What is the integral from 0 to 1 of 16sin^4(x) ?

The integral from 0 to 1 of 16sin^4(x) is 16(-1/4 sin^3(1)cos(1)+3/8 (1-1/2 sin(2)))

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integral from 0 to 1 of 16sin^4(x)

\int_{0}^{1} 16 sin^{4} (x) d x

16 (- \frac{1}{4} sin^{3} (1) cos (1) + \frac{3}{8} (1 - \frac{1}{2} sin (2)))

Decimal Notation

1.98440 \dots

Solution steps

\int_{0}^{1} 16 sin^{4} (x) d x

Take the constant out:

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

= 16 \cdot \int_{0}^{1} sin^{4} (x) d x

Apply Integral Reduction

= 16 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{1} + \frac{3}{4} \cdot \int_{0}^{1} sin^{2} (x) d x)

\int_{0}^{1} sin^{2} (x) d x = \frac{1}{2} (1 - \frac{1}{2} sin (2))

= 16 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{1} + \frac{3}{4} \cdot \frac{1}{2} (1 - \frac{1}{2} sin (2)))

Simplify

16 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{1} + \frac{3}{4} \cdot \frac{1}{2} (1 - \frac{1}{2} sin (2))) : 16 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{1} + \frac{3}{8} (1 - \frac{1}{2} sin (2)))

= 16 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{1} + \frac{3}{8} (1 - \frac{1}{2} sin (2)))

Compute the boundaries:

\frac{1}{4} sin^{3} (1) cos (1)

= 16 (- \frac{1}{4} sin^{3} (1) cos (1) + \frac{3}{8} (1 - \frac{1}{2} sin (2)))

x \to 3 lim (\frac{x - 3}{( x - 3 )})

y^{^{''}} - 2 y^{^{'}} + 10 y = 0

\frac{d}{d x} (e^{\frac{4}{x}})

x \to 2 lim (5 x - 6)

\frac{d}{d x} (\frac{4 x - 8}{x - 5})

What is the integral from 0 to 1 of 16sin^4(x) ?
The integral from 0 to 1 of 16sin^4(x) is 16(-1/4 sin^3(1)cos(1)+3/8 (1-1/2 sin(2)))