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

The integral from 0 to pi of-sin^3(x) is -4/3

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integral from 0 to pi of-sin^3(x)

\int_{0}^{π} - sin^{3} (x) d x

- \frac{4}{3}

Decimal Notation

- 1.33333 \dots

Solution steps

\int_{0}^{π} - sin^{3} (x) d x

Take the constant out:

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

= - \int_{0}^{π} sin^{3} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

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

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

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

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

= - (- (- 2 + \frac{2}{3}))

Simplify

- (- (- 2 + \frac{2}{3})) : - \frac{4}{3}

= - \frac{4}{3}

\int_{0}^{L} x \cdot sin (\frac{nπ x}{L}) d x

\int_{0}^{x} \frac{2}{3} e^{- \frac{2 x}{3}} d x

\int_{0}^{1} 2 - 2 cos (t) d t

\int_{1}^{4} (x^{3}) d x

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

What is the integral from 0 to pi of-sin^3(x) ?
The integral from 0 to pi of-sin^3(x) is -4/3