What is the integral from 0 to pi/2 of (sin(x))^3 ?

The integral from 0 to pi/2 of (sin(x))^3 is 2/3

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/2 of (sin(x))^3

\int_{0}^{\frac{π}{2}} (sin (x))^{3} d x

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{0}^{\frac{π}{2}} (sin (x))^{3} d x

Simplify

= \int_{0}^{\frac{π}{2}} sin^{3} (x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int_{1}^{0} - 1 + u^{2} d u

= \int_{1}^{0} - 1 + u^{2} d u

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

= - \int_{0}^{1} - 1 + u^{2} 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_{0}^{1} 1 d u + \int_{0}^{1} u^{2} d u)

\int_{0}^{1} 1 d u = 1

\int_{0}^{1} u^{2} d u = \frac{1}{3}

= - (- 1 + \frac{1}{3})

Simplify

- (- 1 + \frac{1}{3}) : \frac{2}{3}

= \frac{2}{3}

\int_{0}^{1} 6 cos (\frac{π x}{2}) d x

\int_{0}^{3} (x^{2} + 2) d x

\int_{0}^{4} 3 x^{\frac{1}{2}} - (2 x - 2) d x

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

\int_{- 6}^{6} 36 - x^{2} d x

What is the integral from 0 to pi/2 of (sin(x))^3 ?
The integral from 0 to pi/2 of (sin(x))^3 is 2/3