Integral of $\sin\left(8 x\right)$

Q: What is the integral of sin(8x)?

The integral of ≤ft(8 x) is (-≤ft(8 x))/(8) + C.

The integral of $\sin\left(8 x\right)$ is $\frac{-\cos\left(8 x\right)}{8} + C$ .

\int \sin\left(8 x\right)\,dx = \frac{-\cos\left(8 x\right)}{8} + C

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Graph of f(x) = sin(8x) with the area from 0 to π shaded. The shaded area equals the definite integral, about 0.

How to work it out

1
We want an antiderivative of f(x) = sin(8x).
$\int \sin\left(8 x\right)\,dx$
2
Method: ∫sin(kx) dx = −cos(kx)/k.
3
That gives the antiderivative; add the constant C because any constant disappears when you differentiate:
$\int \sin\left(8 x\right)\,dx = \frac{-\cos\left(8 x\right)}{8} + C$
4
Check by differentiating the answer; you must get the original function back:
$\frac{d}{dx}\left[\frac{-\cos\left(8 x\right)}{8}\right] = \frac{64 \sin\left(8 x\right)}{64}$
5
As a definite integral, evaluate the antiderivative at both ends and subtract. Between 0 and π:
$\int_{0}^{\pi} \sin\left(8 x\right)\,dx = \Big[\frac{-\cos\left(8 x\right)}{8}\Big]_{0}^{\pi} \approx 0$

LaTeX source of the formulas on this page

\[ \int \sin\left(8 x\right)\,dx = \frac{-\cos\left(8 x\right)}{8} + C \]
\[ \int \sin\left(8 x\right)\,dx \]
\[ \frac{d}{dx}\left[\frac{-\cos\left(8 x\right)}{8}\right] = \frac{64 \sin\left(8 x\right)}{64} \]
\[ \int_{0}^{\pi} \sin\left(8 x\right)\,dx = \Big[\frac{-\cos\left(8 x\right)}{8}\Big]_{0}^{\pi} \approx 0 \]

Definite integrals of the same function

Interval	Value
from 0 to π	0
from 0 to π/2	0

Questions

What is the integral of sin(8x)?

The integral of $\sin\left(8 x\right)$ is $\frac{-\cos\left(8 x\right)}{8} + C$ .

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(8x) from 0 to π?

About 0.

Integral of sin⁡(8x)\sin\left(8 x\right)sin(8x)

How to work it out

Definite integrals of the same function

Questions

More integrals

Integral of $\sin\left(8 x\right)$