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

The integral from 0 to pi/2 of 8cos(t)sin(t) is 4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\int_{0}^{\frac{π}{2}} 8 cos (t) sin (t) d t

4

Solution steps

\int_{0}^{\frac{π}{2}} 8 cos (t) sin (t) d t

Take the constant out:

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

= 8 \cdot \int_{0}^{\frac{π}{2}} cos (t) sin (t) d t

Apply u-substitution

= 8 \cdot \int_{0}^{1} u d u

Apply the Power Rule

= 8 [\frac{u ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

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

Simplify

= 4

What is the integral from 0 to pi/2 of 8cos(t)sin(t) ?
The integral from 0 to pi/2 of 8cos(t)sin(t) is 4