What is the integral from 0 to pi/2 of 5sin^2(2x) ?

The integral from 0 to pi/2 of 5sin^2(2x) is (5pi)/4

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integral from 0 to pi/2 of 5sin^2(2x)

\int_{0}^{\frac{π}{2}} 5 sin^{2} (2 x) d x

\frac{5 π}{4}

Decimal Notation

3.92699 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 5 sin^{2} (2 x) d x

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (2 x) d x

Rewrite using trig identities

= 5 \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

= 5 \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

= 5 \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

= 5 \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

5 \cdot \frac{1}{2} (\frac{π}{2} - 0) : \frac{5 π}{4}

= \frac{5 π}{4}

f (x) = 5 - 2 x^{2}, at x = - 3, 13

4 x^{\frac{3}{2}}

\int \frac{x}{1 + x} d x

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\int_{0}^{π} 6 sin (x) + sin (6 x) d x

What is the integral from 0 to pi/2 of 5sin^2(2x) ?
The integral from 0 to pi/2 of 5sin^2(2x) is (5pi)/4