What is the integral from 0 to pi/2 of 19sin^2(2x) ?

The integral from 0 to pi/2 of 19sin^2(2x) is (19pi)/4

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integral from 0 to pi/2 of 19sin^2(2x)

\int_{0}^{\frac{π}{2}} 19 sin^{2} (2 x) d x

\frac{19 π}{4}

Decimal

14.92256 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 19 sin^{2} (2 x) d x

Take the constant out:

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

= 19 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (2 x) d x

Rewrite using trig identities

= 19 \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

= 19 \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

= 19 \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

= 19 \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

19 \cdot \frac{1}{2} (\frac{π}{2} - 0) : \frac{19 π}{4}

= \frac{19 π}{4}

5 sin (\frac{1}{5} x)

x \to 4 lim (\frac{x - 4}{x + 4})

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

\frac{\partial}{\partial x} (3 x^{2} + y^{3})

y = x^{\frac{2}{5}}

What is the integral from 0 to pi/2 of 19sin^2(2x) ?
The integral from 0 to pi/2 of 19sin^2(2x) is (19pi)/4