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

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

\int_{0}^{2 π} cos (θ) + sin (θ) d θ

0

Solution steps

\int_{0}^{2 π} cos (θ) + sin (θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{2 π} cos (θ) d θ = 0

\int_{0}^{2 π} sin (θ) d θ = 0

= 0 + 0

Simplify

= 0

\int_{- 2}^{1} x^{2} ∣ x ∣ d x

\int_{- 2}^{3} (3 x^{2} - 5 x) d x

\int_{4}^{9} 2 x^{\frac{3}{2}} d x

\int_{2}^{6} x^{2} - 2 x - 8 d x

\int_{1}^{16} \frac{5}{4} x^{\frac{1}{4}} d x

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