What is the integral from 0 to pi/8 of sin(8x) ?

The integral from 0 to pi/8 of sin(8x) is 1/4

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integral from 0 to pi/8 of sin(8x)

\int_{0}^{\frac{π}{8}} sin (8 x) d x

\frac{1}{4}

Decimal Notation

0.25

Solution steps

\int_{0}^{\frac{π}{8}} sin (8 x) d x

Apply u-substitution

: \int_{0}^{π} sin (u) \frac{1}{8} d u

= \int_{0}^{π} sin (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{0}^{π} sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{8} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= \frac{1}{8} \cdot 2

Simplify

= \frac{1}{4}

\frac{d ^{2}}{d x ^{2}} (2 x^{\frac{3}{2}})

x \to - 41 lim (39 + x)

\int (\frac{1}{x ^{3}} - x^{3} - \frac{1}{8}) d x

\frac{\partial}{\partial x} (x \cdot ln (x^{2} + y^{2}))

x^{3} \frac{d y}{d x} - x^{2} y - y^{3} = 0

What is the integral from 0 to pi/8 of sin(8x) ?
The integral from 0 to pi/8 of sin(8x) is 1/4