Integration using trig identities

sin² x, cos² x and tan² x cannot be integrated directly. Rewrite them with an identity first, then integrate.

Identities to use

The first two come from cos 2x ≡ 1 - 2sin² x ≡ 2cos² x - 1.

Double the angle inside: cos² 3x ≡ (1 + cos 6x)/2 and sin² 4x ≡ (1 - cos 8x)/2.

∫ tan² x dx = ∫ (sec² x - 1) dx = tan x - x + c.

Find ∫ sin² x dx.

Answer: x/2 - (1/4)sin 2x + c

Key idea: Rewrite sin², cos² and tan² with double angle or Pythagorean identities before integrating.

Find ∫ cos² x dx.
- x/2 + (1/4)sin 2x + c
- (1/3)cos³ x + c
- x/2 - (1/4)sin 2x + c
- sin² x + c
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x/2 + (1/4)sin 2x + c. cos² x = 1/2 + (1/2)cos 2x, which integrates to x/2 + (1/4)sin 2x.
Which is an integral of tan² x with respect to x?
- tan x + x + c
- (1/3)tan³ x + c
- tan x - x + c
- sec² x - x + c
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tan x - x + c. tan² x = sec² x - 1.
Evaluate ∫₀^{π} sin² x dx, to 3 decimal places.
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1.571. The integral is x/2 - (1/4)sin 2x. From 0 to π: π/2 = 1.571.
Using tan² x = sec² x - 1, show that ∫₀^{π/4} tan² x dx = 1 - π/4 and give its value to 3 decimal places.
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0.215. The integral is tan x - x. From 0 to π/4: 1 - π/4 = 0.215.
The region under y = cos² 3x between x = 0 and x = π/6 has what area? Give it to 3 decimal places.
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0.262. cos² 3x = (1 + cos 6x)/2, which integrates to x/2 + (1/12)sin 6x. From 0 to π/6: π/12 + 0 = 0.262.
Evaluate ∫₀^{π/4} sin x cos x dx. Give your answer as a fraction in its simplest form.
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1/4. sin x cos x = (1/2)sin 2x, which integrates to -(1/4)cos 2x. From 0 to π/4: 0 - (-1/4) = 1/4.
Which identity would you use to find ∫ sin² 3x dx?
- sin² 3x ≡ (1 - cos 3x)/2
- sin² 3x ≡ (1 - cos 6x)/2
- sin² 3x ≡ 1 + cos² 3x
- sin² 3x ≡ (1 + cos 6x)/2
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sin² 3x ≡ (1 - cos 6x)/2. Replace x by 3x in sin² x ≡ (1 - cos 2x)/2.

Last updated 7 November 2026. Spotted a mistake? Tell us.