What is the integral from 0 to pi/8 of sin(2x) ?

The integral from 0 to pi/8 of sin(2x) is (-1+sqrt(2))/(2sqrt(2))

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integral from 0 to pi/8 of sin(2x)

\int_{0}^{\frac{π}{8}} sin (2 x) d x

\frac{- 1 + 2}{2 2}

Decimal Notation

0.14644 \dots

Solution steps

\int_{0}^{\frac{π}{8}} sin (2 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{4}} sin (u) \frac{1}{2} d u

= \int_{0}^{\frac{π}{4}} sin (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{0}^{\frac{π}{4}}

Compute the boundaries:

- \frac{1}{2} + 1

= \frac{1}{2} (- \frac{1}{2} + 1)

Simplify

= \frac{- 1 + 2}{2 2}

\int \frac{3}{2 - x} d x

\int_{1}^{3} 7 r^{3} ln (r) d r

\frac{4}{2 ( s - 3 ) ^{2} - 50}

f (x) = (x^{3} - 4 x + 1) e^{x}, at x = 0

x^{2} + 2 x + 2, 2 x^{2} + 3 x - 4

What is the integral from 0 to pi/8 of sin(2x) ?
The integral from 0 to pi/8 of sin(2x) is (-1+sqrt(2))/(2sqrt(2))