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

The integral from 0 to pi/4 of sin^4(x) is -1/16+(3(pi-2))/(32)

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

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

- \frac{1}{16} + \frac{3 ( π - 2 )}{32}

Decimal

0.04452 \dots

Solution steps

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

Apply Integral Reduction

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

\int_{0}^{\frac{π}{4}} sin^{2} (x) d x = \frac{π - 2}{8}

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

Simplify

- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{\frac{π}{4}} + \frac{3}{4} \cdot \frac{π - 2}{8} : - [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{\frac{π}{4}} + \frac{3 ( π - 2 )}{32}

= - [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{\frac{π}{4}} + \frac{3 ( π - 2 )}{32}

Compute the boundaries:

\frac{1}{16}

= - \frac{1}{16} + \frac{3 ( π - 2 )}{32}

2 y^{^{'}} = 4 t cos (t) y^{2}, y (0) = 1

\int 25 x^{\frac{3}{2}} d x

\frac{d}{d x} ((x^{2}) (x + 1)^{\frac{1}{3}})

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

\frac{5}{x ^{4}}

What is the integral from 0 to pi/4 of sin^4(x) ?
The integral from 0 to pi/4 of sin^4(x) is -1/16+(3(pi-2))/(32)