What is the integral from 0 to 1/2 of sin^4(x) ?

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

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integral from 0 to 1/2 of sin^4(x)

\int_{0}^{\frac{1}{2}} sin^{4} (x) d x

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

Decimal

0.00554 \dots

Solution steps

\int_{0}^{\frac{1}{2}} sin^{4} (x) d x

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{1}{4} sin^{3} (\frac{1}{2}) cos (\frac{1}{2})

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

\int_{- 2}^{2} (x^{2} 4 - x^{2}) d x

\int_{- 3}^{2} x^{3} + x^{2} - 6 x d x

\int_{1}^{8} \frac{2}{x} d x

\int_{x}^{1} (1 - y) cos^{2} (z - y) d z

\int_{0}^{1} 4 x^{3} e^{x^{2}} d x

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