What is the integral from 0 to pi/2 of-6sin^2(3x) ?

The integral from 0 to pi/2 of-6sin^2(3x) is -(3pi)/2

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integral from 0 to pi/2 of-6sin^2(3x)

\int_{0}^{\frac{π}{2}} - 6 sin^{2} (3 x) d x

- \frac{3 π}{2}

Decimal Notation

- 4.71238 \dots

Solution steps

\int_{0}^{\frac{π}{2}} - 6 sin^{2} (3 x) d x

Take the constant out:

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

= - 6 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (3 x) d x

Rewrite using trig identities

= - 6 \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

= - 6 \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

= - 6 \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

= - 6 \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

- 6 \cdot \frac{1}{2} (\frac{π}{2} - 0) : - \frac{3 π}{2}

= - \frac{3 π}{2}

\frac{\partial}{\partial x} (2 x y (9 - x - y))

x^{2} y^{2}

\int 2 x y e^{x^{2} y} d x

\frac{d}{d t} (3 t^{2})

\frac{d}{d x} (- 4 e^{- x^{2}} x)

What is the integral from 0 to pi/2 of-6sin^2(3x) ?
The integral from 0 to pi/2 of-6sin^2(3x) is -(3pi)/2