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

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

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

0

Solution steps

\int_{0}^{2 π} sin (θ) cos (θ) d θ

Rewrite using trig identities

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

Take the constant out:

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

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

Apply u-substitution

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

= \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

= \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)

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

Simplify

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

= \frac{1}{4} [- cos (u)]_{0}^{4 π}

Compute the boundaries:

0

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

Simplify

= 0

\int_{0}^{l n (10)} cosh (t) d t

\int_{0}^{3} \frac{9}{x - 3} d x

\int_{x}^{y} cos (t^{2}) d t

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

\int_{- 2}^{0} 3 x^{2} d x

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