What is the integral from 0 to pi/2 of-10sin^2(3x) ?

The integral from 0 to pi/2 of-10sin^2(3x) is -(5pi)/2

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integral from 0 to pi/2 of-10sin^2(3x)

\int_{0}^{\frac{π}{2}} - 10 sin^{2} (3 x) d x

- \frac{5 π}{2}

Decimal Notation

- 7.85398 \dots

Solution steps

\int_{0}^{\frac{π}{2}} - 10 sin^{2} (3 x) d x

Take the constant out:

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

= - 10 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (3 x) d x

Rewrite using trig identities

= - 10 \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

= - 10 \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

= - 10 \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

= - 10 \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

- 10 \cdot \frac{1}{2} (\frac{π}{2} - 0) : - \frac{5 π}{2}

= - \frac{5 π}{2}

\int \frac{x}{( 1 + 2 x )} d x

\frac{d}{d x} (tan (6 x^{2}))

\frac{\partial}{\partial x} (cos^{2} (x - y))

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

\int_{4}^{6} 2 x d x

What is the integral from 0 to pi/2 of-10sin^2(3x) ?
The integral from 0 to pi/2 of-10sin^2(3x) is -(5pi)/2