What is the integral from 0 to pi of sin^7(x) ?

The integral from 0 to pi of sin^7(x) is 32/35

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi of sin^7(x)

\int_{0}^{π} sin^{7} (x) d x

\frac{32}{35}

Decimal

0.91428 \dots

Solution steps

\int_{0}^{π} sin^{7} (x) d x

Rewrite using trig identities

= \int_{0}^{π} (1 - cos^{2} (x))^{3} sin (x) d x

Apply u-substitution

: \int_{1}^{- 1} - (1 - u^{2})^{3} d u

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

Expand

- (1 - u^{2})^{3} : - 1 + 3 u^{2} - 3 u^{4} + u^{6}

= \int_{1}^{- 1} - 1 + 3 u^{2} - 3 u^{4} + u^{6} d u

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

= - \int_{- 1}^{1} - 1 + 3 u^{2} - 3 u^{4} + u^{6} 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}^{1} 1 d u + \int_{- 1}^{1} 3 u^{2} d u - \int_{- 1}^{1} 3 u^{4} d u + \int_{- 1}^{1} u^{6} d u)

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

\int_{- 1}^{1} 3 u^{2} d u = 2

\int_{- 1}^{1} 3 u^{4} d u = \frac{6}{5}

\int_{- 1}^{1} u^{6} d u = \frac{2}{7}

= - (- 2 + 2 - \frac{6}{5} + \frac{2}{7})

Simplify

- (- 2 + 2 - \frac{6}{5} + \frac{2}{7}) : \frac{32}{35}

= \frac{32}{35}

\int_{0}^{2} \frac{x}{( 1 - x ) ^{2}} d x

\int_{0}^{1} sin (n x) d x

\int_{- \infty}^{0} x e^{- 9 x} d x

\int_{0}^{2 π} (\frac{9 θ}{2} + \frac{81}{4}) d θ

\int_{0}^{1} - 4 t d t

What is the integral from 0 to pi of sin^7(x) ?
The integral from 0 to pi of sin^7(x) is 32/35