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

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

\int_{0}^{π} - sin (t) cos (t) d t

0

Solution steps

\int_{0}^{π} - sin (t) cos (t) d t

Take the constant out:

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

= - \int_{0}^{π} sin (t) cos (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

\int_{e}^{- 2 x} 1 d x

\int_{- 5}^{0} 2 x - x^{2} - 7 x d x

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

\int_{- 1}^{1} x^{\frac{3}{2}} d x

\int_{1}^{3} \frac{1}{t - 2} d t

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