What is the integral from 0 to pi/2 of 8sin^2(3x) ?

The integral from 0 to pi/2 of 8sin^2(3x) is 2pi

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integral from 0 to pi/2 of 8sin^2(3x)

\int_{0}^{\frac{π}{2}} 8 sin^{2} (3 x) d x

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 8 sin^{2} (3 x) d x

Take the constant out:

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

= 8 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (3 x) d x

Rewrite using trig identities

= 8 \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

= 8 \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

= 8 \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

= 8 \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

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

= 2 π

\int x (x^{2} + 1) 4 - 2 x^{2} - x^{4} d x

\int 16 + 25 x^{2} d x

x \to 2 lim (\frac{x - 3}{x - 2})

\int 2 t - 5 d t

\frac{d y}{d x} = \frac{x}{y}

What is the integral from 0 to pi/2 of 8sin^2(3x) ?
The integral from 0 to pi/2 of 8sin^2(3x) is 2pi