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

The integral from 0 to pi/4 of sin^4(4x) is (3pi)/(32)

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integral from 0 to pi/4 of sin^4(4x)

\int_{0}^{\frac{π}{4}} sin^{4} (4 x) d x

\frac{3 π}{32}

Decimal Notation

0.29452 \dots

Solution steps

\int_{0}^{\frac{π}{4}} sin^{4} (4 x) d x

Apply u-substitution

: \int_{0}^{π} sin^{4} (u) \frac{1}{4} d u

= \int_{0}^{π} sin^{4} (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_{0}^{π} sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{4} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \int_{0}^{π} sin^{2} (u) d u)

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

= \frac{1}{4} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2})

Simplify

\frac{1}{4} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : \frac{1}{4} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{π} + \frac{3 π}{8})

= \frac{1}{4} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{π} + \frac{3 π}{8})

Compute the boundaries:

0

= \frac{1}{4} (- 0 + \frac{3 π}{8})

Simplify

= \frac{3 π}{32}

x \to 2 lim (x^{3} + 4)

\int (x - 3) (x + 11)^{15} d x

y = \frac{x ^{6}}{f ( x )}

\int x^{2} x - 1 d x

f (x) = x + 2

What is the integral from 0 to pi/4 of sin^4(4x) ?
The integral from 0 to pi/4 of sin^4(4x) is (3pi)/(32)