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Integral of sin⁡(6x)\sin\left(6 x\right)

The integral of sin⁡(6x)\sin\left(6 x\right) is −cos⁡(6x)6+C\frac{-\cos\left(6 x\right)}{6} + C.

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Graph of f(x) = sin(6x) with the area from 0 to π shaded-112345-1-0.50.51f(x)
Graph of f(x) = sin(6x) 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(6x).

    ∫sin⁡(6x) dx\int \sin\left(6 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⁡(6x) dx=−cos⁡(6x)6+C\int \sin\left(6 x\right)\,dx = \frac{-\cos\left(6 x\right)}{6} + C
  4. 4

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

    ddx[−cos⁡(6x)6]=36sin⁡(6x)36\frac{d}{dx}\left[\frac{-\cos\left(6 x\right)}{6}\right] = \frac{36 \sin\left(6 x\right)}{36}
  5. 5

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

    ∫0πsin⁡(6x) dx=[−cos⁡(6x)6]0π≈0\int_{0}^{\pi} \sin\left(6 x\right)\,dx = \Big[\frac{-\cos\left(6 x\right)}{6}\Big]_{0}^{\pi} \approx 0
LaTeX source of the formulas on this page
\[ \int \sin\left(6 x\right)\,dx = \frac{-\cos\left(6 x\right)}{6} + C \]
\[ \int \sin\left(6 x\right)\,dx \]
\[ \frac{d}{dx}\left[\frac{-\cos\left(6 x\right)}{6}\right] = \frac{36 \sin\left(6 x\right)}{36} \]
\[ \int_{0}^{\pi} \sin\left(6 x\right)\,dx = \Big[\frac{-\cos\left(6 x\right)}{6}\Big]_{0}^{\pi} \approx 0 \]

Definite integrals of the same function

IntervalValue
from 0 to π0
from 0 to π/20.33333333

Questions

What is the integral of sin(6x)?

The integral of sin⁡(6x)\sin\left(6 x\right) is −cos⁡(6x)6+C\frac{-\cos\left(6 x\right)}{6} + 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(6x) from 0 to π?

About 0.

More integrals

    Integral of sin(6x): -cos(6x) / 6 + C (with steps) | WordGPT