AP Calculus AB and BC

Integral of sin 2x cos 2x: Proof and Steps

The integral of sin(2x)cos(2x) is -(cos(4x))/8 + C. The double-angle identity sin(2A) = 2 sin(A)cos(A) turns sin(2x)cos(2x) into (1/2)sin(4x), which integrates in one step.

∫sin⁡2xcos⁡2x dx=−cos⁡4x8+C\int \sin 2x\cos 2x\,dx = -\frac{\cos 4x}{8} + C

Collapse the product with a double angle

The identity sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A turns a product of sine and cosine into a single sine at twice the angle. With A=2xA = 2x, the factor sin⁡2xcos⁡2x\sin 2x\cos 2x is half of sin⁡4x\sin 4x.

sin⁡2xcos⁡2x=12sin⁡4x\sin 2x\cos 2x = \frac{1}{2}\sin 4x

One sine at a stretched angle is far easier to integrate than a product.

Integrate the single sine

Integrate 12sin⁡4x\frac{1}{2}\sin 4x, dividing by the inner coefficient 4 because ddxcos⁡4x=−4sin⁡4x\frac{d}{dx}\cos 4x = -4\sin 4x.

∫12sin⁡4x dx=12⋅(−cos⁡4x4)+C=−cos⁡4x8+C\int \frac{1}{2}\sin 4x\,dx = \frac{1}{2}\cdot\left(-\frac{\cos 4x}{4}\right) + C = -\frac{\cos 4x}{8} + C

Differentiating −cos⁡4x8-\frac{\cos 4x}{8} gives −18⋅(−sin⁡4x)⋅4=12sin⁡4x=sin⁡2xcos⁡2x-\frac{1}{8}\cdot(-\sin 4x)\cdot 4 = \frac{1}{2}\sin 4x = \sin 2x\cos 2x, back to the start.

The mistake students make

A common attempt substitutes u=sin⁡2xu = \sin 2x, giving du=2cos⁡2x dxdu = 2\cos 2x\,dx and the answer sin⁡22x4+C\frac{\sin^2 2x}{4} + C. That is also correct, but it looks nothing like −cos⁡4x8-\frac{\cos 4x}{8}, so students often assume one of them is wrong. The two differ only by a constant, since sin⁡22x4=1−cos⁡4x8\frac{\sin^2 2x}{4} = \frac{1 - \cos 4x}{8}.

Two right answers, one constant apart

Antiderivatives are unique only up to +C+C. sin⁡22x4\frac{\sin^2 2x}{4} and −cos⁡4x8-\frac{\cos 4x}{8} describe the same family; expanding the first with sin⁡2θ=1−cos⁡2θ2\sin^2\theta = \frac{1-\cos 2\theta}{2} shows the gap is the constant 18\frac{1}{8}.

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An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

Is -(cos 4x)/8 the only correct answer?

No. Substituting u=sin⁡2xu = \sin 2x gives sin⁡22x4+C\frac{\sin^2 2x}{4} + C, and u=cos⁡2xu = \cos 2x gives −cos⁡22x4+C-\frac{\cos^2 2x}{4} + C. All three agree once the arbitrary constant absorbs the fixed difference between them.

Where does the 8 in the denominator come from?

From two halvings: the identity contributes 12\frac{1}{2}, and integrating sin⁡4x\sin 4x divides by the inner coefficient 4. Together 12⋅14=18\frac{1}{2}\cdot\frac{1}{4} = \frac{1}{8}.

Does the same trick work for sin 3x cos 3x?

Yes. sin⁡3xcos⁡3x=12sin⁡6x\sin 3x\cos 3x = \frac{1}{2}\sin 6x, so the integral is −cos⁡6x12+C-\frac{\cos 6x}{12} + C. Any sin⁡kxcos⁡kx\sin kx\cos kx becomes 12sin⁡2kx\frac{1}{2}\sin 2kx.