What is the integral from 1 to 6 of sin^4(4x)cos(4x) ?

The integral from 1 to 6 of sin^4(4x)cos(4x) is -(sin^5(4)-sin^5(24))/(20)

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integral from 1 to 6 of sin^4(4x)cos(4x)

\int_{1}^{6} sin^{4} (4 x) cos (4 x) d x

- \frac{sin ^{5} ( 4 ) - sin ^{5} ( 24 )}{20}

Decimal

- 0.01803 \dots

Solution steps

\int_{1}^{6} sin^{4} (4 x) cos (4 x) d x

Apply u-substitution

: \int_{4}^{24} sin^{4} (u) cos (u) \frac{1}{4} d u

= \int_{4}^{24} sin^{4} (u) cos (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{4}^{24} sin^{4} (u) cos (u) d u

Apply u-substitution

: \int_{s i n (4)}^{s i n (24)} v^{4} d v

= \frac{1}{4} \cdot \int_{s i n (4)}^{s i n (24)} v^{4} d v

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

= \frac{1}{4} (- \int_{s i n (24)}^{s i n (4)} v^{4} d v)

Apply the Power Rule

: [\frac{v ^{5}}{5}]_{s i n (24)}^{s i n (4)}

= \frac{1}{4} (- [\frac{v ^{5}}{5}]_{s i n (24)}^{s i n (4)})

Simplify

= - \frac{1}{4} [\frac{v ^{5}}{5}]_{s i n (24)}^{s i n (4)}

Compute the boundaries:

\frac{sin ^{5} ( 4 ) - sin ^{5} ( 24 )}{5}

= - \frac{1}{4} \cdot \frac{sin ^{5} ( 4 ) - sin ^{5} ( 24 )}{5}

Simplify

= - \frac{sin ^{5} ( 4 ) - sin ^{5} ( 24 )}{20}

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What is the integral from 1 to 6 of sin^4(4x)cos(4x) ?
The integral from 1 to 6 of sin^4(4x)cos(4x) is -(sin^5(4)-sin^5(24))/(20)