What is the integral from pi/4 to pi of sin^5(x) ?

The integral from pi/4 to pi of sin^5(x) is 8/15+(43)/(60sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from pi/4 to pi of sin^5(x)

\int_{\frac{π}{4}}^{π} sin^{5} (x) d x

\frac{8}{15} + \frac{43}{60 2}

Decimal Notation

1.04009 \dots

Solution steps

\int_{\frac{π}{4}}^{π} sin^{5} (x) d x

Rewrite using trig identities

= \int_{\frac{π}{4}}^{π} (1 - cos^{2} (x))^{2} sin (x) d x

Apply u-substitution

: \int_{\frac{2}{2}}^{- 1} - 1 + 2 u^{2} - u^{4} d u

= \int_{\frac{2}{2}}^{- 1} - 1 + 2 u^{2} - u^{4} d u

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

= - \int_{- 1}^{\frac{2}{2}} - 1 + 2 u^{2} - u^{4} d u

Apply the Sum Rule:

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

= - (- \int_{- 1}^{\frac{2}{2}} 1 d u + \int_{- 1}^{\frac{2}{2}} 2 u^{2} d u - \int_{- 1}^{\frac{2}{2}} u^{4} d u)

\int_{- 1}^{\frac{2}{2}} 1 d u = \frac{1}{2} + 1

\int_{- 1}^{\frac{2}{2}} 2 u^{2} d u = 2 (\frac{1}{6 2} + \frac{1}{3})

\int_{- 1}^{\frac{2}{2}} u^{4} d u = \frac{1}{20 2} + \frac{1}{5}

= - (- (\frac{1}{2} + 1) + 2 (\frac{1}{6 2} + \frac{1}{3}) - (\frac{1}{20 2} + \frac{1}{5}))

Simplify

- (- (\frac{1}{2} + 1) + 2 (\frac{1}{6 2} + \frac{1}{3}) - (\frac{1}{20 2} + \frac{1}{5})) : \frac{8}{15} + \frac{43}{60 2}

= \frac{8}{15} + \frac{43}{60 2}

\int_{0}^{1} x^{4} - x^{6} d x

\int_{7}^{14} x^{2} - 16 d x

\int_{1}^{10} \frac{x - 1}{x} d x

\int_{0}^{2 π} 5 t \cdot 2 d t

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

What is the integral from pi/4 to pi of sin^5(x) ?
The integral from pi/4 to pi of sin^5(x) is 8/15+(43)/(60sqrt(2))