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

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

\int_{0}^{2 π} cos (4 x) d x

0

Solution steps

\int_{0}^{2 π} cos (4 x) d x

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_{0}^{\frac{π}{2}} 2 x cos (2 x) d x

\int_{\frac{1}{2}}^{1} \frac{8}{x ^{3}} d x

\int_{3}^{6} 8 x - 24 d x

\int_{1}^{2} (2 x^{2} + 3) d x

\int_{0}^{π} sin (x) x d x

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