What is the integral from 0 to 2pi of 2cos(x)sin(x) ?

The integral from 0 to 2pi of 2cos(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 2pi of 2cos(x)sin(x)

\int_{0}^{2 π} 2 cos (x) sin (x) d x

0

Solution steps

\int_{0}^{2 π} 2 cos (x) sin (x) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2 π} cos (x) sin (x) d x

Rewrite using trig identities

= 2 \cdot \int_{0}^{2 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

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

Apply u-substitution

: \int_{0}^{4 π} sin (u) \frac{1}{2} d u

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{4 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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{4 π} sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

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

Simplify

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

= \frac{1}{2} [- cos (u)]_{0}^{4 π}

Compute the boundaries:

0

= \frac{1}{2} \cdot 0

Simplify

= 0

\int_{- 2}^{4} x^{2} - 1 d x

\int_{0}^{2} \frac{x ^{2}}{x ^{3} + 8} d x

\int_{0}^{\frac{π}{6}} sin (2 x) sin (x) d x

\int_{1}^{10} 2 d x

\int_{0}^{1} - 2 x + 1 d x

What is the integral from 0 to 2pi of 2cos(x)sin(x) ?
The integral from 0 to 2pi of 2cos(x)sin(x) is 0