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

The integral from 0 to pi of cos(t)sin(t) is 0

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 of cos(t)sin(t)

\int_{0}^{π} cos (t) sin (t) d t

0

Solution steps

\int_{0}^{π} cos (t) sin (t) d t

Rewrite using trig identities

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

Take the constant out:

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

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

Apply u-substitution

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

= \frac{1}{2} \cdot \int_{0}^{2 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}^{2 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}^{2 π}

Simplify

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

= \frac{1}{4} [- cos (u)]_{0}^{2 π}

Compute the boundaries:

0

= \frac{1}{4} \cdot 0

Simplify

= 0

n = 0 \sum \infty \frac{x ^{n}}{n}

\frac{d}{d x} (3, x)

x \to + 0 lim (\frac{x}{x})

\frac{d}{d x} ((x^{3} - 3 x) ln (2 x + 1))

y^{^{''}} - 2 y^{^{'}} + 10 y = x^{2} \cdot e^{x}

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