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

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}

x \to 1 + lim (x ln (x))

\frac{d}{d x} (3 3 x^{2})

x \to 8 lim (1 + 3 x)

x^{2} y^{^{'}} - x (x + 2) y = e^{x}

ln (t^{2} + 1)

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