What is the integral from-pi/2 to pi/2 of sin^2(2x) ?

The integral from-pi/2 to pi/2 of sin^2(2x) is pi/2

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integral from-pi/2 to pi/2 of sin^2(2x)

\int_{- \frac{π}{2}}^{\frac{π}{2}} sin^{2} (2 x) d x

\frac{π}{2}

Solution steps

\int_{- \frac{π}{2}}^{\frac{π}{2}} sin^{2} (2 x) d x

Rewrite using trig identities

= \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1}{2} (1 - cos (4 x)) 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_{- \frac{π}{2}}^{\frac{π}{2}} 1 - cos (4 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_{- \frac{π}{2}}^{\frac{π}{2}} 1 d x - \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (4 x) d x)

\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d x = π

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (4 x) d x = 0

= \frac{1}{2} (π - 0)

Simplify

\frac{1}{2} (π - 0) : \frac{π}{2}

= \frac{π}{2}

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What is the integral from-pi/2 to pi/2 of sin^2(2x) ?
The integral from-pi/2 to pi/2 of sin^2(2x) is pi/2