What is the integral from 0 to pi of 15sin^2(2x) ?

The integral from 0 to pi of 15sin^2(2x) is (15pi)/2

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integral from 0 to pi of 15sin^2(2x)

\int_{0}^{π} 15 sin^{2} (2 x) d x

\frac{15 π}{2}

Decimal

23.56194 \dots

Solution steps

\int_{0}^{π} 15 sin^{2} (2 x) d x

Take the constant out:

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

= 15 \cdot \int_{0}^{π} sin^{2} (2 x) d x

Rewrite using trig identities

= 15 \cdot \int_{0}^{π} \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

= 15 \cdot \frac{1}{2} \cdot \int_{0}^{π} 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

= 15 \cdot \frac{1}{2} (\int_{0}^{π} 1 d x - \int_{0}^{π} cos (4 x) d x)

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (4 x) d x = 0

= 15 \cdot \frac{1}{2} (π - 0)

Simplify

15 \cdot \frac{1}{2} (π - 0) : \frac{15 π}{2}

= \frac{15 π}{2}

f (x) = - \frac{1}{2} x^{2} - 2 x + 3, at x = 0

f (t) = t - sin (t)

\frac{d}{d y} (\frac{t ^{2}}{4 y ^{4}})

\frac{d}{d x} (10 x e^{x^{2}})

\frac{d}{d x} (4^{\frac{x}{2}})

What is the integral from 0 to pi of 15sin^2(2x) ?
The integral from 0 to pi of 15sin^2(2x) is (15pi)/2