What is the integral from 0 to 2pi of 16cos^2(x) ?

The integral from 0 to 2pi of 16cos^2(x) is 16pi

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 16cos^2(x)

\int_{0}^{2 π} 16 cos^{2} (x) d x

16 π

Decimal

50.26548 \dots

Solution steps

\int_{0}^{2 π} 16 cos^{2} (x) d x

Take the constant out:

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

= 16 \cdot \int_{0}^{2 π} cos^{2} (x) d x

Rewrite using trig identities

= 16 \cdot \int_{0}^{2 π} \frac{1 + cos ( 2 x )}{2} d x

Take the constant out:

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

= 16 \cdot \frac{1}{2} \cdot \int_{0}^{2 π} 1 + cos (2 x) d x

Apply the Sum Rule:

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

= 16 \cdot \frac{1}{2} (\int_{0}^{2 π} 1 d x + \int_{0}^{2 π} cos (2 x) d x)

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (2 x) d x = 0

= 16 \cdot \frac{1}{2} (2 π + 0)

Simplify

16 \cdot \frac{1}{2} (2 π + 0) : 16 π

= 16 π

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

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

\int_{- 1}^{1} π (1 - y^{2})^{2} d y

\int_{0}^{4} 2 y d y

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

What is the integral from 0 to 2pi of 16cos^2(x) ?
The integral from 0 to 2pi of 16cos^2(x) is 16pi