Integral of sin²x

∫sin⁡2(x) dx=12x−14sin⁡(2x)+C\int \sin^{2}\left(x\right)\,dx = \frac{1}{2}x - \frac{1}{4}\sin\left(2 x\right) + C

Use the power-reduction identity sin²x = (1 − cos 2x)/2 first, then each piece integrates directly.

Step by step

  1. Trig identity

    Rewrite powers/products of sin and cos with the power-reduction, double-angle or Pythagorean identities (sin²u = (1 − cos 2u)/2, cos²u = (1 + cos 2u)/2, sin²u + cos²u = 1).

    sin⁡2(x)=−12cos⁡(2x)+12\sin^{2}\left(x\right) = -\frac{1}{2}\cos\left(2 x\right) + \frac{1}{2}
  2. Sum rule

    Integrate term by term.

    ∫(−12cos⁡(2x)+12) dx=∫−12cos⁡(2x) dx+∫12 dx\int \left(-\frac{1}{2}\cos\left(2 x\right) + \frac{1}{2}\right)\,dx = \int -\frac{1}{2}\cos\left(2 x\right)\,dx + \int \frac{1}{2}\,dx
  3. Constant multiple rule

    Move the constant outside the integral.

    ∫−12cos⁡(2x) dx=−12∫cos⁡(2x) dx\int -\frac{1}{2}\cos\left(2 x\right)\,dx = -\frac{1}{2} \int \cos\left(2 x\right)\,dx
  4. Trig rule

    ∫cos u du = sin u. Divide by the inner slope 2 (reverse chain rule).

    ∫cos⁡(2x) dx=12sin⁡(2x)\int \cos\left(2 x\right)\,dx = \frac{1}{2}\sin\left(2 x\right)
  5. Constant rule

    The integral of a constant k is k·x.

    ∫12 dx=12x\int \frac{1}{2}\,dx = \frac{1}{2}x
-4-3-2-112340.20.40.60.81
  • f(x) = sin²x
  • F(x) + C

Every choice of C gives a valid antiderivative, since shifting up or down doesn’t change the slope.

Open in the integral calculator

Questions students ask

What is the integral of sin²x?

∫sin⁡2(x) dx=12x−14sin⁡(2x)+C\int \sin^{2}\left(x\right)\,dx = \frac{1}{2}x - \frac{1}{4}\sin\left(2 x\right) + C.

How do you integrate sin²x?

Use the power-reduction identity sin²x = (1 − cos 2x)/2 first, then each piece integrates directly.

How can I check the answer?

Differentiate it: the derivative of 12x−14sin⁡(2x)\frac{1}{2}x - \frac{1}{4}\sin\left(2 x\right) is sin⁡2(x)\sin^{2}\left(x\right). CalcViz checks every antiderivative this way before showing it.