What is the integral from 0 to pi of sin^6(x/2) ?

The integral from 0 to pi of sin^6(x/2) is (5pi)/(16)

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integral from 0 to pi of sin^6(x/2)

\int_{0}^{π} sin^{6} (\frac{x}{2}) d x

\frac{5 π}{16}

Decimal Notation

0.98174 \dots

Solution steps

\int_{0}^{π} sin^{6} (\frac{x}{2}) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} sin^{6} (u) \cdot 2 d u

= \int_{0}^{\frac{π}{2}} sin^{6} (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int_{0}^{\frac{π}{2}} sin^{6} (u) d u

Apply Integral Reduction

= 2 (- [\frac{cos ( u ) sin ^{5} ( u )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \int_{0}^{\frac{π}{2}} sin^{4} (u) d u)

\int_{0}^{\frac{π}{2}} sin^{4} (u) d u = \frac{3 π}{16}

= 2 (- [\frac{cos ( u ) sin ^{5} ( u )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \frac{3 π}{16})

Simplify

2 (- [\frac{cos ( u ) sin ^{5} ( u )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \frac{3 π}{16}) : 2 (- [\frac{1}{6} sin^{5} (u) cos (u)]_{0}^{\frac{π}{2}} + \frac{5 π}{32})

= 2 (- [\frac{1}{6} sin^{5} (u) cos (u)]_{0}^{\frac{π}{2}} + \frac{5 π}{32})

Compute the boundaries:

0

= 2 (- 0 + \frac{5 π}{32})

Simplify

= \frac{5 π}{16}

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What is the integral from 0 to pi of sin^6(x/2) ?
The integral from 0 to pi of sin^6(x/2) is (5pi)/(16)