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

The integral from 0 to pi of cos(x)*sin(x) 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(x)*sin(x)

\int_{0}^{π} cos (x) \cdot sin (x) d x

0

Solution steps

\int_{0}^{π} cos (x) sin (x) d x

Rewrite using trig identities

= \int_{0}^{π} \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}^{π} sin (2 x) d x

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_{- 2}^{0} \frac{2}{x ^{2} + 4 x + 8} d x

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

\int_{2}^{4} \frac{x ^{3}}{x + 2} d x

\int_{1}^{3} e^{x} \cdot ln (x) d x

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

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