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}

x \cdot \frac{d y}{d x} - y = x^{2} + y^{2}

\frac{d}{d x} (64 sin (4 x))

\frac{d}{d t} (\frac{2 t}{2 t ^{2} + 1})

\int \frac{sin ( x )}{1 + cos ( x )} d x

f (x) = (x + 2) \cdot (2 \cdot x^{2} - 2)

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