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

The integral from 0 to pi/2 of 11sin^2(2x) is (11pi)/4

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

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

\frac{11 π}{4}

Decimal

8.63937 \dots

Solution steps

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

Rewrite using trig identities

= 11 \cdot \int_{0}^{\frac{π}{2}} \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}^{\frac{π}{2}} 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}^{\frac{π}{2}} 1 d x - \int_{0}^{\frac{π}{2}} cos (4 x) d x)

\int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (4 x) d x = 0

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

Simplify

11 \cdot \frac{1}{2} (\frac{π}{2} - 0) : \frac{11 π}{4}

= \frac{11 π}{4}

\int \frac{a ^{2} - x ^{2}}{x} d x

x y^{^{'}} + 2 y = 4 x^{2}

\frac{d}{d x} (\frac{- e ^{\frac{1}{x}}}{x ^{2}})

y^{^{''}} + 5 y^{^{'}} + 6 y = 2 x

\frac{d}{d x} (x^{3} \cdot sin^{2} (5 x))

What is the integral from 0 to pi/2 of 11sin^2(2x) ?
The integral from 0 to pi/2 of 11sin^2(2x) is (11pi)/4