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

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

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

\int_{0}^{\frac{π}{2}} 4 sin^{2} (3 x) d x

π

Decimal Notation

3.14159 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 4 sin^{2} (3 x) d x

Take the constant out:

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

= 4 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (3 x) d x

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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

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

Simplify

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

= π

\frac{3}{x ( x + 1 ) ^{2}}

\int \frac{( 8 x ^{3} + 3 )}{( x ^{2} + 2 ) ^{2}} d x

\int (3 x + 2)^{2} d x

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

y = 4 x - x^{2}, y = - 6 x

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