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

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

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

\int_{0}^{π} 3 sin^{3} (x) d x

4

Solution steps

\int_{0}^{π} 3 sin^{3} (x) d x

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

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

= 3 (- \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

= 3 (- (- \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}

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

Simplify

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

= 4

\frac{d}{d x} ((ln (x))^{2} cos (x))

\frac{d}{d t} (e^{t^{2}})

\int \frac{( 7 x ^{2} + 2 x - 7 )}{x ^{3} - x} d x

\frac{\partial}{\partial x} (3 sin (x y))

\frac{d y}{d x} - 2 y = 4

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