What is the integral from 0 to pi of 9sin^2(4x) ?

The integral from 0 to pi of 9sin^2(4x) is (9pi)/2

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integral from 0 to pi of 9sin^2(4x)

\int_{0}^{π} 9 sin^{2} (4 x) d x

\frac{9 π}{2}

Decimal Notation

14.13716 \dots

Solution steps

\int_{0}^{π} 9 sin^{2} (4 x) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{π} sin^{2} (4 x) d x

Rewrite using trig identities

= 9 \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

= 9 \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

= 9 \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

= 9 \cdot \frac{1}{2} (π - 0)

Simplify

9 \cdot \frac{1}{2} (π - 0) : \frac{9 π}{2}

= \frac{9 π}{2}

x \to 0 lim (sin^{3} (2 x))

\frac{d}{d x} (\frac{- 8 x}{( x ^{2} - 16 ) ^{2}})

\frac{\partial}{\partial x} (e^{- x - 3 y})

x^{2} - 6, ∣ x ∣

f (x) 5 x + y = 6

What is the integral from 0 to pi of 9sin^2(4x) ?
The integral from 0 to pi of 9sin^2(4x) is (9pi)/2