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

The integral from 0 to pi/2 of (sin(2x))/2 is 1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

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}{2}} sin (2 x) d x

Apply u-substitution

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

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

Use the common integral:

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

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

Simplify

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

= \frac{1}{4} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= \frac{1}{4} \cdot 2

Simplify

= \frac{1}{2}

\int_{- 4}^{1} (- x^{2} - 3 x + 4) d x

\int_{- 1}^{7} x 2 + x d x

\int_{0}^{3} 25 - x^{2} d x

\int_{- 2}^{3} (3 e^{- 2 x - 1}) d x

\int_{0}^{a} (a - x) d x

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