What is the integral from 0 to 2pi of sin^8(x) ?

The integral from 0 to 2pi of sin^8(x) is (35pi)/(64)

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integral from 0 to 2pi of sin^8(x)

\int_{0}^{2 π} sin^{8} (x) d x

\frac{35 π}{64}

Decimal Notation

1.71805 \dots

Solution steps

\int_{0}^{2 π} sin^{8} (x) d x

Apply Integral Reduction

= - [\frac{cos ( x ) sin ^{7} ( x )}{8}]_{0}^{2 π} + \frac{7}{8} \cdot \int_{0}^{2 π} sin^{6} (x) d x

\int_{0}^{2 π} sin^{6} (x) d x = \frac{5 π}{8}

= - [\frac{cos ( x ) sin ^{7} ( x )}{8}]_{0}^{2 π} + \frac{7}{8} \cdot \frac{5 π}{8}

Simplify

- [\frac{cos ( x ) sin ^{7} ( x )}{8}]_{0}^{2 π} + \frac{7}{8} \cdot \frac{5 π}{8} : - [\frac{1}{8} sin^{7} (x) cos (x)]_{0}^{2 π} + \frac{35 π}{64}

= - [\frac{1}{8} sin^{7} (x) cos (x)]_{0}^{2 π} + \frac{35 π}{64}

Compute the boundaries:

0

= - 0 + \frac{35 π}{64}

Simplify

= \frac{35 π}{64}

\int_{- 1}^{1} 5 x^{2} - 1 d x

\int_{- 3}^{0} 5 e^{- jω t} d t

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

\int_{0}^{\frac{x}{4}} e^{- x^{2}} d x

What is the integral from 0 to 2pi of sin^8(x) ?
The integral from 0 to 2pi of sin^8(x) is (35pi)/(64)