What is the integral from 0 to pi of 13sin^2(4x) ?

The integral from 0 to pi of 13sin^2(4x) is (13pi)/2

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integral from 0 to pi of 13sin^2(4x)

\int_{0}^{π} 13 sin^{2} (4 x) d x

\frac{13 π}{2}

Decimal Notation

20.42035 \dots

Solution steps

\int_{0}^{π} 13 sin^{2} (4 x) d x

Take the constant out:

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

= 13 \cdot \int_{0}^{π} sin^{2} (4 x) d x

Rewrite using trig identities

= 13 \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

= 13 \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

= 13 \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

= 13 \cdot \frac{1}{2} (π - 0)

Simplify

13 \cdot \frac{1}{2} (π - 0) : \frac{13 π}{2}

= \frac{13 π}{2}

\frac{( s + 1 )}{( s ^{2} + s - 1 )}

(\frac{d y}{d x}) + \frac{1}{y} = \frac{y}{2} + \frac{x}{y}

f (x) = c y^{- 5}

\frac{\partial}{\partial x} (ln (k t - m x))

\frac{\partial}{\partial x} (\frac{x}{y ^{2}} - \frac{y}{x ^{2}})

What is the integral from 0 to pi of 13sin^2(4x) ?
The integral from 0 to pi of 13sin^2(4x) is (13pi)/2