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

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

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

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

\frac{3 π}{16}

Decimal Notation

0.58904 \dots

Solution steps

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

0

= - 0 + \frac{3 π}{16}

Simplify

= \frac{3 π}{16}

\int \frac{( ln ( x ) - 1 )}{x ^{2}} d x

\frac{d}{d x} (9 x^{5})

\int (ln (17 x))^{2} d x

x \to \infty lim (\frac{8}{10})

f (x) = \frac{x ^{2} + 5}{x - 2}

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