Integration · Trigonometric Integrals

Trigonometric Integrals — Powers of Sin, Cos, Tan & Sec

Integrals involving powers of trig functions follow systematic strategies based on whether the powers are odd or even, and which Pythagorean identity to apply. Master these patterns and a whole class of otherwise-difficult integrals becomes algorithmic.

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§ 01Essential Trigonometric Identities

Three Pythagorean identities and two half-angle identities are the engine behind all trig integral strategies. Internalise these before anything else.

Pythagorean Identities

sin²x + cos²x = 1
sin²x = 1 − cos²x
tan²x + 1 = sec²x
tan²x = sec²x − 1
1 + cot²x = csc²x
cot²x = csc²x − 1

Half-Angle (Power-Reducing) Identities

sin²x = (1 − cos 2x) / 2
cos²x = (1 + cos 2x) / 2
sin x cos x = sin(2x) / 2
cos²x − sin²x = cos(2x)
Prerequisites U-substitution must be fluent. You should recognise that ∫f′(x)/f(x) dx = ln|f(x)| + C and know the derivatives of all six trig functions including sec x and tan x.

§ 02Integrals of sinᵐx cosⁿx

The strategy depends on the parities of the exponents m and n.

Case A
n is odd (cosine power is odd)

Save one cos x for du. Convert remaining cos²x to (1−sin²x). Let u = sin x.

∫sinᵐx cos²ᵏ⁺¹x dx → u = sin x, remaining factors become (1−u²)ᵏ.

Case B
m is odd (sine power is odd)

Save one sin x for du. Convert remaining sin²x to (1−cos²x). Let u = cos x.

∫sin²ᵏ⁺¹x cosⁿx dx → u = cos x, du = −sin x dx.

Case C
Both m and n are even

Use half-angle identities to reduce powers: sin²x = (1−cos2x)/2, cos²x = (1+cos2x)/2. Repeat as needed.

Case D
Products: sin(mx)cos(nx)

Use product-to-sum: sin A cos B = ½[sin(A+B)+sin(A−B)]. Integrate each term separately.

Case A∫sin²x cos³x dx — cosine odd
  1. n = 3 (odd). Save one cos x: ∫sin²x · cos²x · cos x dx.
  2. cos²x = 1−sin²x. So: ∫sin²x(1−sin²x)cos x dx.
  3. u = sin x, du = cos x dx: ∫u²(1−u²) du = ∫(u²−u⁴) du.
  4. = u³/3 − u⁵/5 + C.
∫sin²x cos³x dx = sin³x/3 − sin⁵x/5 + C
Case B∫sin³x cos²x dx — sine odd
  1. m = 3 (odd). Save one sin x: ∫sin²x · cos²x · sin x dx.
  2. sin²x = 1−cos²x: ∫(1−cos²x)cos²x sin x dx.
  3. u = cos x, du = −sin x dx: ∫(1−u²)u²(−du) = −∫(u²−u⁴) du.
  4. = −u³/3 + u⁵/5 + C.
∫sin³x cos²x dx = −cos³x/3 + cos⁵x/5 + C
Case C∫sin²x dx — both even
  1. Use half-angle: sin²x = (1 − cos 2x)/2.
  2. ∫(1 − cos 2x)/2 dx = x/2 − sin(2x)/4 + C.
∫sin²x dx = x/2 − sin(2x)/4 + C
Case C∫cos⁴x dx — even power
  1. cos⁴x = (cos²x)² = [(1+cos2x)/2]² = (1 + 2cos2x + cos²2x)/4.
  2. cos²2x = (1+cos4x)/2. So cos⁴x = (1 + 2cos2x + (1+cos4x)/2)/4 = 3/8 + cos2x/2 + cos4x/8.
  3. ∫cos⁴x dx = 3x/8 + sin(2x)/4 + sin(4x)/32 + C.
∫cos⁴x dx = 3x/8 + sin(2x)/4 + sin(4x)/32 + C

§ 03Integrals of tanᵐx secⁿx

The key identity is tan²x = sec²x − 1. The strategy mirrors the sin/cos case.

Case A
n is even (secant power even)

Save sec²x for du. Convert remaining sec²x to (1+tan²x). Let u = tan x, du = sec²x dx.

Case B
m is odd (tangent power odd)

Save sec x tan x for du. Convert remaining tan²x to (sec²x−1). Let u = sec x, du = sec x tan x dx.

Case C
m even, n odd — harder

No clean substitution exists. Use tan²x = sec²x − 1 repeatedly to reduce, then use the reduction formula for ∫secⁿx dx or integration by parts.

Special
∫sec x dx and ∫tan x dx

∫tan x dx = −ln|cos x| + C = ln|sec x| + C

∫sec x dx = ln|sec x + tan x| + C

tan/sec∫tan²x dx
  1. tan²x = sec²x − 1.
  2. ∫(sec²x − 1) dx = tan x − x + C.
∫tan²x dx = tan x − x + C
tan/sec∫tan³x sec⁴x dx — secant even
  1. n = 4 (even). Save sec²x: ∫tan³x · sec²x · sec²x dx.
  2. sec²x = 1+tan²x: ∫tan³x(1+tan²x)sec²x dx.
  3. u = tan x, du = sec²x dx: ∫u³(1+u²) du = ∫(u³+u⁵) du.
  4. = u⁴/4 + u⁶/6 + C.
∫tan³x sec⁴x dx = tan⁴x/4 + tan⁶x/6 + C
tan/sec∫tan³x sec x dx — tangent odd
  1. m = 3 (odd). Save sec x tan x: ∫tan²x · sec x · (sec x tan x) dx.
  2. tan²x = sec²x−1: ∫(sec²x−1)sec x · (sec x tan x) dx.
  3. u = sec x, du = sec x tan x dx: ∫(u²−1)u du = ∫(u³−u) du.
  4. = u⁴/4 − u²/2 + C.
∫tan³x sec x dx = sec⁴x/4 − sec²x/2 + C
tan/sec∫sec³x dx — reduction formula needed

This classic integral uses IBP and the Pythagorean identity.

  1. Write sec³x = sec x · sec²x. IBP: u = sec x, dv = sec²x dx → du = sec x tan x dx, v = tan x.
  2. ∫sec³x dx = sec x tan x − ∫tan²x sec x dx.
  3. tan²x = sec²x−1: ∫sec³x dx = sec x tan x − ∫(sec²x−1)sec x dx = sec x tan x − ∫sec³x dx + ∫sec x dx.
  4. Let I = ∫sec³x dx: 2I = sec x tan x + ln|sec x + tan x|.
∫sec³x dx = ½[sec x tan x + ln|sec x + tan x|] + C

§ 04Standard Results and Definite Integrals

IntegralResultMethod
∫sin x dx−cos x + CDirect
∫cos x dxsin x + CDirect
∫tan x dxln|sec x| + CSub: u = cos x
∫sec x dxln|sec x + tan x| + CMultiply by (sec x+tan x)/(sec x+tan x)
∫sec²x dxtan x + CDirect
∫csc²x dx−cot x + CDirect
∫sec x tan x dxsec x + CDirect
∫sin²x dxx/2 − sin(2x)/4 + CHalf-angle
∫cos²x dxx/2 + sin(2x)/4 + CHalf-angle
∫tan²x dxtan x − x + Ctan²x = sec²x−1
Definite∫₀^(π/2) sin²x dx — Wallis-type integral
  1. Antiderivative: x/2 − sin(2x)/4.
  2. [x/2 − sin(2x)/4]₀^(π/2) = (π/4 − sin(π)/4) − (0 − 0) = π/4 − 0 = π/4.
∫₀^(π/2) sin²x dx = π/4
Definite∫₀^(π/2) sin³x dx
  1. Odd power, save one sin x, convert sin²x = 1−cos²x.
  2. u = cos x, du = −sin x dx. Limits: x=0→u=1; x=π/2→u=0.
  3. ∫₁⁰ (1−u²)(−du) = ∫₀¹ (1−u²) du = [u − u³/3]₀¹ = 1 − 1/3 = 2/3.
∫₀^(π/2) sin³x dx = 2/3
Wallis's Formula — Shortcut for ∫₀^(π/2) sinⁿx dx For even n: π/2 · (n−1)!!/n!! where !! is double factorial. For odd n: (n−1)!!/n!!. For instance ∫₀^(π/2) sin⁴x dx = (3/4)(1/2)(π/2) = 3π/16.

§ 05Common Mistakes

Mistake 1 — Using the wrong Pythagorean identityFor sin/cos integrals use sin²+cos²=1. For tan/sec use tan²+1=sec². Swapping them creates unsolvable integrals. Always identify which trig functions are present before choosing an identity.
Mistake 2 — Forgetting the sign when u = cos xIf u = cos x then du = −sin x dx, so sin x dx = −du. The negative sign must be tracked. A missing minus sign is the most frequent algebraic error in Case B integrals.
Mistake 3 — Trying substitution when both powers are even∫sin²x cos²x dx cannot be simplified with u = sin x or u = cos x — no derivative is present to serve as du. The only method is half-angle identities. Recognise that even-even integrals always require half-angle reduction.
Mistake 4 — Missing the ½ in half-angle identitiessin²x = (1−cos2x)/2, not (1−cos2x). The division by 2 is essential. Writing (1−cos2x) gives a result twice too large.

§ 06Test Yourself — 10-Question Quiz

Score: 0 / 0
Q1 of 10

What is the half-angle identity for sin²x?

sin²x = (1 − cos2x)/2. This follows from the cosine double angle formula: cos2x = 1 − 2sin²x, rearranged gives sin²x = (1−cos2x)/2.

Q2 of 10

For ∫sin³x cos²x dx, which substitution is correct?

m = 3 is odd (Case B). Save one sin x as part of du. With u = cos x, du = −sin x dx. The remaining sin² x = 1−cos²x = 1−u².

Q3 of 10

Evaluate ∫sin²x dx. The answer has form ax + b·sin(2x) + C. What is a?

  1. sin²x = (1−cos2x)/2. ∫(1−cos2x)/2 dx = x/2 − sin(2x)/4 + C.
  2. a = 1/2 = 0.5.
Q4 of 10

What is ∫tan²x dx?

Use tan²x = sec²x − 1. ∫(sec²x−1)dx = tan x − x + C.

Q5 of 10

Evaluate ∫₀^(π/2) sin²x dx. Enter exact decimal.

  1. [x/2 − sin2x/4]₀^(π/2) = π/4 − 0 = π/4 ≈ 0.785.
Q6 of 10

For ∫tan³x sec⁴x dx, which substitution is most efficient?

n = 4 is even (Case A for tan/sec). Save sec²x as du: u = tan x, du = sec²x dx. Replace remaining sec²x with 1+tan²x = 1+u².

Q7 of 10

What is the antiderivative of sec x?

The standard result ∫sec x dx = ln|sec x + tan x| + C is derived by multiplying numerator and denominator by (sec x + tan x), giving an integrand of the form f′/f.

Q8 of 10

Evaluate ∫₀^(π/2) sin³x dx. Enter exact decimal.

  1. Odd sine power. u = cos x, limits 1→0.
  2. ∫₀¹(1−u²)du = [u−u³/3]₀¹ = 1−1/3 = 2/3 ≈ 0.667.
Q9 of 10

Which strategy handles ∫sin⁴x cos⁴x dx (both exponents even)?

Both powers even: Case C. Write sin⁴x cos⁴x = (sin x cos x)⁴ = (sin 2x/2)⁴ = sin⁴(2x)/16, then use half-angle again on sin⁴(2x). No substitution is possible since no derivative factor is present.

Q10 of 10

In ∫sin²x cos³x dx with u = sin x, the factor cos³x is split as cos²x · cos x. What does cos²x become in terms of u?

u = sin x, so sin²x = u². From sin²x + cos²x = 1: cos²x = 1 − sin²x = 1 − u². The remaining cos x provides du = cos x dx, and cos²x = 1 − u² enters the integrand.

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