What is the integral from 0 to \sqrt[3]{pi/2 of}21x^2sin^3(x^3) ?

The integral from 0 to \sqrt[3]{pi/2 of}21x^2sin^3(x^3) is 14/3

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integral from 0 to \sqrt[3]{pi/2 of}21x^2sin^3(x^3)

\frac{14}{3}

Decimal Notation

4.66666 \dots

Solution steps

Take the constant out:

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

Apply u-substitution

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

= 21 \cdot \int_{0}^{\frac{π}{2}} \frac{sin ^{3} ( u )}{3} d u

Take the constant out:

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

= 21 \cdot \frac{1}{3} \cdot \int_{0}^{\frac{π}{2}} sin^{3} (u) d u

Rewrite using trig identities

= 21 \cdot \frac{1}{3} \cdot \int_{0}^{\frac{π}{2}} (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

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

= 21 \cdot \frac{1}{3} \cdot \int_{1}^{0} - 1 + v^{2} d v

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

= 21 \cdot \frac{1}{3} (- \int_{0}^{1} - 1 + v^{2} d v)

Apply the Sum Rule:

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

= 21 \cdot \frac{1}{3} (- (- \int_{0}^{1} 1 d v + \int_{0}^{1} v^{2} d v))

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

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

= 21 \cdot \frac{1}{3} (- (- 1 + \frac{1}{3}))

Simplify

21 \cdot \frac{1}{3} (- (- 1 + \frac{1}{3})) : \frac{14}{3}

= \frac{14}{3}

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t \to - 1 lim (t^{2} - x^{2})

\int \frac{( x e ^{x} )}{( 1 + x ) ^{2}} d x

\frac{\partial}{\partial z} (z - x)

\frac{d}{d x} (\frac{1}{x ^{5}} + \frac{1}{x ^{6}})

What is the integral from 0 to \sqrt[3]{pi/2 of}21x^2sin^3(x^3) ?
The integral from 0 to \sqrt[3]{pi/2 of}21x^2sin^3(x^3) is 14/3