What is the integral from 0 to pi/2 of 17sin^2(2x) ?

The integral from 0 to pi/2 of 17sin^2(2x) is (17pi)/4

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integral from 0 to pi/2 of 17sin^2(2x)

\int_{0}^{\frac{π}{2}} 17 sin^{2} (2 x) d x

\frac{17 π}{4}

Decimal

13.35176 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 17 sin^{2} (2 x) d x

Take the constant out:

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

= 17 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (2 x) d x

Rewrite using trig identities

= 17 \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

= 17 \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

= 17 \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

= 17 \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

17 \cdot \frac{1}{2} (\frac{π}{2} - 0) : \frac{17 π}{4}

= \frac{17 π}{4}

x^{4} + 2 x, x = 1

y = \frac{4 x ^{2} + 8 x + 8}{x}

\frac{\partial}{\partial y} (ln (x - 4 y))

x \to 1 lim (2 x^{3} + x - 1)

f (x) =, \frac{1}{x}

What is the integral from 0 to pi/2 of 17sin^2(2x) ?
The integral from 0 to pi/2 of 17sin^2(2x) is (17pi)/4