What is the integral from 0 to pi/2 of-8sin^2(3x) ?

The integral from 0 to pi/2 of-8sin^2(3x) is -2pi

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integral from 0 to pi/2 of-8sin^2(3x)

\int_{0}^{\frac{π}{2}} - 8 sin^{2} (3 x) d x

- 2 π

Decimal

- 6.28318 \dots

Solution steps

\int_{0}^{\frac{π}{2}} - 8 sin^{2} (3 x) d x

Take the constant out:

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

= - 8 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (3 x) d x

Rewrite using trig identities

= - 8 \cdot \int_{0}^{\frac{π}{2}} \frac{1}{2} (1 - cos (6 x)) d x

Take the constant out:

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

= - 8 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (6 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

= - 8 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d x - \int_{0}^{\frac{π}{2}} cos (6 x) d x)

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

\int_{0}^{\frac{π}{2}} cos (6 x) d x = 0

= - 8 \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

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

= - 2 π

\frac{d y}{d x} = cos (x)

\frac{d}{d x} (\frac{4}{x} + 2 sec (x))

(2 y + 2 t - 7) d t + (4 y + 2 t - 1) d y = 0, y (- 1) = 2

\frac{\partial}{\partial y} (x^{6 y})

f (x) = 2 x - 3

What is the integral from 0 to pi/2 of-8sin^2(3x) ?
The integral from 0 to pi/2 of-8sin^2(3x) is -2pi