What is the integral from 0 to pi/2 of 9sin^2(3x) ?

The integral from 0 to pi/2 of 9sin^2(3x) is (9pi)/4

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integral from 0 to pi/2 of 9sin^2(3x)

\int_{0}^{\frac{π}{2}} 9 sin^{2} (3 x) d x

\frac{9 π}{4}

Decimal Notation

7.06858 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 9 sin^{2} (3 x) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (3 x) d x

Rewrite using trig identities

= 9 \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

= 9 \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

= 9 \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

= 9 \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

9 \cdot \frac{1}{2} (\frac{π}{2} - 0) : \frac{9 π}{4}

= \frac{9 π}{4}

n = 1 \sum \infty n e^{- 2 n}

(x + y) d x - (x + y + 3) d y = 0

y = x^{3} - 3 x + 2, (4, 54)

\frac{s ^{2} + 12}{s ^{4} - 16}

What is the integral from 0 to pi/2 of 9sin^2(3x) ?
The integral from 0 to pi/2 of 9sin^2(3x) is (9pi)/4