What is the integral from 0 to pi of 15sin^2(4x) ?

The integral from 0 to pi of 15sin^2(4x) is (15pi)/2

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integral from 0 to pi of 15sin^2(4x)

\int_{0}^{π} 15 sin^{2} (4 x) d x

\frac{15 π}{2}

Decimal Notation

23.56194 \dots

Solution steps

\int_{0}^{π} 15 sin^{2} (4 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} (4 x) d x

Rewrite using trig identities

= 15 \cdot \int_{0}^{π} \frac{1}{2} (1 - cos (8 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 (8 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 (8 x) d x)

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (8 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}

\int 3 x^{2} e^{2 x} d x

n = 1 \sum \infty \frac{1}{n} \cdot x^{n}

\frac{d}{d x} (y x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the integral from 0 to pi of 15sin^2(4x) ?
The integral from 0 to pi of 15sin^2(4x) is (15pi)/2