 3π/16
A numerical decimal is unnecessary.
12 · A square over a half period
Evaluate ∫₀^π cos²x dx.
Hint
Use x/2 + sin2x/4.
Worked solution
π/2.
13 · A product of squares
Evaluate ∫₀^(π/2) sin²x cos²x dx.
Hint
Use x/8 − sin4x/32.
Worked solution
π/16.
10 / Check identities, factors and domains
A correct rewrite must remain correct after integration.
Differentiate your answer and use an identity to simplify it back to the original integrand. This catches lost factors from double angles and sign errors. For tan or cot, also check the whole interval for poles; an algebraic rewrite does not remove them.
14 · A missing chain factor
A student writes ∫sin²x dx = x/2 − sin2x/2 + C. What is wrong?
Hint
Differentiate the sine term.
Worked solution
The derivative is 1/2 − cos2x, not sin²x. The sine coefficient should be −1/4.
15 · A singular definite integral
Can ∫₀^π tan²x dx be evaluated as [tan x − x]₀^π?
Hint
tan x has a pole at π/2.
Worked solution
No. The interval crosses a singularity; the ordinary improper integral diverges. Endpoint subtraction would give a meaningless negative result for this nonnegative integrand.
11 / Match the identity to the shape
Squares, products and fourth powers suggest different rewrites.
- sin² and cos²: use half-angle identities.
- tan² and cot²: use reciprocal Pythagorean identities.
- Different-angle products: use sum/product formulas.
- Fourth powers: reduce squares twice.
- Expand brackets carefully, integrate each term and differentiate to check.
16 · Choose a useful first step
Which first steps suit ∫cos⁴(2x) dx and ∫sin5x cos2x dx?
Hint
One is a fourth power; the other is a different-angle product.
Worked solution
For the first, reduce cos⁴u to (3 + 4cos2u + cos4u)/8 with u = 2x, then integrate. For the second, rewrite the product as (sin7x + sin3x)/2. Keep the separate inner-angle factors.
Section 1 of 11 · Rewrite the integrand first