AP Calculus AB and BC

Integral of sin(2x): Answer, Proof, and Steps

The integral of sin(2x) is -cos(2x)/2 + C. The antiderivative of sine is negative cosine, and because the inside is 2x you divide by the inner coefficient 2. Differentiating -cos(2x)/2 returns sin(2x).

∫sin⁡2x dx=−cos⁡2x2+C\int \sin 2x\,dx = -\frac{\cos 2x}{2} + C

Setting up the antiderivative

Two facts combine here. First, the antiderivative of sine is negative cosine. Second, the inside function is 2x2x, so a u-substitution divides by the inner coefficient.

∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C

Let u=2xu = 2x, so du=2 dxdu = 2\,dx and dx=du2dx = \frac{du}{2}.

∫sin⁡2x dx=∫sin⁡u⋅du2=12∫sin⁡u du=−12cos⁡u+C\int \sin 2x\,dx = \int \sin u \cdot \frac{du}{2} = \frac{1}{2}\int \sin u\,du = -\frac{1}{2}\cos u + C

Substituting u=2xu = 2x back gives the final antiderivative.

∫sin⁡2x dx=−cos⁡2x2+C\int \sin 2x\,dx = -\frac{\cos 2x}{2} + C

Verifying the sign and the 1/2

Differentiate the answer and watch both the sign and the coefficient fall into place.

ddx(−cos⁡2x2)=−12⋅(−sin⁡2x)⋅2=sin⁡2x\frac{d}{dx}\left(-\frac{\cos 2x}{2}\right) = -\frac{1}{2}\cdot\left(-\sin 2x\right)\cdot 2 = \sin 2x

The chain rule contributes the inner factor 22, which cancels the 12\frac{1}{2}. The derivative of cosine is negative sine, and that second negative cancels the leading minus, leaving +sin⁡2x+\sin 2x.

Two sign flips

The minus in front comes from integrating sine, and the derivative of cosine supplies a second minus when you check. The two negatives cancel.

Common mistakes

  • Dropping the negative sign and writing cos⁡2x2\frac{\cos 2x}{2}. Integrating sine gives negative cosine.
  • Forgetting to divide by 22. The inside coefficient from u=2xu = 2x produces the 12\frac{1}{2}; without it, differentiating gives 2sin⁡2x2\sin 2x.
  • Multiplying by 22 instead of dividing. Differentiating the inner function multiplies by 22, so integrating divides by it.

Every answer on this page is machine checked

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

Why is the integral of sin(2x) negative?

Because the antiderivative of sin⁡x\sin x is −cos⁡x-\cos x. That leading minus carries through the substitution, giving −cos⁡2x2+C-\frac{\cos 2x}{2} + C.

What is the integral of sin(2x) from 0 to pi/2?

∫0π2sin⁡2x dx=[−cos⁡2x2]0π2=−cos⁡π2+cos⁡02=12+12=1\int_0^{\frac{\pi}{2}} \sin 2x\,dx = \left[-\frac{\cos 2x}{2}\right]_0^{\frac{\pi}{2}} = -\frac{\cos \pi}{2} + \frac{\cos 0}{2} = \frac{1}{2} + \frac{1}{2} = 1.

Does integrating sin(2x) need the chain rule?

It needs the reverse of the chain rule, which is u-substitution. The substitution u=2xu = 2x introduces the 12\frac{1}{2} that a plain −cos⁡2x-\cos 2x would miss.