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

The integral from pi to 0 of (sin(x)-2)^2 is -pi/2+8-4pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from pi to 0 of (sin(x)-2)^2

\int_{π}^{0} (sin (x) - 2)^{2} d x

- \frac{π}{2} + 8 - 4 π

Decimal Notation

- 6.13716 \dots

Solution steps

\int_{π}^{0} (sin (x) - 2)^{2} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{π} (sin (x) - 2)^{2} d x

Expand

(sin (x) - 2)^{2} : sin^{2} (x) - 4 sin (x) + 4

= - \int_{0}^{π} sin^{2} (x) - 4 sin (x) + 4 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

= - (\int_{0}^{π} sin^{2} (x) d x - \int_{0}^{π} 4 sin (x) d x + \int_{0}^{π} 4 d x)

\int_{0}^{π} sin^{2} (x) d x = \frac{π}{2}

\int_{0}^{π} 4 sin (x) d x = 8

\int_{0}^{π} 4 d x = 4 π

= - (\frac{π}{2} - 8 + 4 π)

Simplify

= - \frac{π}{2} + 8 - 4 π

\int_{0}^{1} \frac{1}{3 x + 2} d x

\int_{0}^{16} \frac{x}{4 + 2 x} d x

\int_{7}^{14} 0.05 e^{- 0.05 x} d x

\int_{- \infty}^{0} e^{\frac{x}{4}} d x

\int_{1}^{5} (6 n^{- 2} - n^{- 3}) d n

What is the integral from pi to 0 of (sin(x)-2)^2 ?
The integral from pi to 0 of (sin(x)-2)^2 is -pi/2+8-4pi