What is the integral from 0 to pi/2 of 15sin^2(2x) ?

The integral from 0 to pi/2 of 15sin^2(2x) is (15pi)/4

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integral from 0 to pi/2 of 15sin^2(2x)

\int_{0}^{\frac{π}{2}} 15 sin^{2} (2 x) d x

\frac{15 π}{4}

Decimal Notation

11.78097 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 15 sin^{2} (2 x) d x

Take the constant out:

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

= 15 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (2 x) d x

Rewrite using trig identities

= 15 \cdot \int_{0}^{\frac{π}{2}} \frac{1}{2} (1 - cos (4 x)) d x

Take the constant out:

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

= 15 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 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

= 15 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d x - \int_{0}^{\frac{π}{2}} cos (4 x) d x)

\int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (4 x) d x = 0

= 15 \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

15 \cdot \frac{1}{2} (\frac{π}{2} - 0) : \frac{15 π}{4}

= \frac{15 π}{4}

tan^{2} (6 x)

\int_{1}^{3} \frac{x ^{2} + 3}{4 x - x ^{2}} d x

\frac{\partial}{\partial y} (38 - x^{2} - 4 y^{2})

x^{4} + y^{3} x^{5} + 4 \frac{d y}{d x} = 0, y (1) = 0

\int \frac{t ^{5}}{t ^{2} + 8} d t

What is the integral from 0 to pi/2 of 15sin^2(2x) ?
The integral from 0 to pi/2 of 15sin^2(2x) is (15pi)/4