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

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

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

\int_{0}^{π} 17 sin^{2} (4 x) d x

\frac{17 π}{2}

Decimal Notation

26.70353 \dots

Solution steps

\int_{0}^{π} 17 sin^{2} (4 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}^{π} sin^{2} (4 x) d x

Rewrite using trig identities

= 17 \cdot \int_{0}^{π} \frac{1}{2} (1 - cos (8 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}^{π} 1 - cos (8 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}^{π} 1 d x - \int_{0}^{π} cos (8 x) d x)

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (8 x) d x = 0

= 17 \cdot \frac{1}{2} (π - 0)

Simplify

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

= \frac{17 π}{2}

y^{^{'''}} + 4 y^{^{'}} = 3 x - 1

y = \frac{1}{5 + 2 x}, (- 1, \frac{1}{3})

x \frac{d y}{d x} = x^{2} - y^{2} + y

\int - t d t

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