What is the integral from 0 to pi of sin^6(x) ?

The integral from 0 to pi of sin^6(x) is (5pi)/(16)

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integral from 0 to pi of sin^6(x)

\int_{0}^{π} sin^{6} (x) d x

\frac{5 π}{16}

Decimal Notation

0.98174 \dots

Solution steps

\int_{0}^{π} sin^{6} (x) d x

Apply Integral Reduction

= - [\frac{cos ( x ) sin ^{5} ( x )}{6}]_{0}^{π} + \frac{5}{6} \cdot \int_{0}^{π} sin^{4} (x) d x

\int_{0}^{π} sin^{4} (x) d x = \frac{3 π}{8}

= - [\frac{cos ( x ) sin ^{5} ( x )}{6}]_{0}^{π} + \frac{5}{6} \cdot \frac{3 π}{8}

Simplify

- [\frac{cos ( x ) sin ^{5} ( x )}{6}]_{0}^{π} + \frac{5}{6} \cdot \frac{3 π}{8} : - [\frac{1}{6} sin^{5} (x) cos (x)]_{0}^{π} + \frac{5 π}{16}

= - [\frac{1}{6} sin^{5} (x) cos (x)]_{0}^{π} + \frac{5 π}{16}

Compute the boundaries:

0

= - 0 + \frac{5 π}{16}

Simplify

= \frac{5 π}{16}

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

\int \frac{1}{t} d t

\frac{d}{d x} (- \frac{1}{( x + 4 ) ^{2}})

f (x) = 4 (x^{2} + 4 x)^{3}

y^{^{'}} = - y^{3}

What is the integral from 0 to pi of sin^6(x) ?
The integral from 0 to pi of sin^6(x) is (5pi)/(16)