What is the integral from x-2t to x+2t of sin^2(x) ?

The integral from x-2t to x+2t of sin^2(x) is 1/2 (4t-1/2 (sin(2(x+2t))-sin(2(x-2t))))

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integral from x-2t to x+2t of sin^2(x)

\int_{x - 2 t}^{x + 2 t} sin^{2} (x) d x

\frac{1}{2} (4 t - \frac{1}{2} (sin (2 (x + 2 t)) - sin (2 (x - 2 t))))

Solution steps

\int_{x - 2 t}^{x + 2 t} sin^{2} (x) d x

Rewrite using trig identities

= \int_{x - 2 t}^{x + 2 t} \frac{1 - cos ( 2 x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{x - 2 t}^{x + 2 t} 1 - cos (2 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{x - 2 t}^{x + 2 t} 1 d x - \int_{x - 2 t}^{x + 2 t} cos (2 x) d x)

\int_{x - 2 t}^{x + 2 t} 1 d x = 4 t

\int_{x - 2 t}^{x + 2 t} cos (2 x) d x = \frac{1}{2} (sin (2 (x + 2 t)) - sin (2 (x - 2 t)))

= \frac{1}{2} (4 t - \frac{1}{2} (sin (2 (x + 2 t)) - sin (2 (x - 2 t))))

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What is the integral from x-2t to x+2t of sin^2(x) ?
The integral from x-2t to x+2t of sin^2(x) is 1/2 (4t-1/2 (sin(2(x+2t))-sin(2(x-2t))))