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

The integral from 0 to \sqrt[3]{pi of}t^2cos^2(t^3)sin(t^3) is 2/9

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integral from 0 to \sqrt[3]{pi of}t^2cos^2(t^3)sin(t^3)

\int_{0}^{3 π} t^{2} cos^{2} (t^{3}) sin (t^{3}) d t

\frac{2}{9}

Decimal

0.22222 \dots

Solution steps

\int_{0}^{3 π} t^{2} cos^{2} (t^{3}) sin (t^{3}) d t

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{v ^{3}}{3}]_{- 1}^{1}

= \frac{1}{3} (- (- [\frac{v ^{3}}{3}]_{- 1}^{1}))

Simplify

= \frac{1}{3} [\frac{v ^{3}}{3}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{3}

= \frac{1}{3} \cdot \frac{2}{3}

Simplify

= \frac{2}{9}

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What is the integral from 0 to \sqrt[3]{pi of}t^2cos^2(t^3)sin(t^3) ?
The integral from 0 to \sqrt[3]{pi of}t^2cos^2(t^3)sin(t^3) is 2/9