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Integral of sin⁡(x)\sin\left(x\right)

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

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

How to work it out

  1. 1

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

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

    Method: standard integral.

  3. 3

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

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

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

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

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

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

Definite integrals of the same function

IntervalValue
from 0 to π2
from 0 to π/21

Questions

What is the integral of sin(x)?

The integral of sin⁡(x)\sin\left(x\right) is −cos⁡(x)+C-\cos\left(x\right) + 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(x) from 0 to π?

About 2.

More integrals

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