What is the integral from 0 to 2pi of-98sin^2(t) ?

The integral from 0 to 2pi of-98sin^2(t) is -98pi

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-98sin^2(t)

\int_{0}^{2 π} - 98 sin^{2} (t) d t

- 98 π

Decimal

- 307.87608 \dots

Solution steps

\int_{0}^{2 π} - 98 sin^{2} (t) d t

Take the constant out:

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

= - 98 \cdot \int_{0}^{2 π} sin^{2} (t) d t

Rewrite using trig identities

= - 98 \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

= - 98 \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

= - 98 \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

= - 98 \cdot \frac{1}{2} (2 π - 0)

Simplify

- 98 \cdot \frac{1}{2} (2 π - 0) : - 98 π

= - 98 π

\int_{0}^{x} 2 - t d t

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

\int_{0}^{1} 10 x^{2} + 1 d x

\int_{0}^{4} 3 x - x^{2} d x

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

What is the integral from 0 to 2pi of-98sin^2(t) ?
The integral from 0 to 2pi of-98sin^2(t) is -98pi