What is the integral from 0 to 2pi of 2cos^2(t) ?

The integral from 0 to 2pi of 2cos^2(t) is 2pi

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^2(t)

\int_{0}^{2 π} 2 cos^{2} (t) d t

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{0}^{2 π} 2 cos^{2} (t) d t

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^{2} (t) d t

Rewrite using trig identities

= 2 \cdot \int_{0}^{2 π} \frac{1 + cos ( 2 t )}{2} d t

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} 1 + cos (2 t) d t

Apply the Sum Rule:

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

= 2 \cdot \frac{1}{2} (\int_{0}^{2 π} 1 d t + \int_{0}^{2 π} cos (2 t) d t)

\int_{0}^{2 π} 1 d t = 2 π

\int_{0}^{2 π} cos (2 t) d t = 0

= 2 \cdot \frac{1}{2} (2 π + 0)

Simplify

2 \cdot \frac{1}{2} (2 π + 0) : 2 π

= 2 π

\int_{1}^{4} \frac{ln ( x )}{x} d x

\int_{- 6}^{4} - x^{2} - 3 x d x

\int_{1}^{32} x^{- \frac{8}{5}} d x

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

\int_{0}^{\frac{1}{2}} x e^{2 x} d x

What is the integral from 0 to 2pi of 2cos^2(t) ?
The integral from 0 to 2pi of 2cos^2(t) is 2pi