Practice Maths

← Back to Integration Techniques›Specialist Maths

Integration Using Trigonometric Identities

Key Terms

Power-reduction identities
convert even powers to double-angle forms:
   sin²x = (1 − cos 2x)/2    cos²x = (1 + cos 2x)/2
Pythagorean identity
sin²x + cos²x = 1. Use to reduce odd powers: sin³x = sin x(1−cos²x).
Double angle
sin 2x = 2 sin x cos x, so sin x cos x = sin(2x)/2.
Integrating sinnx or cosnx
   — Even n: use power-reduction identities repeatedly.
   — Odd n: peel off one factor, use Pythagorean identity, then substitute.
Product-to-sum
sin A cos B = (1/2)[sin(A+B) + sin(A−B)].   Similar for cos·cos and sin·sin.
Integrating tan2x
tan²x = sec²x − 1, so ∫tan²x dx = tan x − x + C.
Essential Identities for Integration:
  • sin²x = (1 − cos 2x)/2
  • cos²x = (1 + cos 2x)/2
  • sin 2x = 2 sin x cos x   ⇒   sin x cos x = (sin 2x)/2
  • sin²x cos²x = (sin² 2x)/4 = (1 − cos 4x)/8
  • tan²x = sec²x − 1
  • sin A cos B = (1/2)[sin(A+B) + sin(A−B)]
  • cos A cos B = (1/2)[cos(A−B) + cos(A+B)]
  • sin A sin B = (1/2)[cos(A−B) − cos(A+B)]
Hot Tip For even powers of sin or cos, apply the power-reduction formula once to get cos 2x terms, then apply again if needed. For odd powers, always peel off exactly one factor of sin x (or cos x) to pair with the dx, then rewrite the even-power remainder using sin²x + cos²x = 1, and substitute u = cos x (or u = sin x). This always works.

Worked Example 1 — Integrating sin²x

Find ∫sin²(x) dx.

Use sin²x = (1 − cos 2x)/2:

∫sin²x dx = ∫(1 − cos 2x)/2 dx = (1/2)∫1 dx − (1/2)∫cos 2x dx

= (1/2)x − (1/2) × (1/2)sin 2x + C

= x/2 − (sin 2x)/4 + C

Worked Example 2 — Integrating sin³x (odd power)

Find ∫sin³(x) dx.

Write sin³x = sin x × sin²x = sin x(1 − cos²x). Let u = cos x, du = −sin x dx.

∫sin³x dx = ∫(1 − cos²x) sin x dx = ∫(1 − u²)(−du) = −u + u³/3 + C

= −cos x + cos³x/3 + C

Worked Example 3 — Product-to-sum

Find ∫sin(2x)cos(x) dx.

Use sin A cos B = (1/2)[sin(A+B) + sin(A−B)] with A = 2x, B = x:

sin(2x)cos(x) = (1/2)[sin(3x) + sin(x)]

∫sin(2x)cos(x) dx = (1/2)∫[sin 3x + sin x] dx = (1/2)[−cos(3x)/3 − cos x] + C

= −cos(3x)/6 − cos(x)/2 + C

Why Direct Integration Fails for Powers of Trig Functions

The standard integral ∫sin x dx = −cos x + C is simple. But ∫sin²x dx cannot be found by the same method, because sin²x is not itself a derivative of any obvious function. The power-reduction identity sin²x = (1 − cos 2x)/2 transforms the integrand into a form involving cos 2x, which integrates to (sin 2x)/2 — a standard result. This is the core strategy: use identities to rewrite the integrand in terms of integrable pieces.

Even Powers: Power-Reduction Applied Repeatedly

For ∫cos4x dx, apply the reduction cos²x = (1+cos 2x)/2 twice:

cos4x = (cos²x)² = [(1+cos 2x)/2]² = (1 + 2cos 2x + cos²2x)/4.

Then cos²2x = (1+cos 4x)/2, so cos4x = (1 + 2cos 2x + (1+cos 4x)/2)/4 = 3/8 + (cos 2x)/2 + (cos 4x)/8.

This integrates term by term: ∫cos4x dx = 3x/8 + (sin 2x)/4 + (sin 4x)/32 + C.

The pattern is clear: each application of power-reduction introduces a double-angle argument, and you may need to apply the formula again for higher even powers.

Odd Powers: Peel Off One Factor

For odd powers like sin5x, write sin5x = sin x × sin4x = sin x × (sin²x)² = sin x(1 − cos²x)². Then substitute u = cos x, du = −sin x dx, converting the entire integral to a polynomial in u. The result integrates easily.

The key structural point: peeling off one sin x gives the du for the substitution u = cos x, and the remaining sinevenx becomes a polynomial in cos²x, which rewrites as (1−u²) terms after the identity.

Mixed Powers: sinmx·cosnx

  • If m (the power of sin) is odd: peel off one sin x, substitute u = cos x.
  • If n (the power of cos) is odd: peel off one cos x, substitute u = sin x.
  • If both are even: use power-reduction on both.

For sin²x cos²x, the most efficient method is: sin²x cos²x = (sin x cos x)² = (sin 2x/2)² = sin²2x/4. Then apply power-reduction to sin²2x.

Product-to-Sum Formulas

When integrating a product of trig functions with different arguments, such as sin(3x)cos(x) or cos(5x)cos(2x), the product-to-sum formulas transform the product into a sum of single-argument trig functions, which integrate directly. These identities come directly from the sum and difference formulas for sin and cos.

The tan²x Integral

Using the Pythagorean identity in the form tan²x + 1 = sec²x gives tan²x = sec²x − 1. Since ∫sec²x dx = tan x + C, we get ∫tan²x dx = tan x − x + C. This approach generalises: higher powers of tan can always be reduced using tan²x = sec²x − 1.

Exam Tip: When you see ∫sin²x cos²x dx, the fastest path is to use sin x cos x = (sin 2x)/2 first, so sin²x cos²x = sin²(2x)/4, then apply sin²(2x) = (1−cos 4x)/2. The result is immediate. Expanding and using separate power-reduction formulas takes much longer.
Exam Tip: For area questions involving sin²x and cos²x on [0, π/2], use symmetry: ∫0π/2 sin²x dx = ∫0π/2 cos²x dx = π/4. This follows from sin²x + cos²x = 1 and the half-period symmetry.

Mastery Practice

  1. Find ∫sin²(x) dx. Fluency

  2. Find ∫cos²(x) dx. Fluency

  3. Find ∫sin²(x)cos²(x) dx. Fluency

  4. Find ∫sin³(x) dx using the Pythagorean identity. Fluency

  5. Find ∫cos&sup4;(x) dx. Understanding

  6. Evaluate ∫0π/4 tan²(x) dx. Understanding

  7. Find ∫sin(3x)cos(x) dx using a product-to-sum formula. Understanding

  8. Find ∫cos²(2x)sin²(2x) dx. Understanding

  9. Find the area enclosed between y = sin²(x) and y = cos²(x) over the interval [0, π/2]. Problem Solving

  10. Show that ∫0π sin²(nx) dx = π/2 for any positive integer n. Hence evaluate ∫0π cos²(nx) dx. Problem Solving

See Answers ➔