What is the integral from 0 to 2pi of cos(4θ) ?

The integral from 0 to 2pi of cos(4θ) 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 cos(4θ)

\int_{0}^{2 π} cos (4 θ) d θ

0

Solution steps

\int_{0}^{2 π} cos (4 θ) d θ

Apply u-substitution

: \int_{0}^{8 π} cos (u) \frac{1}{4} d u

= \int_{0}^{8 π} cos (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{8 π} cos (u) d u

Use the common integral:

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

= \frac{1}{4} [sin (u)]_{0}^{8 π}

Compute the boundaries:

0

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

Simplify

= 0

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

\int_{4}^{9} 4 d x

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

\int_{0}^{2} e^{- s x} d x

\int_{0}^{3} t^{2} e^{- s t} d t

What is the integral from 0 to 2pi of cos(4θ) ?
The integral from 0 to 2pi of cos(4θ) is 0