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

The integral from 0 to pi/2 of-3sin^2(3x) is -(3pi)/4

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

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

- \frac{3 π}{4}

Decimal Notation

- 2.35619 \dots

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

= - 3 \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

= - 3 \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

= - 3 \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

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

Simplify

- 3 \cdot \frac{1}{2} (\frac{π}{2} - 0) : - \frac{3 π}{4}

= - \frac{3 π}{4}

\frac{( 3 - 2 s )}{( s ^{2} + 4 s + 5 )}

\int x^{2} - 7 x + 5 d x

\int_{\frac{9}{8}}^{4} \frac{1}{1 - x ^{2}} d x

\int - \frac{x}{1 + x ^{2}} d x

\int \frac{1}{3 x 9 x ^{2} - 12} d x

What is the integral from 0 to pi/2 of-3sin^2(3x) ?
The integral from 0 to pi/2 of-3sin^2(3x) is -(3pi)/4