What is the integral from 0 to pi of 7sin^2(4x) ?

The integral from 0 to pi of 7sin^2(4x) is (7pi)/2

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integral from 0 to pi of 7sin^2(4x)

\int_{0}^{π} 7 sin^{2} (4 x) d x

\frac{7 π}{2}

Decimal Notation

10.99557 \dots

Solution steps

\int_{0}^{π} 7 sin^{2} (4 x) d x

Take the constant out:

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

= 7 \cdot \int_{0}^{π} sin^{2} (4 x) d x

Rewrite using trig identities

= 7 \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

= 7 \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

= 7 \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

= 7 \cdot \frac{1}{2} (π - 0)

Simplify

7 \cdot \frac{1}{2} (π - 0) : \frac{7 π}{2}

= \frac{7 π}{2}

\frac{d}{d t} (2 cos (2 t))

\frac{d}{d x} (\frac{x}{5 x + 8})

x \to - 2 + lim (\frac{1 - x}{x + 2})

\frac{\partial}{\partial x} (x^{5})

\int cos (4 x) sin^{5} (4 x) d x

What is the integral from 0 to pi of 7sin^2(4x) ?
The integral from 0 to pi of 7sin^2(4x) is (7pi)/2