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Integral of sin⁡(2x)\sin\left(2 x\right)

The integral of sin⁡(2x)\sin\left(2 x\right) is −cos⁡(2x)2+C\frac{-\cos\left(2 x\right)}{2} + C.

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Graph of f(x) = sin(2x) with the area from 0 to π shaded-112345-1-0.50.51f(x)
Graph of f(x) = sin(2x) with the area from 0 to π shaded. The shaded area equals the definite integral, about 0.

How to work it out

  1. 1

    We want an antiderivative of f(x) = sin(2x).

    ∫sin⁡(2x) dx\int \sin\left(2 x\right)\,dx
  2. 2

    Method: ∫sin(kx) dx = −cos(kx)/k.

  3. 3

    That gives the antiderivative; add the constant C because any constant disappears when you differentiate:

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

    Check by differentiating the answer; you must get the original function back:

    ddx[−cos⁡(2x)2]=4sin⁡(2x)4\frac{d}{dx}\left[\frac{-\cos\left(2 x\right)}{2}\right] = \frac{4 \sin\left(2 x\right)}{4}
  5. 5

    As a definite integral, evaluate the antiderivative at both ends and subtract. Between 0 and π:

    ∫0πsin⁡(2x) dx=[−cos⁡(2x)2]0π≈0\int_{0}^{\pi} \sin\left(2 x\right)\,dx = \Big[\frac{-\cos\left(2 x\right)}{2}\Big]_{0}^{\pi} \approx 0
LaTeX source of the formulas on this page
\[ \int \sin\left(2 x\right)\,dx = \frac{-\cos\left(2 x\right)}{2} + C \]
\[ \int \sin\left(2 x\right)\,dx \]
\[ \frac{d}{dx}\left[\frac{-\cos\left(2 x\right)}{2}\right] = \frac{4 \sin\left(2 x\right)}{4} \]
\[ \int_{0}^{\pi} \sin\left(2 x\right)\,dx = \Big[\frac{-\cos\left(2 x\right)}{2}\Big]_{0}^{\pi} \approx 0 \]

Definite integrals of the same function

IntervalValue
from 0 to π0
from 0 to π/21

Questions

What is the integral of sin(2x)?

The integral of sin⁡(2x)\sin\left(2 x\right) is −cos⁡(2x)2+C\frac{-\cos\left(2 x\right)}{2} + C.

Why do we add + C?

Every function F(x) + C has the same derivative, so the antiderivative is only defined up to a constant.

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

About 0.

More integrals

    Integral of sin(2x): -cos(2x) / 2 + C (with steps) | WordGPT