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Trigonometric integrals

Integral of sin 2x: Formula, Derivation and Examples

The integral of sin 2x is -cos(2x)/2 + C. See the u-substitution, why sin² x + C and -cos² x + C are also correct, and worked definite integral examples.

Trigonometric integrals Updated 2 min read

At a glance

\int \sin 2x\,dx = -\frac{1}{2}\cos 2x + C
Antiderivative
-\dfrac{1}{2}\cos 2x + C
Also correct
\sin^2 x + C or -\cos^2 x + C
Method
u = 2x, du = 2\,dx
From 0 to π/2
1
General form
\int\sin(kx)\,dx = -\dfrac{1}{k}\cos(kx) + C
On this page
  1. Graph of sin 2x and its antiderivative
  2. Derivation with u-substitution
  3. Check by differentiating
  4. Three correct answers
  5. Definite integrals
  6. Worked examples
  7. Common mistakes

The integral of \sin 2x is -\dfrac{1}{2}\cos 2x + C. It is the integral of sin x with one extra step to handle the 2 inside.

\int \sin 2x\,dx = -\frac{1}{2}\cos 2x + C

Graph of sin 2x and its antiderivative

Area under y = sin 2x from 0 to π/2, with the antiderivative y = -cos(2x)/2
The shaded arch has area 1, half of the area under one arch of \sin x, because it is squeezed to half the width. The dashed antiderivative rises from -\frac{1}{2} to \frac{1}{2}.

Derivation with u-substitution

  1. Let u = 2x. Then du = 2\,dx, so dx = \frac{1}{2}du.
  2. Substitute: \displaystyle\int\sin 2x\,dx = \frac{1}{2}\int\sin u\,du.
  3. Integrate: \frac{1}{2}(-\cos u) + C.
  4. Put u = 2x back: -\frac{1}{2}\cos 2x + C.

The same steps give the general rule \int\sin(kx)\,dx = -\frac{1}{k}\cos(kx) + C. See the u-substitution rule.

Check by differentiating

\frac{d}{dx}\left(-\frac{1}{2}\cos 2x\right) = -\frac{1}{2}\cdot(-\sin 2x)\cdot 2 = \sin 2x \checkmark

The chain rule factor 2 cancels the \frac{1}{2}. The forward direction is on the derivative of sin 2x page.

Three correct answers

Using the double-angle identity \sin 2x = 2\sin x\cos x:

  • With u = \sin x, du = \cos x\,dx: \int 2u\,du = \sin^2 x + C.
  • With u = \cos x, du = -\sin x\,dx: \int -2u\,du = -\cos^2 x + C.

These look different from -\frac{1}{2}\cos 2x, but they are not. Since \cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1:

-\frac{1}{2}\cos 2x = \sin^2 x - \frac{1}{2} = -\cos^2 x + \frac{1}{2}

So the three answers differ only by constants. That is allowed: the +C absorbs the difference. This also explains why the derivative of sin² x is \sin 2x.

Definite integrals

\int_0^{\pi/2}\sin 2x\,dx = -\frac{1}{2}\cos\pi + \frac{1}{2}\cos 0 = \frac{1}{2} + \frac{1}{2} = 1
\int_0^{\pi/4}\sin 2x\,dx = -\frac{1}{2}\cos\frac{\pi}{2} + \frac{1}{2} = \frac{1}{2}
\int_0^{\pi}\sin 2x\,dx = 0

For a definite integral, all three forms of the antiderivative give the same number, because the constant cancels when you subtract.

Worked examples

Example 1. \int(\sin 2x + \cos 2x)\,dx = -\frac{1}{2}\cos 2x + \frac{1}{2}\sin 2x + C.

Example 2. \int 6\sin 3x\,dx = 6\cdot\left(-\frac{1}{3}\cos 3x\right) + C = -2\cos 3x + C.

Example 3: sin 2x cos x. Rewrite as 2\sin x\cos^2 x and let u = \cos x:

\int\sin 2x\cos x\,dx = -\frac{2}{3}\cos^3 x + C

Check: \frac{d}{dx}\left(-\frac{2}{3}\cos^3 x\right) = 2\cos^2 x\sin x = \sin 2x\cos x.

Example 4: sin 2x cos 2x. Use \sin 2x\cos 2x = \frac{1}{2}\sin 4x, so the integral is -\frac{1}{8}\cos 4x + C.

Common mistakes

  1. Answering -\cos 2x + C. You forgot the \frac{1}{2}. Its derivative is 2\sin 2x.
  2. Answering -2\cos 2x + C. That multiplies by 2, which is what differentiating would do.
  3. Thinking \sin^2 x is wrong because it differs from the book. Differentiate it: you get \sin 2x, so it is right.
  4. Writing 2\sin x for \sin 2x. \sin 2x = 2\sin x\cos x, not 2\sin x.

Practise on u-substitution problems.

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Frequently asked questions

What is the integral of sin 2x?
The integral of \sin 2x is -\frac{1}{2}\cos 2x + C.

Why do I get sin² x when I integrate sin 2x?
If you write \sin 2x = 2\sin x\cos x and use u = \sin x, you get \sin^2 x + C. This is also correct, because \sin^2 x = \frac{1}{2} - \frac{1}{2}\cos 2x, which differs from -\frac{1}{2}\cos 2x only by the constant \frac{1}{2}.

What is the integral of sin 2x from 0 to π/2?
It is \left[-\frac{1}{2}\cos 2x\right]_0^{\pi/2} = \frac{1}{2} + \frac{1}{2} = 1.

What is the integral of 2 sin x cos x?
It is the same integral, since 2\sin x\cos x = \sin 2x. Any of -\frac{1}{2}\cos 2x, \sin^2 x or -\cos^2 x, plus C.

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Related integrals

Integral of sin x \int \sin x\,dx = -\cos x + C The integral of sin x is -cos x + C. See where the minus sign comes from, the area of one arch, sin(ax + b) by substitution and worked definite integrals. Integral of sin² x \int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C The integral of sin² x is x/2 - sin(2x)/4 + C. See the power-reducing identity step by step, a check by differentiating, sin²(ax) and definite integrals. Integral of cos x \int \cos x\,dx = \sin x + C The integral of cos x is sin x + C. See why there is no minus sign, the area of one hump, cos(ax + b) by substitution and worked definite integral examples.