What is the integral from 0 to pi of 11sin^2(2x) ?

The integral from 0 to pi of 11sin^2(2x) is (11pi)/2

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integral from 0 to pi of 11sin^2(2x)

\int_{0}^{π} 11 sin^{2} (2 x) d x

\frac{11 π}{2}

Decimal

17.27875 \dots

Solution steps

\int_{0}^{π} 11 sin^{2} (2 x) d x

Take the constant out:

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

= 11 \cdot \int_{0}^{π} sin^{2} (2 x) d x

Rewrite using trig identities

= 11 \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

= 11 \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

= 11 \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

= 11 \cdot \frac{1}{2} (π - 0)

Simplify

11 \cdot \frac{1}{2} (π - 0) : \frac{11 π}{2}

= \frac{11 π}{2}

d er i v a t i v e y = ln (8 x^{2} - 7 x + 9)

d er i v a t i v e f (t) = e^{9 t s i n (2 t)}

d er i v a t i v e \frac{x ^{2}}{x + 8}

\int 3 sec^{- 2} (x t a) n^{3} x d x

x \to 9 lim (2 x^{2} - 3)

What is the integral from 0 to pi of 11sin^2(2x) ?
The integral from 0 to pi of 11sin^2(2x) is (11pi)/2