What is the integral from 0 to 2pi of sin^7(x) ?

The integral from 0 to 2pi of sin^7(x) is 0

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integral from 0 to 2pi of sin^7(x)

\int_{0}^{2 π} sin^{7} (x) d x

0

Solution steps

\int_{0}^{2 π} sin^{7} (x) d x

Simplify

sin^{7} (x) : sin^{6} (x) sin (x)

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

Apply Integration By Parts

: [- sin^{6} (x) cos (x) - \int - 6 sin^{5} (x) cos^{2} (x) d x]_{0}^{2 π}

= [- sin^{6} (x) cos (x) - \int - 6 sin^{5} (x) cos^{2} (x) d x]_{0}^{2 π}

\int - 6 sin^{5} (x) cos^{2} (x) d x = - 6 (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7})

= [- sin^{6} (x) cos (x) - (- 6 (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7}))]_{0}^{2 π}

Simplify

[- sin^{6} (x) cos (x) - (- 6 (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7}))]_{0}^{2 π} : [- sin^{6} (x) cos (x) + 6 (- \frac{1}{3} cos^{3} (x) + \frac{2}{5} cos^{5} (x) - \frac{1}{7} cos^{7} (x))]_{0}^{2 π}

= [- sin^{6} (x) cos (x) + 6 (- \frac{1}{3} cos^{3} (x) + \frac{2}{5} cos^{5} (x) - \frac{1}{7} cos^{7} (x))]_{0}^{2 π}

Compute the boundaries:

0

= 0

\int_{0}^{1} x (1 - x^{2}) d x

\int_{1}^{\infty} n^{- 3}

\int_{0}^{1} x (x - 1)^{4} d x

\int_{0}^{π} 1 sin (x) d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} 5 cos (t) d t

What is the integral from 0 to 2pi of sin^7(x) ?
The integral from 0 to 2pi of sin^7(x) is 0