What is the integral from 0 to 2pi of 16sin^2(t) ?

The integral from 0 to 2pi of 16sin^2(t) 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 16sin^2(t)

\int_{0}^{2 π} 16 sin^{2} (t) d t

16 π

Decimal

50.26548 \dots

Solution steps

\int_{0}^{2 π} 16 sin^{2} (t) d t

Take the constant out:

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

= 16 \cdot \int_{0}^{2 π} sin^{2} (t) d t

Rewrite using trig identities

= 16 \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

= 16 \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

= 16 \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

= 16 \cdot \frac{1}{2} (2 π - 0)

Simplify

16 \cdot \frac{1}{2} (2 π - 0) : 16 π

= 16 π

\int_{- 1}^{5} 3 x^{2} + 8 x d x

\int_{0}^{0.25} e^{- x^{2}} d x

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

\int_{0}^{0.5} 5 x^{4} d x

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

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