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

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

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

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

- \frac{π}{2}

Decimal Notation

- 1.57079 \dots

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

= - 2 \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

= - 2 \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

= - 2 \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

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

Simplify

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

= - \frac{π}{2}

(\frac{d y}{d x}) = 0.011 \cdot y \cdot (2 - (y - 25))

\int \frac{5}{( 3 x - 1 ) ^{7}} d x

\frac{d ^{2}}{d x ^{2}} (4 x cos (x^{2}))

\frac{4 s}{s ^{2} + 6 s + 9}

x \to 0 lim ((1 + 3 x)^{\frac{7}{x}})

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