What is the integral from 0 to pi of-0.1cos(x)sin(x) ?

The integral from 0 to pi of-0.1cos(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-0.1cos(x)sin(x)

\int_{0}^{π} - 0.1 cos (x) sin (x) d x

0

Solution steps

\int_{0}^{π} - 0.1 cos (x) sin (x) d x

Take the constant out:

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

= - 0.1 \cdot \int_{0}^{π} cos (x) sin (x) d x

Rewrite using trig identities

= - 0.1 \cdot \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

= - 0.1 \cdot \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

= - 0.1 \cdot \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

= - 0.1 \cdot \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)

= - 0.1 \cdot \frac{1}{2} \cdot \frac{1}{2} [- cos (u)]_{0}^{2 π}

Simplify

- 0.1 \cdot \frac{1}{2} \cdot \frac{1}{2} [- cos (u)]_{0}^{2 π} : - 0.025 [- cos (u)]_{0}^{2 π}

= - 0.025 [- cos (u)]_{0}^{2 π}

Compute the boundaries:

0

= - 0.025 \cdot 0

Simplify

= 0

\int_{- 2}^{4} x^{2} - 2 x d x

\int_{0}^{100} 2 ((x^{2}) - a x) d x

\int_{0}^{1} 2 x \cdot e^{x} d x

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

\int_{2}^{7} (- \frac{x}{7} + \frac{9}{7}) - (\frac{2}{x}) d x

What is the integral from 0 to pi of-0.1cos(x)sin(x) ?
The integral from 0 to pi of-0.1cos(x)sin(x) is 0